Limit of this definite integral [closed]Can't compute definite integralHow to evaluate this definite integral in Mathematica?Evaluation of this definite integralAny reason why this definite integral is so slow to compute and the indefinite fast?Strange result of parameter-dependent definite integralEvaluate an Exponential involving an Integral OperatorDifficult definite integralHow to calculate this integral symbolically?Solving integral equationApproximating integral with small parameter
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Limit of this definite integral [closed]
Can't compute definite integralHow to evaluate this definite integral in Mathematica?Evaluation of this definite integralAny reason why this definite integral is so slow to compute and the indefinite fast?Strange result of parameter-dependent definite integralEvaluate an Exponential involving an Integral OperatorDifficult definite integralHow to calculate this integral symbolically?Solving integral equationApproximating integral with small parameter
$begingroup$
How can I compute this integral for $yrightarrow0$ ?
$$ int_0^1fracy(1-x)^2(1+x)x+(1-x^2)y dx $$
calculus-and-analysis
asked May 1 at 8:36
Luca RossiLuca Rossi
41
$endgroup$
closed as off-topic by corey979, Roman, Michael E2, xzczd, AccidentalFourierTransform May 1 at 14:39
- The question does not concern the technical computing software Mathematica by Wolfram Research. Please see the help center to find out about the topics that can be asked here.
add a comment |
$begingroup$
How can I compute this integral for $yrightarrow0$ ?
$$ int_0^1fracy(1-x)^2(1+x)x+(1-x^2)y dx $$
calculus-and-analysis
asked May 1 at 8:36
Luca RossiLuca Rossi
41
$endgroup$
closed as off-topic by corey979, Roman, Michael E2, xzczd, AccidentalFourierTransform May 1 at 14:39
- The question does not concern the technical computing software Mathematica by Wolfram Research. Please see the help center to find out about the topics that can be asked here.
$begingroup$
Did you really mean to ask this here since you also posted on Math.SE?
$endgroup$
– StubbornAtom
May 1 at 10:19
$begingroup$
Am I missing something? The integrand clearly goes to 0 as y approaches zero.
$endgroup$
– infinitezero
May 1 at 13:26
1
$begingroup$
@infinitezero It seems there are some doubts about order of limits iny->0andxclose to zero. Doingxintegral first and then takingy->0does give the expected zero of course.
$endgroup$
– Kagaratsch
May 1 at 13:30
add a comment |
$begingroup$
How can I compute this integral for $yrightarrow0$ ?
$$ int_0^1fracy(1-x)^2(1+x)x+(1-x^2)y dx $$
calculus-and-analysis
asked May 1 at 8:36
Luca RossiLuca Rossi
41
$endgroup$
How can I compute this integral for $yrightarrow0$ ?
$$ int_0^1fracy(1-x)^2(1+x)x+(1-x^2)y dx $$
calculus-and-analysis
calculus-and-analysis
asked May 1 at 8:36
Luca RossiLuca Rossi
41
asked May 1 at 8:36
Luca RossiLuca Rossi
41
asked May 1 at 8:36
Luca RossiLuca Rossi
41
asked May 1 at 8:36
Luca RossiLuca Rossi
41
asked May 1 at 8:36
Luca RossiLuca Rossi
41
41
closed as off-topic by corey979, Roman, Michael E2, xzczd, AccidentalFourierTransform May 1 at 14:39
- The question does not concern the technical computing software Mathematica by Wolfram Research. Please see the help center to find out about the topics that can be asked here.
closed as off-topic by corey979, Roman, Michael E2, xzczd, AccidentalFourierTransform May 1 at 14:39
- The question does not concern the technical computing software Mathematica by Wolfram Research. Please see the help center to find out about the topics that can be asked here.
$begingroup$
Did you really mean to ask this here since you also posted on Math.SE?
$endgroup$
– StubbornAtom
May 1 at 10:19
$begingroup$
Am I missing something? The integrand clearly goes to 0 as y approaches zero.
$endgroup$
– infinitezero
May 1 at 13:26
1
$begingroup$
@infinitezero It seems there are some doubts about order of limits iny->0andxclose to zero. Doingxintegral first and then takingy->0does give the expected zero of course.
$endgroup$
– Kagaratsch
May 1 at 13:30
add a comment |
$begingroup$
Did you really mean to ask this here since you also posted on Math.SE?
$endgroup$
– StubbornAtom
May 1 at 10:19
$begingroup$
Am I missing something? The integrand clearly goes to 0 as y approaches zero.
$endgroup$
– infinitezero
May 1 at 13:26
1
$begingroup$
@infinitezero It seems there are some doubts about order of limits iny->0andxclose to zero. Doingxintegral first and then takingy->0does give the expected zero of course.
$endgroup$
– Kagaratsch
May 1 at 13:30
$begingroup$
Did you really mean to ask this here since you also posted on Math.SE?
$endgroup$
– StubbornAtom
May 1 at 10:19
$begingroup$
Did you really mean to ask this here since you also posted on Math.SE?
$endgroup$
– StubbornAtom
May 1 at 10:19
$begingroup$
Am I missing something? The integrand clearly goes to 0 as y approaches zero.
$endgroup$
– infinitezero
May 1 at 13:26
$begingroup$
Am I missing something? The integrand clearly goes to 0 as y approaches zero.
$endgroup$
– infinitezero
May 1 at 13:26
1
1
$begingroup$
@infinitezero It seems there are some doubts about order of limits in
y->0 and x close to zero. Doing x integral first and then taking y->0 does give the expected zero of course.$endgroup$
– Kagaratsch
May 1 at 13:30
$begingroup$
@infinitezero It seems there are some doubts about order of limits in
y->0 and x close to zero. Doing x integral first and then taking y->0 does give the expected zero of course.$endgroup$
– Kagaratsch
May 1 at 13:30
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Assuming[y > 0,
AsymptoticIntegrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1, y, 0, 1]]
1/6 y (-7 - 6 Log[y])
So the limit for $yto0^+$ is zero:
Limit[%, y -> 0, Direction -> "FromAbove"]
0
For $y<0$ the integral does not converge.
answered May 1 at 9:24
RomanRoman
7,93411237
$endgroup$
$begingroup$
Is there a way to achieve the same result but analytically?
$endgroup$
– Luca Rossi
May 1 at 9:30
$begingroup$
Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1, Assumptions -> y > 0]maybe, although the above is already analytic.
$endgroup$
– Roman
May 1 at 9:33
add a comment |
$begingroup$
Another approach if you want to avoid the use of AsymptoticIntegrate (whose very presence I learnt today, thanks @Roman:-)!).
Timing[
FullSimplify[
Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1,
Assumptions -> y > 0], y > 0]]
(* 6.64063, ((-2 + y) y Sqrt[1 + 4 y^2] - (-1 + y) Sqrt[
1 + 4 y^2]
Log[y] + (1 - 2 y) y Log[(1 + 2 y - Sqrt[1 + 4 y^2])/(
1 + 2 y + Sqrt[1 + 4 y^2])] +
Log[(1 + 2 y + Sqrt[1 + 4 y^2])/(1 + 2 y - Sqrt[1 + 4 y^2])])/(
2 y^2 Sqrt[1 + 4 y^2]) *)
Normal[
Series[(1/(
2 y^2 Sqrt[
1 + 4 y^2]))((-2 + y) y Sqrt[1 + 4 y^2] - (-1 + y) Sqrt[1 + 4 y^2]
Log[y] + (1 - 2 y) y Log[(1 + 2 y - Sqrt[1 + 4 y^2])/(
1 + 2 y + Sqrt[1 + 4 y^2])] +
Log[(1 + 2 y + Sqrt[1 + 4 y^2])/(
1 + 2 y - Sqrt[1 + 4 y^2])]), y, 0, 1]] //
FullSimplify[#, y > 0] &
(* -(1/6) y (7 + 6 Log[y]) *)
answered May 1 at 10:56
DimitrisDimitris
2,5131332
$endgroup$
add a comment |
$begingroup$
The OP asked whether there is a way to achieve the result "analytically", which I assume implies "by steps that can be considered a proof by humans". Here is an attempt.
Define integrand
integrand = (y (1 - x)^2 (1 + x))/(x + (1 - x^2) y);
For any y>0 we see that there is no pole on the interval 1>x>0, which means we can obtain the integral over this region using the fundamental theorem of calculus by taking a difference of anti-derivatives evaluated at the boundary points.
First we may "guess" an anti-derivative:
antiderivative = Assuming[y > 0, Integrate[integrand, x] // Simplify]
to prove that this is in fact a valid anti-derivative, simply take the derivative which is a simple mechanical process and can be done either by hand or as:
D[antiderivative, x] - integrand // Simplify
0
Having proven that we in fact have a correct anti-derivative, we evaluate it at the boundary points and take a difference:
integral = (antiderivative /. x -> 1) - (antiderivative /. x -> 0) // FullSimplify
Finally, we take a series expansion of this result around the point y=0 to figure out its leading behavior:
Assuming[y > 0, Series[integral, y, 0, 0]]
This means that the integral result vanishes at least linearly in y->0 so that the integral is zero in that limit.
Note that the derivative to prove the anti-derivative, the evaluation at x=1 and x=0, and the Taylor expansion around y=0 all can be done by hand on a piece of paper, which makes the steps humanly traceable and therefore constitutes a proof.
answered May 1 at 13:17
KagaratschKagaratsch
5,03231350
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming[y > 0,
AsymptoticIntegrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1, y, 0, 1]]
1/6 y (-7 - 6 Log[y])
So the limit for $yto0^+$ is zero:
Limit[%, y -> 0, Direction -> "FromAbove"]
0
For $y<0$ the integral does not converge.
answered May 1 at 9:24
RomanRoman
7,93411237
$endgroup$
$begingroup$
Is there a way to achieve the same result but analytically?
$endgroup$
– Luca Rossi
May 1 at 9:30
$begingroup$
Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1, Assumptions -> y > 0]maybe, although the above is already analytic.
$endgroup$
– Roman
May 1 at 9:33
add a comment |
$begingroup$
Assuming[y > 0,
AsymptoticIntegrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1, y, 0, 1]]
1/6 y (-7 - 6 Log[y])
So the limit for $yto0^+$ is zero:
Limit[%, y -> 0, Direction -> "FromAbove"]
0
For $y<0$ the integral does not converge.
answered May 1 at 9:24
RomanRoman
7,93411237
$endgroup$
$begingroup$
Is there a way to achieve the same result but analytically?
$endgroup$
– Luca Rossi
May 1 at 9:30
$begingroup$
Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1, Assumptions -> y > 0]maybe, although the above is already analytic.
$endgroup$
– Roman
May 1 at 9:33
add a comment |
$begingroup$
Assuming[y > 0,
AsymptoticIntegrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1, y, 0, 1]]
1/6 y (-7 - 6 Log[y])
So the limit for $yto0^+$ is zero:
Limit[%, y -> 0, Direction -> "FromAbove"]
0
For $y<0$ the integral does not converge.
answered May 1 at 9:24
RomanRoman
7,93411237
$endgroup$
Assuming[y > 0,
AsymptoticIntegrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1, y, 0, 1]]
1/6 y (-7 - 6 Log[y])
So the limit for $yto0^+$ is zero:
Limit[%, y -> 0, Direction -> "FromAbove"]
0
For $y<0$ the integral does not converge.
answered May 1 at 9:24
RomanRoman
7,93411237
edited May 1 at 11:00
answered May 1 at 9:24
RomanRoman
7,93411237
answered May 1 at 9:24
RomanRoman
7,93411237
answered May 1 at 9:24
RomanRoman
7,93411237
7,93411237
$begingroup$
Is there a way to achieve the same result but analytically?
$endgroup$
– Luca Rossi
May 1 at 9:30
$begingroup$
Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1, Assumptions -> y > 0]maybe, although the above is already analytic.
$endgroup$
– Roman
May 1 at 9:33
add a comment |
$begingroup$
Is there a way to achieve the same result but analytically?
$endgroup$
– Luca Rossi
May 1 at 9:30
$begingroup$
Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1, Assumptions -> y > 0]maybe, although the above is already analytic.
$endgroup$
– Roman
May 1 at 9:33
$begingroup$
Is there a way to achieve the same result but analytically?
$endgroup$
– Luca Rossi
May 1 at 9:30
$begingroup$
Is there a way to achieve the same result but analytically?
$endgroup$
– Luca Rossi
May 1 at 9:30
$begingroup$
Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1, Assumptions -> y > 0] maybe, although the above is already analytic.$endgroup$
– Roman
May 1 at 9:33
$begingroup$
Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1, Assumptions -> y > 0] maybe, although the above is already analytic.$endgroup$
– Roman
May 1 at 9:33
add a comment |
$begingroup$
Another approach if you want to avoid the use of AsymptoticIntegrate (whose very presence I learnt today, thanks @Roman:-)!).
Timing[
FullSimplify[
Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1,
Assumptions -> y > 0], y > 0]]
(* 6.64063, ((-2 + y) y Sqrt[1 + 4 y^2] - (-1 + y) Sqrt[
1 + 4 y^2]
Log[y] + (1 - 2 y) y Log[(1 + 2 y - Sqrt[1 + 4 y^2])/(
1 + 2 y + Sqrt[1 + 4 y^2])] +
Log[(1 + 2 y + Sqrt[1 + 4 y^2])/(1 + 2 y - Sqrt[1 + 4 y^2])])/(
2 y^2 Sqrt[1 + 4 y^2]) *)
Normal[
Series[(1/(
2 y^2 Sqrt[
1 + 4 y^2]))((-2 + y) y Sqrt[1 + 4 y^2] - (-1 + y) Sqrt[1 + 4 y^2]
Log[y] + (1 - 2 y) y Log[(1 + 2 y - Sqrt[1 + 4 y^2])/(
1 + 2 y + Sqrt[1 + 4 y^2])] +
Log[(1 + 2 y + Sqrt[1 + 4 y^2])/(
1 + 2 y - Sqrt[1 + 4 y^2])]), y, 0, 1]] //
FullSimplify[#, y > 0] &
(* -(1/6) y (7 + 6 Log[y]) *)
answered May 1 at 10:56
DimitrisDimitris
2,5131332
$endgroup$
add a comment |
$begingroup$
Another approach if you want to avoid the use of AsymptoticIntegrate (whose very presence I learnt today, thanks @Roman:-)!).
Timing[
FullSimplify[
Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1,
Assumptions -> y > 0], y > 0]]
(* 6.64063, ((-2 + y) y Sqrt[1 + 4 y^2] - (-1 + y) Sqrt[
1 + 4 y^2]
Log[y] + (1 - 2 y) y Log[(1 + 2 y - Sqrt[1 + 4 y^2])/(
1 + 2 y + Sqrt[1 + 4 y^2])] +
Log[(1 + 2 y + Sqrt[1 + 4 y^2])/(1 + 2 y - Sqrt[1 + 4 y^2])])/(
2 y^2 Sqrt[1 + 4 y^2]) *)
Normal[
Series[(1/(
2 y^2 Sqrt[
1 + 4 y^2]))((-2 + y) y Sqrt[1 + 4 y^2] - (-1 + y) Sqrt[1 + 4 y^2]
Log[y] + (1 - 2 y) y Log[(1 + 2 y - Sqrt[1 + 4 y^2])/(
1 + 2 y + Sqrt[1 + 4 y^2])] +
Log[(1 + 2 y + Sqrt[1 + 4 y^2])/(
1 + 2 y - Sqrt[1 + 4 y^2])]), y, 0, 1]] //
FullSimplify[#, y > 0] &
(* -(1/6) y (7 + 6 Log[y]) *)
answered May 1 at 10:56
DimitrisDimitris
2,5131332
$endgroup$
add a comment |
$begingroup$
Another approach if you want to avoid the use of AsymptoticIntegrate (whose very presence I learnt today, thanks @Roman:-)!).
Timing[
FullSimplify[
Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1,
Assumptions -> y > 0], y > 0]]
(* 6.64063, ((-2 + y) y Sqrt[1 + 4 y^2] - (-1 + y) Sqrt[
1 + 4 y^2]
Log[y] + (1 - 2 y) y Log[(1 + 2 y - Sqrt[1 + 4 y^2])/(
1 + 2 y + Sqrt[1 + 4 y^2])] +
Log[(1 + 2 y + Sqrt[1 + 4 y^2])/(1 + 2 y - Sqrt[1 + 4 y^2])])/(
2 y^2 Sqrt[1 + 4 y^2]) *)
Normal[
Series[(1/(
2 y^2 Sqrt[
1 + 4 y^2]))((-2 + y) y Sqrt[1 + 4 y^2] - (-1 + y) Sqrt[1 + 4 y^2]
Log[y] + (1 - 2 y) y Log[(1 + 2 y - Sqrt[1 + 4 y^2])/(
1 + 2 y + Sqrt[1 + 4 y^2])] +
Log[(1 + 2 y + Sqrt[1 + 4 y^2])/(
1 + 2 y - Sqrt[1 + 4 y^2])]), y, 0, 1]] //
FullSimplify[#, y > 0] &
(* -(1/6) y (7 + 6 Log[y]) *)
answered May 1 at 10:56
DimitrisDimitris
2,5131332
$endgroup$
Another approach if you want to avoid the use of AsymptoticIntegrate (whose very presence I learnt today, thanks @Roman:-)!).
Timing[
FullSimplify[
Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y), x, 0, 1,
Assumptions -> y > 0], y > 0]]
(* 6.64063, ((-2 + y) y Sqrt[1 + 4 y^2] - (-1 + y) Sqrt[
1 + 4 y^2]
Log[y] + (1 - 2 y) y Log[(1 + 2 y - Sqrt[1 + 4 y^2])/(
1 + 2 y + Sqrt[1 + 4 y^2])] +
Log[(1 + 2 y + Sqrt[1 + 4 y^2])/(1 + 2 y - Sqrt[1 + 4 y^2])])/(
2 y^2 Sqrt[1 + 4 y^2]) *)
Normal[
Series[(1/(
2 y^2 Sqrt[
1 + 4 y^2]))((-2 + y) y Sqrt[1 + 4 y^2] - (-1 + y) Sqrt[1 + 4 y^2]
Log[y] + (1 - 2 y) y Log[(1 + 2 y - Sqrt[1 + 4 y^2])/(
1 + 2 y + Sqrt[1 + 4 y^2])] +
Log[(1 + 2 y + Sqrt[1 + 4 y^2])/(
1 + 2 y - Sqrt[1 + 4 y^2])]), y, 0, 1]] //
FullSimplify[#, y > 0] &
(* -(1/6) y (7 + 6 Log[y]) *)
answered May 1 at 10:56
DimitrisDimitris
2,5131332
edited May 1 at 11:04
answered May 1 at 10:56
DimitrisDimitris
2,5131332
answered May 1 at 10:56
DimitrisDimitris
2,5131332
answered May 1 at 10:56
DimitrisDimitris
2,5131332
2,5131332
add a comment |
add a comment |
$begingroup$
The OP asked whether there is a way to achieve the result "analytically", which I assume implies "by steps that can be considered a proof by humans". Here is an attempt.
Define integrand
integrand = (y (1 - x)^2 (1 + x))/(x + (1 - x^2) y);
For any y>0 we see that there is no pole on the interval 1>x>0, which means we can obtain the integral over this region using the fundamental theorem of calculus by taking a difference of anti-derivatives evaluated at the boundary points.
First we may "guess" an anti-derivative:
antiderivative = Assuming[y > 0, Integrate[integrand, x] // Simplify]
to prove that this is in fact a valid anti-derivative, simply take the derivative which is a simple mechanical process and can be done either by hand or as:
D[antiderivative, x] - integrand // Simplify
0
Having proven that we in fact have a correct anti-derivative, we evaluate it at the boundary points and take a difference:
integral = (antiderivative /. x -> 1) - (antiderivative /. x -> 0) // FullSimplify
Finally, we take a series expansion of this result around the point y=0 to figure out its leading behavior:
Assuming[y > 0, Series[integral, y, 0, 0]]
This means that the integral result vanishes at least linearly in y->0 so that the integral is zero in that limit.
Note that the derivative to prove the anti-derivative, the evaluation at x=1 and x=0, and the Taylor expansion around y=0 all can be done by hand on a piece of paper, which makes the steps humanly traceable and therefore constitutes a proof.
answered May 1 at 13:17
KagaratschKagaratsch
5,03231350
$endgroup$
add a comment |
$begingroup$
The OP asked whether there is a way to achieve the result "analytically", which I assume implies "by steps that can be considered a proof by humans". Here is an attempt.
Define integrand
integrand = (y (1 - x)^2 (1 + x))/(x + (1 - x^2) y);
For any y>0 we see that there is no pole on the interval 1>x>0, which means we can obtain the integral over this region using the fundamental theorem of calculus by taking a difference of anti-derivatives evaluated at the boundary points.
First we may "guess" an anti-derivative:
antiderivative = Assuming[y > 0, Integrate[integrand, x] // Simplify]
to prove that this is in fact a valid anti-derivative, simply take the derivative which is a simple mechanical process and can be done either by hand or as:
D[antiderivative, x] - integrand // Simplify
0
Having proven that we in fact have a correct anti-derivative, we evaluate it at the boundary points and take a difference:
integral = (antiderivative /. x -> 1) - (antiderivative /. x -> 0) // FullSimplify
Finally, we take a series expansion of this result around the point y=0 to figure out its leading behavior:
Assuming[y > 0, Series[integral, y, 0, 0]]
This means that the integral result vanishes at least linearly in y->0 so that the integral is zero in that limit.
Note that the derivative to prove the anti-derivative, the evaluation at x=1 and x=0, and the Taylor expansion around y=0 all can be done by hand on a piece of paper, which makes the steps humanly traceable and therefore constitutes a proof.
answered May 1 at 13:17
KagaratschKagaratsch
5,03231350
$endgroup$
add a comment |
$begingroup$
The OP asked whether there is a way to achieve the result "analytically", which I assume implies "by steps that can be considered a proof by humans". Here is an attempt.
Define integrand
integrand = (y (1 - x)^2 (1 + x))/(x + (1 - x^2) y);
For any y>0 we see that there is no pole on the interval 1>x>0, which means we can obtain the integral over this region using the fundamental theorem of calculus by taking a difference of anti-derivatives evaluated at the boundary points.
First we may "guess" an anti-derivative:
antiderivative = Assuming[y > 0, Integrate[integrand, x] // Simplify]
to prove that this is in fact a valid anti-derivative, simply take the derivative which is a simple mechanical process and can be done either by hand or as:
D[antiderivative, x] - integrand // Simplify
0
Having proven that we in fact have a correct anti-derivative, we evaluate it at the boundary points and take a difference:
integral = (antiderivative /. x -> 1) - (antiderivative /. x -> 0) // FullSimplify
Finally, we take a series expansion of this result around the point y=0 to figure out its leading behavior:
Assuming[y > 0, Series[integral, y, 0, 0]]
This means that the integral result vanishes at least linearly in y->0 so that the integral is zero in that limit.
Note that the derivative to prove the anti-derivative, the evaluation at x=1 and x=0, and the Taylor expansion around y=0 all can be done by hand on a piece of paper, which makes the steps humanly traceable and therefore constitutes a proof.
answered May 1 at 13:17
KagaratschKagaratsch
5,03231350
$endgroup$
The OP asked whether there is a way to achieve the result "analytically", which I assume implies "by steps that can be considered a proof by humans". Here is an attempt.
Define integrand
integrand = (y (1 - x)^2 (1 + x))/(x + (1 - x^2) y);
For any y>0 we see that there is no pole on the interval 1>x>0, which means we can obtain the integral over this region using the fundamental theorem of calculus by taking a difference of anti-derivatives evaluated at the boundary points.
First we may "guess" an anti-derivative:
antiderivative = Assuming[y > 0, Integrate[integrand, x] // Simplify]
to prove that this is in fact a valid anti-derivative, simply take the derivative which is a simple mechanical process and can be done either by hand or as:
D[antiderivative, x] - integrand // Simplify
0
Having proven that we in fact have a correct anti-derivative, we evaluate it at the boundary points and take a difference:
integral = (antiderivative /. x -> 1) - (antiderivative /. x -> 0) // FullSimplify
Finally, we take a series expansion of this result around the point y=0 to figure out its leading behavior:
Assuming[y > 0, Series[integral, y, 0, 0]]
This means that the integral result vanishes at least linearly in y->0 so that the integral is zero in that limit.
Note that the derivative to prove the anti-derivative, the evaluation at x=1 and x=0, and the Taylor expansion around y=0 all can be done by hand on a piece of paper, which makes the steps humanly traceable and therefore constitutes a proof.
answered May 1 at 13:17
KagaratschKagaratsch
5,03231350
answered May 1 at 13:17
KagaratschKagaratsch
5,03231350
answered May 1 at 13:17
KagaratschKagaratsch
5,03231350
answered May 1 at 13:17
KagaratschKagaratsch
5,03231350
5,03231350
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Did you really mean to ask this here since you also posted on Math.SE?
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– StubbornAtom
May 1 at 10:19
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Am I missing something? The integrand clearly goes to 0 as y approaches zero.
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– infinitezero
May 1 at 13:26
1
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@infinitezero It seems there are some doubts about order of limits in
y->0andxclose to zero. Doingxintegral first and then takingy->0does give the expected zero of course.$endgroup$
– Kagaratsch
May 1 at 13:30