Can you compute $int_0^1fraclog(x)log(1-x)xdx$ more precisely than $1.20206$ and do a comparision with $zeta(3)$?Double integral involving zeta function: $int_0^infty frac1-12y^2(1+4y^2)^3int_1/2^inftylog|zeta(x+iy)|~dx ~dy.$Integral $int_0^1fraclog(x)log^2(1-x)log^2(1+x)xmathrm dx$Can be justified this $zeta(3)=int_0^1frac1xsum_n=0^inftyfracx^(n+1)^3/2(n+1)^3/2dx$?On the change $u=x^1+frac1p_n$ in $log zeta(s)=sint_0^inftyfracpi(x)x(x^s-1)dx$, where $p_n$ is the nth prime numberIs there a simple proof for $int_1^inftyfrac2x^2log^2 x(x^2-1)^2dx=frac14(7zeta(3)+pi^2)$?Evaluate $int_0^1 fraclog(1-z)log(1-z^3)z^2dz$Value of the integral $int_0^2pi log|re^it-zeta| dt$Calculate an approximation of $int_0^1int_0^1fraclog(xy)xy-1+log(xy)dxdy$On $int_0^1fraclog^2(x)1+x^3dx$ and $zeta(3)$What's about $-int_0^1fraclog(1+x^10)log xxdx$?

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Can you compute $int_0^1fraclog(x)log(1-x)xdx$ more precisely than $1.20206$ and do a comparision with $zeta(3)$?


Double integral involving zeta function: $int_0^infty frac1-12y^2(1+4y^2)^3int_1/2^inftylog|zeta(x+iy)|~dx ~dy.$Integral $int_0^1fraclog(x)log^2(1-x)log^2(1+x)xmathrm dx$Can be justified this $zeta(3)=int_0^1frac1xsum_n=0^inftyfracx^(n+1)^3/2(n+1)^3/2dx$?On the change $u=x^1+frac1p_n$ in $log zeta(s)=sint_0^inftyfracpi(x)x(x^s-1)dx$, where $p_n$ is the nth prime numberIs there a simple proof for $int_1^inftyfrac2x^2log^2 x(x^2-1)^2dx=frac14(7zeta(3)+pi^2)$?Evaluate $int_0^1 fraclog(1-z)log(1-z^3)z^2dz$Value of the integral $int_0^2pi log|re^it-zeta| dt$Calculate an approximation of $int_0^1int_0^1fraclog(xy)xy-1+log(xy)dxdy$On $int_0^1fraclog^2(x)1+x^3dx$ and $zeta(3)$What's about $-int_0^1fraclog(1+x^10)log xxdx$?













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I know from Wolfram Alpha that $$int_0^1fraclog(x)log(1-x)xdx=1.20206$$ and in the other hand, too from this online tool that
$$intfraclog(x)log(1-x)xdx=mathrmLi_3(x)-mathrmLi_2(x)log(x)+constant.$$




Question. I would like made a comparision, and need obtain $$int_0^1fraclog(x)log(1-x)xdx,$$
more precisely than $1.20206$. I believe that could be $zeta(3)$, but now I don't sure, and I don't know if holding this claim could be deduce easily.



Can you compute $int_0^1fraclog(x)log(1-x)xdx$ more precisely than $1.20206$ to discard that this value is $zeta(3)$, Apéry constant, or claim that the equality $$int_0^1fraclog(x)log(1-x)xdx=zeta(3)$$ holds and is known/easily deduced (perhaps from some of its known formulas involving integrals)?




This definite integral was computed as a summand, when I made some changes of variable in Beuker's integral (see [1]), and now i don't know if too I could be wrong in my computations.



I've searched in this site about this integral $intfraclog(x)log(1-x)xdx$, and in Wikipedia about a possible identity between $zeta(3)$ and particular values of logarithmic integrals $mathrmLi_2(x)$ and $mathrmLi_3(x)$.



References:



[1] https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant






reference-request definite-integrals logarithms improper-integrals online-resources







share|cite|improve this question











$endgroup$











  • $begingroup$
    My changes from Beukers integral (see Refereces) to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense, and there are no mistakes. Too I am assuming that this way is know. Very thanks much all users.
    $endgroup$
    – user243301
    Dec 18 '15 at 18:55






  • 1




    $begingroup$
    Did you try expanding $ln(1-x)$?
    $endgroup$
    – Clayton
    Apr 26 at 19:37










  • $begingroup$
    Taylor series ... expand $ln(1-x)$ in terms of powers of $x$.
    $endgroup$
    – GEdgar
    Apr 26 at 19:41










  • $begingroup$
    Seems like taylor series then by parts is the best way forward here
    $endgroup$
    – George Dewhirst
    Apr 26 at 20:07















5












$begingroup$


I know from Wolfram Alpha that $$int_0^1fraclog(x)log(1-x)xdx=1.20206$$ and in the other hand, too from this online tool that
$$intfraclog(x)log(1-x)xdx=mathrmLi_3(x)-mathrmLi_2(x)log(x)+constant.$$




Question. I would like made a comparision, and need obtain $$int_0^1fraclog(x)log(1-x)xdx,$$
more precisely than $1.20206$. I believe that could be $zeta(3)$, but now I don't sure, and I don't know if holding this claim could be deduce easily.



Can you compute $int_0^1fraclog(x)log(1-x)xdx$ more precisely than $1.20206$ to discard that this value is $zeta(3)$, Apéry constant, or claim that the equality $$int_0^1fraclog(x)log(1-x)xdx=zeta(3)$$ holds and is known/easily deduced (perhaps from some of its known formulas involving integrals)?




This definite integral was computed as a summand, when I made some changes of variable in Beuker's integral (see [1]), and now i don't know if too I could be wrong in my computations.



I've searched in this site about this integral $intfraclog(x)log(1-x)xdx$, and in Wikipedia about a possible identity between $zeta(3)$ and particular values of logarithmic integrals $mathrmLi_2(x)$ and $mathrmLi_3(x)$.



References:



[1] https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant






reference-request definite-integrals logarithms improper-integrals online-resources







share|cite|improve this question











$endgroup$











  • $begingroup$
    My changes from Beukers integral (see Refereces) to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense, and there are no mistakes. Too I am assuming that this way is know. Very thanks much all users.
    $endgroup$
    – user243301
    Dec 18 '15 at 18:55






  • 1




    $begingroup$
    Did you try expanding $ln(1-x)$?
    $endgroup$
    – Clayton
    Apr 26 at 19:37










  • $begingroup$
    Taylor series ... expand $ln(1-x)$ in terms of powers of $x$.
    $endgroup$
    – GEdgar
    Apr 26 at 19:41










  • $begingroup$
    Seems like taylor series then by parts is the best way forward here
    $endgroup$
    – George Dewhirst
    Apr 26 at 20:07













5












5








5


6



$begingroup$


I know from Wolfram Alpha that $$int_0^1fraclog(x)log(1-x)xdx=1.20206$$ and in the other hand, too from this online tool that
$$intfraclog(x)log(1-x)xdx=mathrmLi_3(x)-mathrmLi_2(x)log(x)+constant.$$




Question. I would like made a comparision, and need obtain $$int_0^1fraclog(x)log(1-x)xdx,$$
more precisely than $1.20206$. I believe that could be $zeta(3)$, but now I don't sure, and I don't know if holding this claim could be deduce easily.



Can you compute $int_0^1fraclog(x)log(1-x)xdx$ more precisely than $1.20206$ to discard that this value is $zeta(3)$, Apéry constant, or claim that the equality $$int_0^1fraclog(x)log(1-x)xdx=zeta(3)$$ holds and is known/easily deduced (perhaps from some of its known formulas involving integrals)?




This definite integral was computed as a summand, when I made some changes of variable in Beuker's integral (see [1]), and now i don't know if too I could be wrong in my computations.



I've searched in this site about this integral $intfraclog(x)log(1-x)xdx$, and in Wikipedia about a possible identity between $zeta(3)$ and particular values of logarithmic integrals $mathrmLi_2(x)$ and $mathrmLi_3(x)$.



References:



[1] https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant






reference-request definite-integrals logarithms improper-integrals online-resources







share|cite|improve this question











$endgroup$




I know from Wolfram Alpha that $$int_0^1fraclog(x)log(1-x)xdx=1.20206$$ and in the other hand, too from this online tool that
$$intfraclog(x)log(1-x)xdx=mathrmLi_3(x)-mathrmLi_2(x)log(x)+constant.$$




Question. I would like made a comparision, and need obtain $$int_0^1fraclog(x)log(1-x)xdx,$$
more precisely than $1.20206$. I believe that could be $zeta(3)$, but now I don't sure, and I don't know if holding this claim could be deduce easily.



Can you compute $int_0^1fraclog(x)log(1-x)xdx$ more precisely than $1.20206$ to discard that this value is $zeta(3)$, Apéry constant, or claim that the equality $$int_0^1fraclog(x)log(1-x)xdx=zeta(3)$$ holds and is known/easily deduced (perhaps from some of its known formulas involving integrals)?




This definite integral was computed as a summand, when I made some changes of variable in Beuker's integral (see [1]), and now i don't know if too I could be wrong in my computations.



I've searched in this site about this integral $intfraclog(x)log(1-x)xdx$, and in Wikipedia about a possible identity between $zeta(3)$ and particular values of logarithmic integrals $mathrmLi_2(x)$ and $mathrmLi_3(x)$.



References:



[1] https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant






reference-request definite-integrals logarithms improper-integrals online-resources




reference-request definite-integrals logarithms improper-integrals online-resources






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 25 '16 at 8:37









wythagoras

21.7k446104




21.7k446104










asked Dec 18 '15 at 18:21







user243301


















  • $begingroup$
    My changes from Beukers integral (see Refereces) to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense, and there are no mistakes. Too I am assuming that this way is know. Very thanks much all users.
    $endgroup$
    – user243301
    Dec 18 '15 at 18:55






  • 1




    $begingroup$
    Did you try expanding $ln(1-x)$?
    $endgroup$
    – Clayton
    Apr 26 at 19:37










  • $begingroup$
    Taylor series ... expand $ln(1-x)$ in terms of powers of $x$.
    $endgroup$
    – GEdgar
    Apr 26 at 19:41










  • $begingroup$
    Seems like taylor series then by parts is the best way forward here
    $endgroup$
    – George Dewhirst
    Apr 26 at 20:07
















  • $begingroup$
    My changes from Beukers integral (see Refereces) to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense, and there are no mistakes. Too I am assuming that this way is know. Very thanks much all users.
    $endgroup$
    – user243301
    Dec 18 '15 at 18:55






  • 1




    $begingroup$
    Did you try expanding $ln(1-x)$?
    $endgroup$
    – Clayton
    Apr 26 at 19:37










  • $begingroup$
    Taylor series ... expand $ln(1-x)$ in terms of powers of $x$.
    $endgroup$
    – GEdgar
    Apr 26 at 19:41










  • $begingroup$
    Seems like taylor series then by parts is the best way forward here
    $endgroup$
    – George Dewhirst
    Apr 26 at 20:07















$begingroup$
My changes from Beukers integral (see Refereces) to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense, and there are no mistakes. Too I am assuming that this way is know. Very thanks much all users.
$endgroup$
– user243301
Dec 18 '15 at 18:55




$begingroup$
My changes from Beukers integral (see Refereces) to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense, and there are no mistakes. Too I am assuming that this way is know. Very thanks much all users.
$endgroup$
– user243301
Dec 18 '15 at 18:55




1




1




$begingroup$
Did you try expanding $ln(1-x)$?
$endgroup$
– Clayton
Apr 26 at 19:37




$begingroup$
Did you try expanding $ln(1-x)$?
$endgroup$
– Clayton
Apr 26 at 19:37












$begingroup$
Taylor series ... expand $ln(1-x)$ in terms of powers of $x$.
$endgroup$
– GEdgar
Apr 26 at 19:41




$begingroup$
Taylor series ... expand $ln(1-x)$ in terms of powers of $x$.
$endgroup$
– GEdgar
Apr 26 at 19:41












$begingroup$
Seems like taylor series then by parts is the best way forward here
$endgroup$
– George Dewhirst
Apr 26 at 20:07




$begingroup$
Seems like taylor series then by parts is the best way forward here
$endgroup$
– George Dewhirst
Apr 26 at 20:07










7 Answers
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$$operatornameLi_s(z) = sum_k=1^infty z^k over k^s = z + z^2 over 2^s + z^3 over 3^s + cdots ,.$$



Therefore, since $log(1)=0$, we have:



$$operatornameLi_3(1) = sum_k=1^infty 1 over k^3 = zeta(3)$$



$$operatornameLi_3(1)-log(1)operatornameLi_2(1) = zeta(3)$$



It remains to show that $$lim_xto0 operatornameLi_3(x)-log(x)operatornameLi_2(x)=0$$



Note that if $z<1$,



$$operatornameLi_2(z) = sum_k=1^infty z^k over k^2 = z + z^2 over 2^2 + z^3 over 3^2 + cdots \< z + z over 2^2 + z over 2^2+ z over 4^2 + z over 4^2 + z over 4^2 + z over 4^2 + z over 8^2+cdots leq zsum_k=0^infty 1 over 2^k = 2z$$



Since $log(z)<z$, for $z>1$, we also have $log(z^2)<2z$ thus $log(x)<2sqrtx$. Thus $log(frac1x)>-2sqrtx$, thus $log(u)>-2sqrtfrac1u$, thus $0>log(u)operatornameLi_2(u)>-2sqrtu$.



We may use the Squeeze Theorem to finish the result.






share|cite|improve this answer











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  • $begingroup$
    Then I understand is a bad question, since is easy to compute the limit (I will try it) too I will try don't make more bad questions. Very thanks much @wythagoras. Now I don't know how comute the limit, but I will try it.
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    – user243301
    Dec 18 '15 at 18:33











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    Very thanks for details @wythagoras
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    – user243301
    Dec 18 '15 at 18:43


















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There is a variety of possibilities how to show that this integral indeed equals $zeta(3)$, i.e. Apéry's Constant. I would like to show some of them



I: Taylor Series Expansion of $log(1-x)$



As it was first suggested within the comments (and done by FDP) we may expand the $log(1-x)$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain



beginalign*
int_0^1fraclog(1-x)log(x)xmathrm dx&=int_0^1fraclog(x)xleft[-sum_n=1^inftyfracx^nnright]mathrm dx\
&=-sum_n=1^inftyfrac1nint_0^1x^n-1log(x)mathrm dx\
&=-sum_n=1^inftyfrac1nleft[-frac1n^2right]\
&=sum_n=1^inftyfrac1n^3\
&=zeta(3)
endalign*



This might be the most straightforward approach possible.



II: Integration By Parts



Choosing $u=log(1-x)$ and $mathrm dv=fraclog(x)x$ we can apply Integration By Parts which gives



beginalign*
int_0^1fraclog(1-x)log(x)x&=underbraceleft[log(1-x)fraclog^2(x)2right]_0^1_to0+frac12int_0^1fraclog^2(x)1-xmathrm dx\
&=frac12int_0^1log^2(x)left[sum_n=0^infty x^nright]mathrm dx\
&=frac12sum_n=0^inftyint_0^1x^nlog^2(x)mathrm dx\
&=frac12sum_n=0^inftyleft[frac2(n+1)^3right]\
&=sum_n=1^inftyfrac1n^3\
&=zeta(3)
endalign*



Again, we utilized a series expansion, this time the one of the geometric series.



III: Integral Representation of the Zeta Function



To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $log(x)mapsto -x$ followed by Integration By Parts again we find



beginalign*
int_0^1fraclog(1-x)log(x)xmathrm dx&=-int_infty^0(-x)log(1-e^-x)mathrm dx\
&=-int_0^infty xlog(1-e^-x)mathrm dx\
&=underbraceleft[fracx^22log(1-e^-x)right]_0^infty_to0+frac12int_0^inftyfracx^21-e^-xe^-xmathrm dx\
&=frac1Gamma(3)int_0^inftyfracx^3-1e^x-1mathrm dx\
&=zeta(3)
endalign*



Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.



IV: The Trilogarithm $operatornameLi_3(1)$



Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $u=log(x)$ and $mathrm dv=fraclog(1-x)x$ to get



beginalign*
int_0^1fraclog(1-x)log(x)xmathrm dx&=underbraceleft[log(x)(-operatornameLi_2(x))right]_0^1_to0+int_0^1fracoperatornameLi_2(x)xmathrm dx\
&=[operatornameLi_3(x)]_0^1\
&=zeta(3)
endalign*



A quick look at the series representation of the Trilogarithm verifies the last line.






share|cite|improve this answer









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  • $begingroup$
    Hey, that's nicely done, but you took them all 😒.
    $endgroup$
    – Number
    Apr 27 at 6:29











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    @Zacky Thank you! I'm sorry, but I couldn't resist ^^
    $endgroup$
    – mrtaurho
    Apr 27 at 7:37


















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beginalignJ&=int_0^1 fracln(1-x)ln xx dx\
&=-int_0^1 left(sum_n=1^infty fracx^n-1nright)ln x,dx\
&=-sum_n=1^infty frac1nint_0^1 x^n-1ln x,dx\
&=sum_n=1^infty frac1n^3\
&=zeta(3)
endalign






share|cite|improve this answer









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    An informal argument: $$log (1-x) = - sum_k=0^infty fracx^k+1k+1, quad |x| < 1.$$ Then $$fraclog x log(1-x)x = -sum_k=0^infty fracx^k log xk+1,$$ and integrating term by term gives $$int_x=0^1 x^k log x , dx = left[ fracx^k+1 log xk+1 right]_x=0^1 - int_x=0^1 fracx^kk+1 , dx = - frac1(k+1)^2.$$ Therefore, $$int_x=0^1 fraclog x log(1-x)x , dx = sum_k=1^infty frac1k^3 = zeta(3).$$






    share|cite|improve this answer









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    • $begingroup$
      Very thanks much @heropup
      $endgroup$
      – user243301
      Dec 18 '15 at 18:34






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      May I ask what is so informal about this solution?
      $endgroup$
      – GohP.iHan
      Jan 1 '16 at 7:48


















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    $begingroup$

    $newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
    newcommandbraces[1]leftlbrace,#1,rightrbrace
    newcommandbracks[1]leftlbrack,#1,rightrbrack
    newcommandddmathrmd
    newcommandds[1]displaystyle#1
    newcommandexpo[1],mathrme^#1,
    newcommandicmathrmi
    newcommandmc[1]mathcal#1
    newcommandmrm[1]mathrm#1
    newcommandpars[1]left(,#1,right)
    newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
    newcommandroot[2][],sqrt[#1],#2,,
    newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
    newcommandverts[1]leftvert,#1,rightvert$



    You can add this "weird" answer to the excellent
    $texttt@mrtaurho$ long list:
    beginalign
    &bbox[10px,#ffd]%
    int_0^1lnpars1 - xlnparsx over x,dd x =
    left.partial^2 over partialmu,partialnuint_0^1
    brackspars1 - x^mu - 1x^nu over x,dd x,rightvert_largemu = 0 atop large,,nu = 0^+
    \[5mm] = &
    partial^2 over partialmu,partialnubracks%
    int_0^1x^nu - 1pars1 - x^mu,dd x -
    int_0^1x^nu - 1,dd x
    _largemu = 0 atop large,,nu = 0^+
    \[5mm] = &
    partial^2 over partialmu,partialnubracks%
    GammaparsnuGammaparsmu + 1 over
    Gammaparsnu + mu + 1 - 1 over nu
    _largemu = 0 atop large,,nu = 0^+
    \[5mm] = &
    partial^2 over partialmu,partialnubraces1 over nubracks%
    Gammaparsnu + 1Gammaparsmu + 1 over
    Gammaparsnu + mu + 1 - 1
    _largemu = 0 atop large,,nu = 0^+
    \[5mm] = &
    1 over 2,partiald[2]nubracesGammaparsnu + 1
    partialdmubracksGammaparsmu + 1 over
    Gammaparsnu + mu + 1
    _largemu = 0 atop large,,nu = 0^+
    \[5mm] = &
    1 over 2,partiald[2]nubracesGammaparsnu + 1
    bracks-,gamma + Psiparsnu + 1 over
    Gammaparsnu + 1 _ nu = 0^+
    \[5mm] = &
    -,1 over 2,Psi,''pars1 = bbxzetapars3
    endalign






    share|cite|improve this answer











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      Let $x = e^-y$, we have
      $$int_0^1 fraclog xlog(1-x)x dx
      = int_0^1 frac(-log x)x sum_n=1^infty fracx^nn dx
      = sum_n=1^infty frac1nint_0^1 (-log x) x^n-1 dx\
      = sum_n=1^infty frac1nint_0^infty y e^-ny dy
      = sum_n=1^infty frac1n^3
      = zeta(3)
      $$
      Please note that we can switch the order of summation and integration because all the individual terms are non-negative.






      share|cite|improve this answer











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      • $begingroup$
        Very thanks much @achillehui incredible. I take notes from your solution.
        $endgroup$
        – user243301
        Dec 18 '15 at 18:37










      • $begingroup$
        @achillehui Are you exploiting Fubini-Tonelli with the counting measure on $mathbbN$ and Lebesgue measure on $[0,1]$ to justify interchanging the series and integration?
        $endgroup$
        – Mark Viola
        Mar 20 '17 at 2:56










      • $begingroup$
        @Dr.MV the last phrase "because all the individual terms are non-negative" is essentially Tonelli theorem.
        $endgroup$
        – achille hui
        Mar 20 '17 at 4:05










      • $begingroup$
        @achillehui Yes, I know. So, you are exploiting it then?
        $endgroup$
        – Mark Viola
        Mar 20 '17 at 4:19


















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      The Beta function and Feynman's trick are another way to go:



      $$I=int_0^1fraclog(x)log(1-x)x,dx =left.fracpartial^2partial a partial bint_0^1x^a-1(1-x)^b,dx,right|_alpha,beta=0^+tag1 $$
      hence:
      $$ I = left.fracpartial^2partial a partial bfracGamma(a)Gamma(b+1)Gamma(a+b+1),right|_alpha,beta=0^+tag2 $$
      and by exploiting $Gamma'(z) = Gamma(z)cdotpsi(z)$ we get:
      $$ I = -frac12psi''(2)=sum_ngeq 0frac1(n+1)^3=colorredzeta(3)tag3 $$
      as wanted.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        I try understand your contribution too, now I am saturated. My changes to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense. Very thanks much @JackD'Aurizio.
        $endgroup$
        – user243301
        Dec 18 '15 at 18:54











      Your Answer








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      7 Answers
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      7 Answers
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      1












      $begingroup$

      $$operatornameLi_s(z) = sum_k=1^infty z^k over k^s = z + z^2 over 2^s + z^3 over 3^s + cdots ,.$$



      Therefore, since $log(1)=0$, we have:



      $$operatornameLi_3(1) = sum_k=1^infty 1 over k^3 = zeta(3)$$



      $$operatornameLi_3(1)-log(1)operatornameLi_2(1) = zeta(3)$$



      It remains to show that $$lim_xto0 operatornameLi_3(x)-log(x)operatornameLi_2(x)=0$$



      Note that if $z<1$,



      $$operatornameLi_2(z) = sum_k=1^infty z^k over k^2 = z + z^2 over 2^2 + z^3 over 3^2 + cdots \< z + z over 2^2 + z over 2^2+ z over 4^2 + z over 4^2 + z over 4^2 + z over 4^2 + z over 8^2+cdots leq zsum_k=0^infty 1 over 2^k = 2z$$



      Since $log(z)<z$, for $z>1$, we also have $log(z^2)<2z$ thus $log(x)<2sqrtx$. Thus $log(frac1x)>-2sqrtx$, thus $log(u)>-2sqrtfrac1u$, thus $0>log(u)operatornameLi_2(u)>-2sqrtu$.



      We may use the Squeeze Theorem to finish the result.






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        Then I understand is a bad question, since is easy to compute the limit (I will try it) too I will try don't make more bad questions. Very thanks much @wythagoras. Now I don't know how comute the limit, but I will try it.
        $endgroup$
        – user243301
        Dec 18 '15 at 18:33











      • $begingroup$
        Very thanks for details @wythagoras
        $endgroup$
        – user243301
        Dec 18 '15 at 18:43















      1












      $begingroup$

      $$operatornameLi_s(z) = sum_k=1^infty z^k over k^s = z + z^2 over 2^s + z^3 over 3^s + cdots ,.$$



      Therefore, since $log(1)=0$, we have:



      $$operatornameLi_3(1) = sum_k=1^infty 1 over k^3 = zeta(3)$$



      $$operatornameLi_3(1)-log(1)operatornameLi_2(1) = zeta(3)$$



      It remains to show that $$lim_xto0 operatornameLi_3(x)-log(x)operatornameLi_2(x)=0$$



      Note that if $z<1$,



      $$operatornameLi_2(z) = sum_k=1^infty z^k over k^2 = z + z^2 over 2^2 + z^3 over 3^2 + cdots \< z + z over 2^2 + z over 2^2+ z over 4^2 + z over 4^2 + z over 4^2 + z over 4^2 + z over 8^2+cdots leq zsum_k=0^infty 1 over 2^k = 2z$$



      Since $log(z)<z$, for $z>1$, we also have $log(z^2)<2z$ thus $log(x)<2sqrtx$. Thus $log(frac1x)>-2sqrtx$, thus $log(u)>-2sqrtfrac1u$, thus $0>log(u)operatornameLi_2(u)>-2sqrtu$.



      We may use the Squeeze Theorem to finish the result.






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        Then I understand is a bad question, since is easy to compute the limit (I will try it) too I will try don't make more bad questions. Very thanks much @wythagoras. Now I don't know how comute the limit, but I will try it.
        $endgroup$
        – user243301
        Dec 18 '15 at 18:33











      • $begingroup$
        Very thanks for details @wythagoras
        $endgroup$
        – user243301
        Dec 18 '15 at 18:43













      1












      1








      1





      $begingroup$

      $$operatornameLi_s(z) = sum_k=1^infty z^k over k^s = z + z^2 over 2^s + z^3 over 3^s + cdots ,.$$



      Therefore, since $log(1)=0$, we have:



      $$operatornameLi_3(1) = sum_k=1^infty 1 over k^3 = zeta(3)$$



      $$operatornameLi_3(1)-log(1)operatornameLi_2(1) = zeta(3)$$



      It remains to show that $$lim_xto0 operatornameLi_3(x)-log(x)operatornameLi_2(x)=0$$



      Note that if $z<1$,



      $$operatornameLi_2(z) = sum_k=1^infty z^k over k^2 = z + z^2 over 2^2 + z^3 over 3^2 + cdots \< z + z over 2^2 + z over 2^2+ z over 4^2 + z over 4^2 + z over 4^2 + z over 4^2 + z over 8^2+cdots leq zsum_k=0^infty 1 over 2^k = 2z$$



      Since $log(z)<z$, for $z>1$, we also have $log(z^2)<2z$ thus $log(x)<2sqrtx$. Thus $log(frac1x)>-2sqrtx$, thus $log(u)>-2sqrtfrac1u$, thus $0>log(u)operatornameLi_2(u)>-2sqrtu$.



      We may use the Squeeze Theorem to finish the result.






      share|cite|improve this answer











      $endgroup$



      $$operatornameLi_s(z) = sum_k=1^infty z^k over k^s = z + z^2 over 2^s + z^3 over 3^s + cdots ,.$$



      Therefore, since $log(1)=0$, we have:



      $$operatornameLi_3(1) = sum_k=1^infty 1 over k^3 = zeta(3)$$



      $$operatornameLi_3(1)-log(1)operatornameLi_2(1) = zeta(3)$$



      It remains to show that $$lim_xto0 operatornameLi_3(x)-log(x)operatornameLi_2(x)=0$$



      Note that if $z<1$,



      $$operatornameLi_2(z) = sum_k=1^infty z^k over k^2 = z + z^2 over 2^2 + z^3 over 3^2 + cdots \< z + z over 2^2 + z over 2^2+ z over 4^2 + z over 4^2 + z over 4^2 + z over 4^2 + z over 8^2+cdots leq zsum_k=0^infty 1 over 2^k = 2z$$



      Since $log(z)<z$, for $z>1$, we also have $log(z^2)<2z$ thus $log(x)<2sqrtx$. Thus $log(frac1x)>-2sqrtx$, thus $log(u)>-2sqrtfrac1u$, thus $0>log(u)operatornameLi_2(u)>-2sqrtu$.



      We may use the Squeeze Theorem to finish the result.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 18 '15 at 18:42

























      answered Dec 18 '15 at 18:29









      wythagoraswythagoras

      21.7k446104




      21.7k446104











      • $begingroup$
        Then I understand is a bad question, since is easy to compute the limit (I will try it) too I will try don't make more bad questions. Very thanks much @wythagoras. Now I don't know how comute the limit, but I will try it.
        $endgroup$
        – user243301
        Dec 18 '15 at 18:33











      • $begingroup$
        Very thanks for details @wythagoras
        $endgroup$
        – user243301
        Dec 18 '15 at 18:43
















      • $begingroup$
        Then I understand is a bad question, since is easy to compute the limit (I will try it) too I will try don't make more bad questions. Very thanks much @wythagoras. Now I don't know how comute the limit, but I will try it.
        $endgroup$
        – user243301
        Dec 18 '15 at 18:33











      • $begingroup$
        Very thanks for details @wythagoras
        $endgroup$
        – user243301
        Dec 18 '15 at 18:43















      $begingroup$
      Then I understand is a bad question, since is easy to compute the limit (I will try it) too I will try don't make more bad questions. Very thanks much @wythagoras. Now I don't know how comute the limit, but I will try it.
      $endgroup$
      – user243301
      Dec 18 '15 at 18:33





      $begingroup$
      Then I understand is a bad question, since is easy to compute the limit (I will try it) too I will try don't make more bad questions. Very thanks much @wythagoras. Now I don't know how comute the limit, but I will try it.
      $endgroup$
      – user243301
      Dec 18 '15 at 18:33













      $begingroup$
      Very thanks for details @wythagoras
      $endgroup$
      – user243301
      Dec 18 '15 at 18:43




      $begingroup$
      Very thanks for details @wythagoras
      $endgroup$
      – user243301
      Dec 18 '15 at 18:43











      9












      $begingroup$

      There is a variety of possibilities how to show that this integral indeed equals $zeta(3)$, i.e. Apéry's Constant. I would like to show some of them



      I: Taylor Series Expansion of $log(1-x)$



      As it was first suggested within the comments (and done by FDP) we may expand the $log(1-x)$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain



      beginalign*
      int_0^1fraclog(1-x)log(x)xmathrm dx&=int_0^1fraclog(x)xleft[-sum_n=1^inftyfracx^nnright]mathrm dx\
      &=-sum_n=1^inftyfrac1nint_0^1x^n-1log(x)mathrm dx\
      &=-sum_n=1^inftyfrac1nleft[-frac1n^2right]\
      &=sum_n=1^inftyfrac1n^3\
      &=zeta(3)
      endalign*



      This might be the most straightforward approach possible.



      II: Integration By Parts



      Choosing $u=log(1-x)$ and $mathrm dv=fraclog(x)x$ we can apply Integration By Parts which gives



      beginalign*
      int_0^1fraclog(1-x)log(x)x&=underbraceleft[log(1-x)fraclog^2(x)2right]_0^1_to0+frac12int_0^1fraclog^2(x)1-xmathrm dx\
      &=frac12int_0^1log^2(x)left[sum_n=0^infty x^nright]mathrm dx\
      &=frac12sum_n=0^inftyint_0^1x^nlog^2(x)mathrm dx\
      &=frac12sum_n=0^inftyleft[frac2(n+1)^3right]\
      &=sum_n=1^inftyfrac1n^3\
      &=zeta(3)
      endalign*



      Again, we utilized a series expansion, this time the one of the geometric series.



      III: Integral Representation of the Zeta Function



      To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $log(x)mapsto -x$ followed by Integration By Parts again we find



      beginalign*
      int_0^1fraclog(1-x)log(x)xmathrm dx&=-int_infty^0(-x)log(1-e^-x)mathrm dx\
      &=-int_0^infty xlog(1-e^-x)mathrm dx\
      &=underbraceleft[fracx^22log(1-e^-x)right]_0^infty_to0+frac12int_0^inftyfracx^21-e^-xe^-xmathrm dx\
      &=frac1Gamma(3)int_0^inftyfracx^3-1e^x-1mathrm dx\
      &=zeta(3)
      endalign*



      Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.



      IV: The Trilogarithm $operatornameLi_3(1)$



      Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $u=log(x)$ and $mathrm dv=fraclog(1-x)x$ to get



      beginalign*
      int_0^1fraclog(1-x)log(x)xmathrm dx&=underbraceleft[log(x)(-operatornameLi_2(x))right]_0^1_to0+int_0^1fracoperatornameLi_2(x)xmathrm dx\
      &=[operatornameLi_3(x)]_0^1\
      &=zeta(3)
      endalign*



      A quick look at the series representation of the Trilogarithm verifies the last line.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        Hey, that's nicely done, but you took them all 😒.
        $endgroup$
        – Number
        Apr 27 at 6:29











      • $begingroup$
        @Zacky Thank you! I'm sorry, but I couldn't resist ^^
        $endgroup$
        – mrtaurho
        Apr 27 at 7:37















      9












      $begingroup$

      There is a variety of possibilities how to show that this integral indeed equals $zeta(3)$, i.e. Apéry's Constant. I would like to show some of them



      I: Taylor Series Expansion of $log(1-x)$



      As it was first suggested within the comments (and done by FDP) we may expand the $log(1-x)$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain



      beginalign*
      int_0^1fraclog(1-x)log(x)xmathrm dx&=int_0^1fraclog(x)xleft[-sum_n=1^inftyfracx^nnright]mathrm dx\
      &=-sum_n=1^inftyfrac1nint_0^1x^n-1log(x)mathrm dx\
      &=-sum_n=1^inftyfrac1nleft[-frac1n^2right]\
      &=sum_n=1^inftyfrac1n^3\
      &=zeta(3)
      endalign*



      This might be the most straightforward approach possible.



      II: Integration By Parts



      Choosing $u=log(1-x)$ and $mathrm dv=fraclog(x)x$ we can apply Integration By Parts which gives



      beginalign*
      int_0^1fraclog(1-x)log(x)x&=underbraceleft[log(1-x)fraclog^2(x)2right]_0^1_to0+frac12int_0^1fraclog^2(x)1-xmathrm dx\
      &=frac12int_0^1log^2(x)left[sum_n=0^infty x^nright]mathrm dx\
      &=frac12sum_n=0^inftyint_0^1x^nlog^2(x)mathrm dx\
      &=frac12sum_n=0^inftyleft[frac2(n+1)^3right]\
      &=sum_n=1^inftyfrac1n^3\
      &=zeta(3)
      endalign*



      Again, we utilized a series expansion, this time the one of the geometric series.



      III: Integral Representation of the Zeta Function



      To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $log(x)mapsto -x$ followed by Integration By Parts again we find



      beginalign*
      int_0^1fraclog(1-x)log(x)xmathrm dx&=-int_infty^0(-x)log(1-e^-x)mathrm dx\
      &=-int_0^infty xlog(1-e^-x)mathrm dx\
      &=underbraceleft[fracx^22log(1-e^-x)right]_0^infty_to0+frac12int_0^inftyfracx^21-e^-xe^-xmathrm dx\
      &=frac1Gamma(3)int_0^inftyfracx^3-1e^x-1mathrm dx\
      &=zeta(3)
      endalign*



      Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.



      IV: The Trilogarithm $operatornameLi_3(1)$



      Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $u=log(x)$ and $mathrm dv=fraclog(1-x)x$ to get



      beginalign*
      int_0^1fraclog(1-x)log(x)xmathrm dx&=underbraceleft[log(x)(-operatornameLi_2(x))right]_0^1_to0+int_0^1fracoperatornameLi_2(x)xmathrm dx\
      &=[operatornameLi_3(x)]_0^1\
      &=zeta(3)
      endalign*



      A quick look at the series representation of the Trilogarithm verifies the last line.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        Hey, that's nicely done, but you took them all 😒.
        $endgroup$
        – Number
        Apr 27 at 6:29











      • $begingroup$
        @Zacky Thank you! I'm sorry, but I couldn't resist ^^
        $endgroup$
        – mrtaurho
        Apr 27 at 7:37













      9












      9








      9





      $begingroup$

      There is a variety of possibilities how to show that this integral indeed equals $zeta(3)$, i.e. Apéry's Constant. I would like to show some of them



      I: Taylor Series Expansion of $log(1-x)$



      As it was first suggested within the comments (and done by FDP) we may expand the $log(1-x)$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain



      beginalign*
      int_0^1fraclog(1-x)log(x)xmathrm dx&=int_0^1fraclog(x)xleft[-sum_n=1^inftyfracx^nnright]mathrm dx\
      &=-sum_n=1^inftyfrac1nint_0^1x^n-1log(x)mathrm dx\
      &=-sum_n=1^inftyfrac1nleft[-frac1n^2right]\
      &=sum_n=1^inftyfrac1n^3\
      &=zeta(3)
      endalign*



      This might be the most straightforward approach possible.



      II: Integration By Parts



      Choosing $u=log(1-x)$ and $mathrm dv=fraclog(x)x$ we can apply Integration By Parts which gives



      beginalign*
      int_0^1fraclog(1-x)log(x)x&=underbraceleft[log(1-x)fraclog^2(x)2right]_0^1_to0+frac12int_0^1fraclog^2(x)1-xmathrm dx\
      &=frac12int_0^1log^2(x)left[sum_n=0^infty x^nright]mathrm dx\
      &=frac12sum_n=0^inftyint_0^1x^nlog^2(x)mathrm dx\
      &=frac12sum_n=0^inftyleft[frac2(n+1)^3right]\
      &=sum_n=1^inftyfrac1n^3\
      &=zeta(3)
      endalign*



      Again, we utilized a series expansion, this time the one of the geometric series.



      III: Integral Representation of the Zeta Function



      To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $log(x)mapsto -x$ followed by Integration By Parts again we find



      beginalign*
      int_0^1fraclog(1-x)log(x)xmathrm dx&=-int_infty^0(-x)log(1-e^-x)mathrm dx\
      &=-int_0^infty xlog(1-e^-x)mathrm dx\
      &=underbraceleft[fracx^22log(1-e^-x)right]_0^infty_to0+frac12int_0^inftyfracx^21-e^-xe^-xmathrm dx\
      &=frac1Gamma(3)int_0^inftyfracx^3-1e^x-1mathrm dx\
      &=zeta(3)
      endalign*



      Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.



      IV: The Trilogarithm $operatornameLi_3(1)$



      Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $u=log(x)$ and $mathrm dv=fraclog(1-x)x$ to get



      beginalign*
      int_0^1fraclog(1-x)log(x)xmathrm dx&=underbraceleft[log(x)(-operatornameLi_2(x))right]_0^1_to0+int_0^1fracoperatornameLi_2(x)xmathrm dx\
      &=[operatornameLi_3(x)]_0^1\
      &=zeta(3)
      endalign*



      A quick look at the series representation of the Trilogarithm verifies the last line.






      share|cite|improve this answer









      $endgroup$



      There is a variety of possibilities how to show that this integral indeed equals $zeta(3)$, i.e. Apéry's Constant. I would like to show some of them



      I: Taylor Series Expansion of $log(1-x)$



      As it was first suggested within the comments (and done by FDP) we may expand the $log(1-x)$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain



      beginalign*
      int_0^1fraclog(1-x)log(x)xmathrm dx&=int_0^1fraclog(x)xleft[-sum_n=1^inftyfracx^nnright]mathrm dx\
      &=-sum_n=1^inftyfrac1nint_0^1x^n-1log(x)mathrm dx\
      &=-sum_n=1^inftyfrac1nleft[-frac1n^2right]\
      &=sum_n=1^inftyfrac1n^3\
      &=zeta(3)
      endalign*



      This might be the most straightforward approach possible.



      II: Integration By Parts



      Choosing $u=log(1-x)$ and $mathrm dv=fraclog(x)x$ we can apply Integration By Parts which gives



      beginalign*
      int_0^1fraclog(1-x)log(x)x&=underbraceleft[log(1-x)fraclog^2(x)2right]_0^1_to0+frac12int_0^1fraclog^2(x)1-xmathrm dx\
      &=frac12int_0^1log^2(x)left[sum_n=0^infty x^nright]mathrm dx\
      &=frac12sum_n=0^inftyint_0^1x^nlog^2(x)mathrm dx\
      &=frac12sum_n=0^inftyleft[frac2(n+1)^3right]\
      &=sum_n=1^inftyfrac1n^3\
      &=zeta(3)
      endalign*



      Again, we utilized a series expansion, this time the one of the geometric series.



      III: Integral Representation of the Zeta Function



      To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $log(x)mapsto -x$ followed by Integration By Parts again we find



      beginalign*
      int_0^1fraclog(1-x)log(x)xmathrm dx&=-int_infty^0(-x)log(1-e^-x)mathrm dx\
      &=-int_0^infty xlog(1-e^-x)mathrm dx\
      &=underbraceleft[fracx^22log(1-e^-x)right]_0^infty_to0+frac12int_0^inftyfracx^21-e^-xe^-xmathrm dx\
      &=frac1Gamma(3)int_0^inftyfracx^3-1e^x-1mathrm dx\
      &=zeta(3)
      endalign*



      Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.



      IV: The Trilogarithm $operatornameLi_3(1)$



      Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $u=log(x)$ and $mathrm dv=fraclog(1-x)x$ to get



      beginalign*
      int_0^1fraclog(1-x)log(x)xmathrm dx&=underbraceleft[log(x)(-operatornameLi_2(x))right]_0^1_to0+int_0^1fracoperatornameLi_2(x)xmathrm dx\
      &=[operatornameLi_3(x)]_0^1\
      &=zeta(3)
      endalign*



      A quick look at the series representation of the Trilogarithm verifies the last line.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Apr 26 at 20:38









      mrtaurhomrtaurho

      6,73071843




      6,73071843











      • $begingroup$
        Hey, that's nicely done, but you took them all 😒.
        $endgroup$
        – Number
        Apr 27 at 6:29











      • $begingroup$
        @Zacky Thank you! I'm sorry, but I couldn't resist ^^
        $endgroup$
        – mrtaurho
        Apr 27 at 7:37
















      • $begingroup$
        Hey, that's nicely done, but you took them all 😒.
        $endgroup$
        – Number
        Apr 27 at 6:29











      • $begingroup$
        @Zacky Thank you! I'm sorry, but I couldn't resist ^^
        $endgroup$
        – mrtaurho
        Apr 27 at 7:37















      $begingroup$
      Hey, that's nicely done, but you took them all 😒.
      $endgroup$
      – Number
      Apr 27 at 6:29





      $begingroup$
      Hey, that's nicely done, but you took them all 😒.
      $endgroup$
      – Number
      Apr 27 at 6:29













      $begingroup$
      @Zacky Thank you! I'm sorry, but I couldn't resist ^^
      $endgroup$
      – mrtaurho
      Apr 27 at 7:37




      $begingroup$
      @Zacky Thank you! I'm sorry, but I couldn't resist ^^
      $endgroup$
      – mrtaurho
      Apr 27 at 7:37











      8












      $begingroup$

      beginalignJ&=int_0^1 fracln(1-x)ln xx dx\
      &=-int_0^1 left(sum_n=1^infty fracx^n-1nright)ln x,dx\
      &=-sum_n=1^infty frac1nint_0^1 x^n-1ln x,dx\
      &=sum_n=1^infty frac1n^3\
      &=zeta(3)
      endalign






      share|cite|improve this answer









      $endgroup$

















        8












        $begingroup$

        beginalignJ&=int_0^1 fracln(1-x)ln xx dx\
        &=-int_0^1 left(sum_n=1^infty fracx^n-1nright)ln x,dx\
        &=-sum_n=1^infty frac1nint_0^1 x^n-1ln x,dx\
        &=sum_n=1^infty frac1n^3\
        &=zeta(3)
        endalign






        share|cite|improve this answer









        $endgroup$















          8












          8








          8





          $begingroup$

          beginalignJ&=int_0^1 fracln(1-x)ln xx dx\
          &=-int_0^1 left(sum_n=1^infty fracx^n-1nright)ln x,dx\
          &=-sum_n=1^infty frac1nint_0^1 x^n-1ln x,dx\
          &=sum_n=1^infty frac1n^3\
          &=zeta(3)
          endalign






          share|cite|improve this answer









          $endgroup$



          beginalignJ&=int_0^1 fracln(1-x)ln xx dx\
          &=-int_0^1 left(sum_n=1^infty fracx^n-1nright)ln x,dx\
          &=-sum_n=1^infty frac1nint_0^1 x^n-1ln x,dx\
          &=sum_n=1^infty frac1n^3\
          &=zeta(3)
          endalign







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 26 at 20:10









          FDPFDP

          6,51712031




          6,51712031





















              4












              $begingroup$

              An informal argument: $$log (1-x) = - sum_k=0^infty fracx^k+1k+1, quad |x| < 1.$$ Then $$fraclog x log(1-x)x = -sum_k=0^infty fracx^k log xk+1,$$ and integrating term by term gives $$int_x=0^1 x^k log x , dx = left[ fracx^k+1 log xk+1 right]_x=0^1 - int_x=0^1 fracx^kk+1 , dx = - frac1(k+1)^2.$$ Therefore, $$int_x=0^1 fraclog x log(1-x)x , dx = sum_k=1^infty frac1k^3 = zeta(3).$$






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                Very thanks much @heropup
                $endgroup$
                – user243301
                Dec 18 '15 at 18:34






              • 1




                $begingroup$
                May I ask what is so informal about this solution?
                $endgroup$
                – GohP.iHan
                Jan 1 '16 at 7:48















              4












              $begingroup$

              An informal argument: $$log (1-x) = - sum_k=0^infty fracx^k+1k+1, quad |x| < 1.$$ Then $$fraclog x log(1-x)x = -sum_k=0^infty fracx^k log xk+1,$$ and integrating term by term gives $$int_x=0^1 x^k log x , dx = left[ fracx^k+1 log xk+1 right]_x=0^1 - int_x=0^1 fracx^kk+1 , dx = - frac1(k+1)^2.$$ Therefore, $$int_x=0^1 fraclog x log(1-x)x , dx = sum_k=1^infty frac1k^3 = zeta(3).$$






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                Very thanks much @heropup
                $endgroup$
                – user243301
                Dec 18 '15 at 18:34






              • 1




                $begingroup$
                May I ask what is so informal about this solution?
                $endgroup$
                – GohP.iHan
                Jan 1 '16 at 7:48













              4












              4








              4





              $begingroup$

              An informal argument: $$log (1-x) = - sum_k=0^infty fracx^k+1k+1, quad |x| < 1.$$ Then $$fraclog x log(1-x)x = -sum_k=0^infty fracx^k log xk+1,$$ and integrating term by term gives $$int_x=0^1 x^k log x , dx = left[ fracx^k+1 log xk+1 right]_x=0^1 - int_x=0^1 fracx^kk+1 , dx = - frac1(k+1)^2.$$ Therefore, $$int_x=0^1 fraclog x log(1-x)x , dx = sum_k=1^infty frac1k^3 = zeta(3).$$






              share|cite|improve this answer









              $endgroup$



              An informal argument: $$log (1-x) = - sum_k=0^infty fracx^k+1k+1, quad |x| < 1.$$ Then $$fraclog x log(1-x)x = -sum_k=0^infty fracx^k log xk+1,$$ and integrating term by term gives $$int_x=0^1 x^k log x , dx = left[ fracx^k+1 log xk+1 right]_x=0^1 - int_x=0^1 fracx^kk+1 , dx = - frac1(k+1)^2.$$ Therefore, $$int_x=0^1 fraclog x log(1-x)x , dx = sum_k=1^infty frac1k^3 = zeta(3).$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 18 '15 at 18:32









              heropupheropup

              66.2k866104




              66.2k866104











              • $begingroup$
                Very thanks much @heropup
                $endgroup$
                – user243301
                Dec 18 '15 at 18:34






              • 1




                $begingroup$
                May I ask what is so informal about this solution?
                $endgroup$
                – GohP.iHan
                Jan 1 '16 at 7:48
















              • $begingroup$
                Very thanks much @heropup
                $endgroup$
                – user243301
                Dec 18 '15 at 18:34






              • 1




                $begingroup$
                May I ask what is so informal about this solution?
                $endgroup$
                – GohP.iHan
                Jan 1 '16 at 7:48















              $begingroup$
              Very thanks much @heropup
              $endgroup$
              – user243301
              Dec 18 '15 at 18:34




              $begingroup$
              Very thanks much @heropup
              $endgroup$
              – user243301
              Dec 18 '15 at 18:34




              1




              1




              $begingroup$
              May I ask what is so informal about this solution?
              $endgroup$
              – GohP.iHan
              Jan 1 '16 at 7:48




              $begingroup$
              May I ask what is so informal about this solution?
              $endgroup$
              – GohP.iHan
              Jan 1 '16 at 7:48











              4












              $begingroup$

              $newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
              newcommandbraces[1]leftlbrace,#1,rightrbrace
              newcommandbracks[1]leftlbrack,#1,rightrbrack
              newcommandddmathrmd
              newcommandds[1]displaystyle#1
              newcommandexpo[1],mathrme^#1,
              newcommandicmathrmi
              newcommandmc[1]mathcal#1
              newcommandmrm[1]mathrm#1
              newcommandpars[1]left(,#1,right)
              newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
              newcommandroot[2][],sqrt[#1],#2,,
              newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
              newcommandverts[1]leftvert,#1,rightvert$



              You can add this "weird" answer to the excellent
              $texttt@mrtaurho$ long list:
              beginalign
              &bbox[10px,#ffd]%
              int_0^1lnpars1 - xlnparsx over x,dd x =
              left.partial^2 over partialmu,partialnuint_0^1
              brackspars1 - x^mu - 1x^nu over x,dd x,rightvert_largemu = 0 atop large,,nu = 0^+
              \[5mm] = &
              partial^2 over partialmu,partialnubracks%
              int_0^1x^nu - 1pars1 - x^mu,dd x -
              int_0^1x^nu - 1,dd x
              _largemu = 0 atop large,,nu = 0^+
              \[5mm] = &
              partial^2 over partialmu,partialnubracks%
              GammaparsnuGammaparsmu + 1 over
              Gammaparsnu + mu + 1 - 1 over nu
              _largemu = 0 atop large,,nu = 0^+
              \[5mm] = &
              partial^2 over partialmu,partialnubraces1 over nubracks%
              Gammaparsnu + 1Gammaparsmu + 1 over
              Gammaparsnu + mu + 1 - 1
              _largemu = 0 atop large,,nu = 0^+
              \[5mm] = &
              1 over 2,partiald[2]nubracesGammaparsnu + 1
              partialdmubracksGammaparsmu + 1 over
              Gammaparsnu + mu + 1
              _largemu = 0 atop large,,nu = 0^+
              \[5mm] = &
              1 over 2,partiald[2]nubracesGammaparsnu + 1
              bracks-,gamma + Psiparsnu + 1 over
              Gammaparsnu + 1 _ nu = 0^+
              \[5mm] = &
              -,1 over 2,Psi,''pars1 = bbxzetapars3
              endalign






              share|cite|improve this answer











              $endgroup$

















                4












                $begingroup$

                $newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
                newcommandbraces[1]leftlbrace,#1,rightrbrace
                newcommandbracks[1]leftlbrack,#1,rightrbrack
                newcommandddmathrmd
                newcommandds[1]displaystyle#1
                newcommandexpo[1],mathrme^#1,
                newcommandicmathrmi
                newcommandmc[1]mathcal#1
                newcommandmrm[1]mathrm#1
                newcommandpars[1]left(,#1,right)
                newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
                newcommandroot[2][],sqrt[#1],#2,,
                newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
                newcommandverts[1]leftvert,#1,rightvert$



                You can add this "weird" answer to the excellent
                $texttt@mrtaurho$ long list:
                beginalign
                &bbox[10px,#ffd]%
                int_0^1lnpars1 - xlnparsx over x,dd x =
                left.partial^2 over partialmu,partialnuint_0^1
                brackspars1 - x^mu - 1x^nu over x,dd x,rightvert_largemu = 0 atop large,,nu = 0^+
                \[5mm] = &
                partial^2 over partialmu,partialnubracks%
                int_0^1x^nu - 1pars1 - x^mu,dd x -
                int_0^1x^nu - 1,dd x
                _largemu = 0 atop large,,nu = 0^+
                \[5mm] = &
                partial^2 over partialmu,partialnubracks%
                GammaparsnuGammaparsmu + 1 over
                Gammaparsnu + mu + 1 - 1 over nu
                _largemu = 0 atop large,,nu = 0^+
                \[5mm] = &
                partial^2 over partialmu,partialnubraces1 over nubracks%
                Gammaparsnu + 1Gammaparsmu + 1 over
                Gammaparsnu + mu + 1 - 1
                _largemu = 0 atop large,,nu = 0^+
                \[5mm] = &
                1 over 2,partiald[2]nubracesGammaparsnu + 1
                partialdmubracksGammaparsmu + 1 over
                Gammaparsnu + mu + 1
                _largemu = 0 atop large,,nu = 0^+
                \[5mm] = &
                1 over 2,partiald[2]nubracesGammaparsnu + 1
                bracks-,gamma + Psiparsnu + 1 over
                Gammaparsnu + 1 _ nu = 0^+
                \[5mm] = &
                -,1 over 2,Psi,''pars1 = bbxzetapars3
                endalign






                share|cite|improve this answer











                $endgroup$















                  4












                  4








                  4





                  $begingroup$

                  $newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
                  newcommandbraces[1]leftlbrace,#1,rightrbrace
                  newcommandbracks[1]leftlbrack,#1,rightrbrack
                  newcommandddmathrmd
                  newcommandds[1]displaystyle#1
                  newcommandexpo[1],mathrme^#1,
                  newcommandicmathrmi
                  newcommandmc[1]mathcal#1
                  newcommandmrm[1]mathrm#1
                  newcommandpars[1]left(,#1,right)
                  newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
                  newcommandroot[2][],sqrt[#1],#2,,
                  newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
                  newcommandverts[1]leftvert,#1,rightvert$



                  You can add this "weird" answer to the excellent
                  $texttt@mrtaurho$ long list:
                  beginalign
                  &bbox[10px,#ffd]%
                  int_0^1lnpars1 - xlnparsx over x,dd x =
                  left.partial^2 over partialmu,partialnuint_0^1
                  brackspars1 - x^mu - 1x^nu over x,dd x,rightvert_largemu = 0 atop large,,nu = 0^+
                  \[5mm] = &
                  partial^2 over partialmu,partialnubracks%
                  int_0^1x^nu - 1pars1 - x^mu,dd x -
                  int_0^1x^nu - 1,dd x
                  _largemu = 0 atop large,,nu = 0^+
                  \[5mm] = &
                  partial^2 over partialmu,partialnubracks%
                  GammaparsnuGammaparsmu + 1 over
                  Gammaparsnu + mu + 1 - 1 over nu
                  _largemu = 0 atop large,,nu = 0^+
                  \[5mm] = &
                  partial^2 over partialmu,partialnubraces1 over nubracks%
                  Gammaparsnu + 1Gammaparsmu + 1 over
                  Gammaparsnu + mu + 1 - 1
                  _largemu = 0 atop large,,nu = 0^+
                  \[5mm] = &
                  1 over 2,partiald[2]nubracesGammaparsnu + 1
                  partialdmubracksGammaparsmu + 1 over
                  Gammaparsnu + mu + 1
                  _largemu = 0 atop large,,nu = 0^+
                  \[5mm] = &
                  1 over 2,partiald[2]nubracesGammaparsnu + 1
                  bracks-,gamma + Psiparsnu + 1 over
                  Gammaparsnu + 1 _ nu = 0^+
                  \[5mm] = &
                  -,1 over 2,Psi,''pars1 = bbxzetapars3
                  endalign






                  share|cite|improve this answer











                  $endgroup$



                  $newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
                  newcommandbraces[1]leftlbrace,#1,rightrbrace
                  newcommandbracks[1]leftlbrack,#1,rightrbrack
                  newcommandddmathrmd
                  newcommandds[1]displaystyle#1
                  newcommandexpo[1],mathrme^#1,
                  newcommandicmathrmi
                  newcommandmc[1]mathcal#1
                  newcommandmrm[1]mathrm#1
                  newcommandpars[1]left(,#1,right)
                  newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
                  newcommandroot[2][],sqrt[#1],#2,,
                  newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
                  newcommandverts[1]leftvert,#1,rightvert$



                  You can add this "weird" answer to the excellent
                  $texttt@mrtaurho$ long list:
                  beginalign
                  &bbox[10px,#ffd]%
                  int_0^1lnpars1 - xlnparsx over x,dd x =
                  left.partial^2 over partialmu,partialnuint_0^1
                  brackspars1 - x^mu - 1x^nu over x,dd x,rightvert_largemu = 0 atop large,,nu = 0^+
                  \[5mm] = &
                  partial^2 over partialmu,partialnubracks%
                  int_0^1x^nu - 1pars1 - x^mu,dd x -
                  int_0^1x^nu - 1,dd x
                  _largemu = 0 atop large,,nu = 0^+
                  \[5mm] = &
                  partial^2 over partialmu,partialnubracks%
                  GammaparsnuGammaparsmu + 1 over
                  Gammaparsnu + mu + 1 - 1 over nu
                  _largemu = 0 atop large,,nu = 0^+
                  \[5mm] = &
                  partial^2 over partialmu,partialnubraces1 over nubracks%
                  Gammaparsnu + 1Gammaparsmu + 1 over
                  Gammaparsnu + mu + 1 - 1
                  _largemu = 0 atop large,,nu = 0^+
                  \[5mm] = &
                  1 over 2,partiald[2]nubracesGammaparsnu + 1
                  partialdmubracksGammaparsmu + 1 over
                  Gammaparsnu + mu + 1
                  _largemu = 0 atop large,,nu = 0^+
                  \[5mm] = &
                  1 over 2,partiald[2]nubracesGammaparsnu + 1
                  bracks-,gamma + Psiparsnu + 1 over
                  Gammaparsnu + 1 _ nu = 0^+
                  \[5mm] = &
                  -,1 over 2,Psi,''pars1 = bbxzetapars3
                  endalign







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Apr 27 at 6:41

























                  answered Apr 27 at 6:36









                  Felix MarinFelix Marin

                  69.6k7111148




                  69.6k7111148





















                      3












                      $begingroup$

                      Let $x = e^-y$, we have
                      $$int_0^1 fraclog xlog(1-x)x dx
                      = int_0^1 frac(-log x)x sum_n=1^infty fracx^nn dx
                      = sum_n=1^infty frac1nint_0^1 (-log x) x^n-1 dx\
                      = sum_n=1^infty frac1nint_0^infty y e^-ny dy
                      = sum_n=1^infty frac1n^3
                      = zeta(3)
                      $$
                      Please note that we can switch the order of summation and integration because all the individual terms are non-negative.






                      share|cite|improve this answer











                      $endgroup$












                      • $begingroup$
                        Very thanks much @achillehui incredible. I take notes from your solution.
                        $endgroup$
                        – user243301
                        Dec 18 '15 at 18:37










                      • $begingroup$
                        @achillehui Are you exploiting Fubini-Tonelli with the counting measure on $mathbbN$ and Lebesgue measure on $[0,1]$ to justify interchanging the series and integration?
                        $endgroup$
                        – Mark Viola
                        Mar 20 '17 at 2:56










                      • $begingroup$
                        @Dr.MV the last phrase "because all the individual terms are non-negative" is essentially Tonelli theorem.
                        $endgroup$
                        – achille hui
                        Mar 20 '17 at 4:05










                      • $begingroup$
                        @achillehui Yes, I know. So, you are exploiting it then?
                        $endgroup$
                        – Mark Viola
                        Mar 20 '17 at 4:19















                      3












                      $begingroup$

                      Let $x = e^-y$, we have
                      $$int_0^1 fraclog xlog(1-x)x dx
                      = int_0^1 frac(-log x)x sum_n=1^infty fracx^nn dx
                      = sum_n=1^infty frac1nint_0^1 (-log x) x^n-1 dx\
                      = sum_n=1^infty frac1nint_0^infty y e^-ny dy
                      = sum_n=1^infty frac1n^3
                      = zeta(3)
                      $$
                      Please note that we can switch the order of summation and integration because all the individual terms are non-negative.






                      share|cite|improve this answer











                      $endgroup$












                      • $begingroup$
                        Very thanks much @achillehui incredible. I take notes from your solution.
                        $endgroup$
                        – user243301
                        Dec 18 '15 at 18:37










                      • $begingroup$
                        @achillehui Are you exploiting Fubini-Tonelli with the counting measure on $mathbbN$ and Lebesgue measure on $[0,1]$ to justify interchanging the series and integration?
                        $endgroup$
                        – Mark Viola
                        Mar 20 '17 at 2:56










                      • $begingroup$
                        @Dr.MV the last phrase "because all the individual terms are non-negative" is essentially Tonelli theorem.
                        $endgroup$
                        – achille hui
                        Mar 20 '17 at 4:05










                      • $begingroup$
                        @achillehui Yes, I know. So, you are exploiting it then?
                        $endgroup$
                        – Mark Viola
                        Mar 20 '17 at 4:19













                      3












                      3








                      3





                      $begingroup$

                      Let $x = e^-y$, we have
                      $$int_0^1 fraclog xlog(1-x)x dx
                      = int_0^1 frac(-log x)x sum_n=1^infty fracx^nn dx
                      = sum_n=1^infty frac1nint_0^1 (-log x) x^n-1 dx\
                      = sum_n=1^infty frac1nint_0^infty y e^-ny dy
                      = sum_n=1^infty frac1n^3
                      = zeta(3)
                      $$
                      Please note that we can switch the order of summation and integration because all the individual terms are non-negative.






                      share|cite|improve this answer











                      $endgroup$



                      Let $x = e^-y$, we have
                      $$int_0^1 fraclog xlog(1-x)x dx
                      = int_0^1 frac(-log x)x sum_n=1^infty fracx^nn dx
                      = sum_n=1^infty frac1nint_0^1 (-log x) x^n-1 dx\
                      = sum_n=1^infty frac1nint_0^infty y e^-ny dy
                      = sum_n=1^infty frac1n^3
                      = zeta(3)
                      $$
                      Please note that we can switch the order of summation and integration because all the individual terms are non-negative.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 18 '15 at 18:39

























                      answered Dec 18 '15 at 18:36









                      achille huiachille hui

                      97.2k5132263




                      97.2k5132263











                      • $begingroup$
                        Very thanks much @achillehui incredible. I take notes from your solution.
                        $endgroup$
                        – user243301
                        Dec 18 '15 at 18:37










                      • $begingroup$
                        @achillehui Are you exploiting Fubini-Tonelli with the counting measure on $mathbbN$ and Lebesgue measure on $[0,1]$ to justify interchanging the series and integration?
                        $endgroup$
                        – Mark Viola
                        Mar 20 '17 at 2:56










                      • $begingroup$
                        @Dr.MV the last phrase "because all the individual terms are non-negative" is essentially Tonelli theorem.
                        $endgroup$
                        – achille hui
                        Mar 20 '17 at 4:05










                      • $begingroup$
                        @achillehui Yes, I know. So, you are exploiting it then?
                        $endgroup$
                        – Mark Viola
                        Mar 20 '17 at 4:19
















                      • $begingroup$
                        Very thanks much @achillehui incredible. I take notes from your solution.
                        $endgroup$
                        – user243301
                        Dec 18 '15 at 18:37










                      • $begingroup$
                        @achillehui Are you exploiting Fubini-Tonelli with the counting measure on $mathbbN$ and Lebesgue measure on $[0,1]$ to justify interchanging the series and integration?
                        $endgroup$
                        – Mark Viola
                        Mar 20 '17 at 2:56










                      • $begingroup$
                        @Dr.MV the last phrase "because all the individual terms are non-negative" is essentially Tonelli theorem.
                        $endgroup$
                        – achille hui
                        Mar 20 '17 at 4:05










                      • $begingroup$
                        @achillehui Yes, I know. So, you are exploiting it then?
                        $endgroup$
                        – Mark Viola
                        Mar 20 '17 at 4:19















                      $begingroup$
                      Very thanks much @achillehui incredible. I take notes from your solution.
                      $endgroup$
                      – user243301
                      Dec 18 '15 at 18:37




                      $begingroup$
                      Very thanks much @achillehui incredible. I take notes from your solution.
                      $endgroup$
                      – user243301
                      Dec 18 '15 at 18:37












                      $begingroup$
                      @achillehui Are you exploiting Fubini-Tonelli with the counting measure on $mathbbN$ and Lebesgue measure on $[0,1]$ to justify interchanging the series and integration?
                      $endgroup$
                      – Mark Viola
                      Mar 20 '17 at 2:56




                      $begingroup$
                      @achillehui Are you exploiting Fubini-Tonelli with the counting measure on $mathbbN$ and Lebesgue measure on $[0,1]$ to justify interchanging the series and integration?
                      $endgroup$
                      – Mark Viola
                      Mar 20 '17 at 2:56












                      $begingroup$
                      @Dr.MV the last phrase "because all the individual terms are non-negative" is essentially Tonelli theorem.
                      $endgroup$
                      – achille hui
                      Mar 20 '17 at 4:05




                      $begingroup$
                      @Dr.MV the last phrase "because all the individual terms are non-negative" is essentially Tonelli theorem.
                      $endgroup$
                      – achille hui
                      Mar 20 '17 at 4:05












                      $begingroup$
                      @achillehui Yes, I know. So, you are exploiting it then?
                      $endgroup$
                      – Mark Viola
                      Mar 20 '17 at 4:19




                      $begingroup$
                      @achillehui Yes, I know. So, you are exploiting it then?
                      $endgroup$
                      – Mark Viola
                      Mar 20 '17 at 4:19











                      1












                      $begingroup$

                      The Beta function and Feynman's trick are another way to go:



                      $$I=int_0^1fraclog(x)log(1-x)x,dx =left.fracpartial^2partial a partial bint_0^1x^a-1(1-x)^b,dx,right|_alpha,beta=0^+tag1 $$
                      hence:
                      $$ I = left.fracpartial^2partial a partial bfracGamma(a)Gamma(b+1)Gamma(a+b+1),right|_alpha,beta=0^+tag2 $$
                      and by exploiting $Gamma'(z) = Gamma(z)cdotpsi(z)$ we get:
                      $$ I = -frac12psi''(2)=sum_ngeq 0frac1(n+1)^3=colorredzeta(3)tag3 $$
                      as wanted.






                      share|cite|improve this answer









                      $endgroup$












                      • $begingroup$
                        I try understand your contribution too, now I am saturated. My changes to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense. Very thanks much @JackD'Aurizio.
                        $endgroup$
                        – user243301
                        Dec 18 '15 at 18:54















                      1












                      $begingroup$

                      The Beta function and Feynman's trick are another way to go:



                      $$I=int_0^1fraclog(x)log(1-x)x,dx =left.fracpartial^2partial a partial bint_0^1x^a-1(1-x)^b,dx,right|_alpha,beta=0^+tag1 $$
                      hence:
                      $$ I = left.fracpartial^2partial a partial bfracGamma(a)Gamma(b+1)Gamma(a+b+1),right|_alpha,beta=0^+tag2 $$
                      and by exploiting $Gamma'(z) = Gamma(z)cdotpsi(z)$ we get:
                      $$ I = -frac12psi''(2)=sum_ngeq 0frac1(n+1)^3=colorredzeta(3)tag3 $$
                      as wanted.






                      share|cite|improve this answer









                      $endgroup$












                      • $begingroup$
                        I try understand your contribution too, now I am saturated. My changes to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense. Very thanks much @JackD'Aurizio.
                        $endgroup$
                        – user243301
                        Dec 18 '15 at 18:54













                      1












                      1








                      1





                      $begingroup$

                      The Beta function and Feynman's trick are another way to go:



                      $$I=int_0^1fraclog(x)log(1-x)x,dx =left.fracpartial^2partial a partial bint_0^1x^a-1(1-x)^b,dx,right|_alpha,beta=0^+tag1 $$
                      hence:
                      $$ I = left.fracpartial^2partial a partial bfracGamma(a)Gamma(b+1)Gamma(a+b+1),right|_alpha,beta=0^+tag2 $$
                      and by exploiting $Gamma'(z) = Gamma(z)cdotpsi(z)$ we get:
                      $$ I = -frac12psi''(2)=sum_ngeq 0frac1(n+1)^3=colorredzeta(3)tag3 $$
                      as wanted.






                      share|cite|improve this answer









                      $endgroup$



                      The Beta function and Feynman's trick are another way to go:



                      $$I=int_0^1fraclog(x)log(1-x)x,dx =left.fracpartial^2partial a partial bint_0^1x^a-1(1-x)^b,dx,right|_alpha,beta=0^+tag1 $$
                      hence:
                      $$ I = left.fracpartial^2partial a partial bfracGamma(a)Gamma(b+1)Gamma(a+b+1),right|_alpha,beta=0^+tag2 $$
                      and by exploiting $Gamma'(z) = Gamma(z)cdotpsi(z)$ we get:
                      $$ I = -frac12psi''(2)=sum_ngeq 0frac1(n+1)^3=colorredzeta(3)tag3 $$
                      as wanted.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 18 '15 at 18:48









                      Jack D'AurizioJack D'Aurizio

                      293k33285675




                      293k33285675











                      • $begingroup$
                        I try understand your contribution too, now I am saturated. My changes to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense. Very thanks much @JackD'Aurizio.
                        $endgroup$
                        – user243301
                        Dec 18 '15 at 18:54
















                      • $begingroup$
                        I try understand your contribution too, now I am saturated. My changes to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense. Very thanks much @JackD'Aurizio.
                        $endgroup$
                        – user243301
                        Dec 18 '15 at 18:54















                      $begingroup$
                      I try understand your contribution too, now I am saturated. My changes to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense. Very thanks much @JackD'Aurizio.
                      $endgroup$
                      – user243301
                      Dec 18 '15 at 18:54




                      $begingroup$
                      I try understand your contribution too, now I am saturated. My changes to obtain the integral as a summand (there were two summands) were first $x=e^z/y$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense. Very thanks much @JackD'Aurizio.
                      $endgroup$
                      – user243301
                      Dec 18 '15 at 18:54

















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