Find limit in use of integralshow to find the limit?Find limit $lim_x rightarrow 0 , fracarctan(3x)tanbig((x+3pi)/3big)$ without using L'Hospital's rulemathematical analysisLimit of a complex sequenceUsing the mean value theorem calculate this limit : $lim_x rightarrow 0 fracarctan(x^2+x-1)+fracpi4x^2+3x$Limit of a definite integral with parameter (2)Limit of $(intlimits_0^n (1+arctan^2x ),dx )^ frac1n$limit of and integral depending on nA tricky limit involving exponential integralsFind limit supremum from 3 sequence theorem
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Find limit in use of integrals
how to find the limit?Find limit $lim_x rightarrow 0 , fracarctan(3x)tanbig((x+3pi)/3big)$ without using L'Hospital's rulemathematical analysisLimit of a complex sequenceUsing the mean value theorem calculate this limit : $lim_x rightarrow 0 fracarctan(x^2+x-1)+fracpi4x^2+3x$Limit of a definite integral with parameter (2)Limit of $(intlimits_0^n (1+arctan^2x ),dx )^ frac1n$limit of and integral depending on nA tricky limit involving exponential integralsFind limit supremum from 3 sequence theorem
$begingroup$
Find limit $lim_n rightarrow infty int_-1^1 x^5 cdot arctan(nx) dx $
From mean-value-theorem we have
$$ frac12 c^5 cdot arctan(nc) mbox for some c in (-1,1) $$
$$ underbracefrac12 cdot arctan(-n)_rightarrow - pi /4 le frac12 c^5 cdot arctan(nc) le underbracefrac12 cdot arctan(n)_rightarrow pi /4 $$
so this bounding doesn't help me. Has somebody better idea how to bound that?
integration limits
asked May 25 at 15:41
Trebacz112Trebacz112
675
$endgroup$
add a comment |
$begingroup$
Find limit $lim_n rightarrow infty int_-1^1 x^5 cdot arctan(nx) dx $
From mean-value-theorem we have
$$ frac12 c^5 cdot arctan(nc) mbox for some c in (-1,1) $$
$$ underbracefrac12 cdot arctan(-n)_rightarrow - pi /4 le frac12 c^5 cdot arctan(nc) le underbracefrac12 cdot arctan(n)_rightarrow pi /4 $$
so this bounding doesn't help me. Has somebody better idea how to bound that?
integration limits
asked May 25 at 15:41
Trebacz112Trebacz112
675
$endgroup$
$begingroup$
Is there a reason to not just do the integration?
$endgroup$
– Clayton
May 25 at 15:45
add a comment |
$begingroup$
Find limit $lim_n rightarrow infty int_-1^1 x^5 cdot arctan(nx) dx $
From mean-value-theorem we have
$$ frac12 c^5 cdot arctan(nc) mbox for some c in (-1,1) $$
$$ underbracefrac12 cdot arctan(-n)_rightarrow - pi /4 le frac12 c^5 cdot arctan(nc) le underbracefrac12 cdot arctan(n)_rightarrow pi /4 $$
so this bounding doesn't help me. Has somebody better idea how to bound that?
integration limits
asked May 25 at 15:41
Trebacz112Trebacz112
675
$endgroup$
Find limit $lim_n rightarrow infty int_-1^1 x^5 cdot arctan(nx) dx $
From mean-value-theorem we have
$$ frac12 c^5 cdot arctan(nc) mbox for some c in (-1,1) $$
$$ underbracefrac12 cdot arctan(-n)_rightarrow - pi /4 le frac12 c^5 cdot arctan(nc) le underbracefrac12 cdot arctan(n)_rightarrow pi /4 $$
so this bounding doesn't help me. Has somebody better idea how to bound that?
integration limits
integration limits
asked May 25 at 15:41
Trebacz112Trebacz112
675
asked May 25 at 15:41
Trebacz112Trebacz112
675
asked May 25 at 15:41
Trebacz112Trebacz112
675
asked May 25 at 15:41
Trebacz112Trebacz112
675
asked May 25 at 15:41
Trebacz112Trebacz112
675
675
$begingroup$
Is there a reason to not just do the integration?
$endgroup$
– Clayton
May 25 at 15:45
add a comment |
$begingroup$
Is there a reason to not just do the integration?
$endgroup$
– Clayton
May 25 at 15:45
$begingroup$
Is there a reason to not just do the integration?
$endgroup$
– Clayton
May 25 at 15:45
$begingroup$
Is there a reason to not just do the integration?
$endgroup$
– Clayton
May 25 at 15:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We'll simplify with $y=nx$. Integration by parts gives $$int y^5arctan ydy=fracy^66arctan y-frac16intfracy^61+y^2dy\=fracy^66arctan y-frac16intleft(y^4-y^2+1-frac11+y^2right)dy\=fracy^6+16arctan y-frac130y^5+frac118y^3-frac16 y+C.$$Hence $$frac1n^6int_-n^n y^5arctan ydy=fracfracn^63arctan n+o(n^6)n^6stackrelntoinftytofrac13arctaninfty=fracpi6.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2lim_ntoinftyint_0^1 x^5arctan nxdx$ (since the integrand is even), which by dominated convergence is $$piint_0^1 x^5dx=fracpi6.$$
answered May 25 at 15:50
J.G.J.G.
39.6k23758
$endgroup$
add a comment |
$begingroup$
Hint: For the integral $$int_-1^1x^5arctan(nx)dx$$ we get
$$frac15 left(n^6+1right) tan ^-1(n)-3 n^5+5
n^3-15 n45 n^6$$
answered May 25 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
82.3k42867
$endgroup$
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
May 25 at 15:46
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
May 25 at 15:47
1
$begingroup$
Hint: $$fracx^6n^2x^2+1=frac x^4n^2-frac x^2n^4+n^-6-frac 1n^6 left( n^2x^2+1 right) $$
$endgroup$
– Dr. Sonnhard Graubner
May 25 at 15:49
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
We'll simplify with $y=nx$. Integration by parts gives $$int y^5arctan ydy=fracy^66arctan y-frac16intfracy^61+y^2dy\=fracy^66arctan y-frac16intleft(y^4-y^2+1-frac11+y^2right)dy\=fracy^6+16arctan y-frac130y^5+frac118y^3-frac16 y+C.$$Hence $$frac1n^6int_-n^n y^5arctan ydy=fracfracn^63arctan n+o(n^6)n^6stackrelntoinftytofrac13arctaninfty=fracpi6.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2lim_ntoinftyint_0^1 x^5arctan nxdx$ (since the integrand is even), which by dominated convergence is $$piint_0^1 x^5dx=fracpi6.$$
answered May 25 at 15:50
J.G.J.G.
39.6k23758
$endgroup$
add a comment |
$begingroup$
We'll simplify with $y=nx$. Integration by parts gives $$int y^5arctan ydy=fracy^66arctan y-frac16intfracy^61+y^2dy\=fracy^66arctan y-frac16intleft(y^4-y^2+1-frac11+y^2right)dy\=fracy^6+16arctan y-frac130y^5+frac118y^3-frac16 y+C.$$Hence $$frac1n^6int_-n^n y^5arctan ydy=fracfracn^63arctan n+o(n^6)n^6stackrelntoinftytofrac13arctaninfty=fracpi6.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2lim_ntoinftyint_0^1 x^5arctan nxdx$ (since the integrand is even), which by dominated convergence is $$piint_0^1 x^5dx=fracpi6.$$
answered May 25 at 15:50
J.G.J.G.
39.6k23758
$endgroup$
add a comment |
$begingroup$
We'll simplify with $y=nx$. Integration by parts gives $$int y^5arctan ydy=fracy^66arctan y-frac16intfracy^61+y^2dy\=fracy^66arctan y-frac16intleft(y^4-y^2+1-frac11+y^2right)dy\=fracy^6+16arctan y-frac130y^5+frac118y^3-frac16 y+C.$$Hence $$frac1n^6int_-n^n y^5arctan ydy=fracfracn^63arctan n+o(n^6)n^6stackrelntoinftytofrac13arctaninfty=fracpi6.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2lim_ntoinftyint_0^1 x^5arctan nxdx$ (since the integrand is even), which by dominated convergence is $$piint_0^1 x^5dx=fracpi6.$$
answered May 25 at 15:50
J.G.J.G.
39.6k23758
$endgroup$
We'll simplify with $y=nx$. Integration by parts gives $$int y^5arctan ydy=fracy^66arctan y-frac16intfracy^61+y^2dy\=fracy^66arctan y-frac16intleft(y^4-y^2+1-frac11+y^2right)dy\=fracy^6+16arctan y-frac130y^5+frac118y^3-frac16 y+C.$$Hence $$frac1n^6int_-n^n y^5arctan ydy=fracfracn^63arctan n+o(n^6)n^6stackrelntoinftytofrac13arctaninfty=fracpi6.$$But it seems such a shame to compute the antiderivative's irrelevant polynomial terms. So for an alternative strategy, let's write the problem as $2lim_ntoinftyint_0^1 x^5arctan nxdx$ (since the integrand is even), which by dominated convergence is $$piint_0^1 x^5dx=fracpi6.$$
answered May 25 at 15:50
J.G.J.G.
39.6k23758
edited May 25 at 15:55
answered May 25 at 15:50
J.G.J.G.
39.6k23758
answered May 25 at 15:50
J.G.J.G.
39.6k23758
answered May 25 at 15:50
J.G.J.G.
39.6k23758
39.6k23758
add a comment |
add a comment |
$begingroup$
Hint: For the integral $$int_-1^1x^5arctan(nx)dx$$ we get
$$frac15 left(n^6+1right) tan ^-1(n)-3 n^5+5
n^3-15 n45 n^6$$
answered May 25 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
82.3k42867
$endgroup$
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
May 25 at 15:46
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
May 25 at 15:47
1
$begingroup$
Hint: $$fracx^6n^2x^2+1=frac x^4n^2-frac x^2n^4+n^-6-frac 1n^6 left( n^2x^2+1 right) $$
$endgroup$
– Dr. Sonnhard Graubner
May 25 at 15:49
add a comment |
$begingroup$
Hint: For the integral $$int_-1^1x^5arctan(nx)dx$$ we get
$$frac15 left(n^6+1right) tan ^-1(n)-3 n^5+5
n^3-15 n45 n^6$$
answered May 25 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
82.3k42867
$endgroup$
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
May 25 at 15:46
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
May 25 at 15:47
1
$begingroup$
Hint: $$fracx^6n^2x^2+1=frac x^4n^2-frac x^2n^4+n^-6-frac 1n^6 left( n^2x^2+1 right) $$
$endgroup$
– Dr. Sonnhard Graubner
May 25 at 15:49
add a comment |
$begingroup$
Hint: For the integral $$int_-1^1x^5arctan(nx)dx$$ we get
$$frac15 left(n^6+1right) tan ^-1(n)-3 n^5+5
n^3-15 n45 n^6$$
answered May 25 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
82.3k42867
$endgroup$
Hint: For the integral $$int_-1^1x^5arctan(nx)dx$$ we get
$$frac15 left(n^6+1right) tan ^-1(n)-3 n^5+5
n^3-15 n45 n^6$$
answered May 25 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
82.3k42867
answered May 25 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
82.3k42867
answered May 25 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
82.3k42867
answered May 25 at 15:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
82.3k42867
82.3k42867
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
May 25 at 15:46
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
May 25 at 15:47
1
$begingroup$
Hint: $$fracx^6n^2x^2+1=frac x^4n^2-frac x^2n^4+n^-6-frac 1n^6 left( n^2x^2+1 right) $$
$endgroup$
– Dr. Sonnhard Graubner
May 25 at 15:49
add a comment |
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
May 25 at 15:46
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
May 25 at 15:47
1
$begingroup$
Hint: $$fracx^6n^2x^2+1=frac x^4n^2-frac x^2n^4+n^-6-frac 1n^6 left( n^2x^2+1 right) $$
$endgroup$
– Dr. Sonnhard Graubner
May 25 at 15:49
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
May 25 at 15:46
$begingroup$
When I am doing integration I stuck there $int x^6/(n^2cdot x^2 +1)$
$endgroup$
– Trebacz112
May 25 at 15:46
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
May 25 at 15:47
$begingroup$
Partial fractions, @Trebacz112
$endgroup$
– Clayton
May 25 at 15:47
1
1
$begingroup$
Hint: $$fracx^6n^2x^2+1=frac x^4n^2-frac x^2n^4+n^-6-frac 1n^6 left( n^2x^2+1 right) $$
$endgroup$
– Dr. Sonnhard Graubner
May 25 at 15:49
$begingroup$
Hint: $$fracx^6n^2x^2+1=frac x^4n^2-frac x^2n^4+n^-6-frac 1n^6 left( n^2x^2+1 right) $$
$endgroup$
– Dr. Sonnhard Graubner
May 25 at 15:49
add a comment |
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$begingroup$
Is there a reason to not just do the integration?
$endgroup$
– Clayton
May 25 at 15:45