How to solve this improper integral $int _{-infty}^{infty} e^{ax}/(e^x+1) dx$? [closed]
$begingroup$
I need to solve the following integral.
$$int _{-infty }^{infty }::dfrac{e^{ax}}{e^x+1}dx$$
where $0<a<1$.
integration improper-integrals contour-integration
$endgroup$
closed as off-topic by Brahadeesh, Shaun, user10354138, mrtaurho, Paul Frost Dec 20 '18 at 18:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, Shaun, user10354138, mrtaurho, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 11 more comments
$begingroup$
I need to solve the following integral.
$$int _{-infty }^{infty }::dfrac{e^{ax}}{e^x+1}dx$$
where $0<a<1$.
integration improper-integrals contour-integration
$endgroup$
closed as off-topic by Brahadeesh, Shaun, user10354138, mrtaurho, Paul Frost Dec 20 '18 at 18:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, Shaun, user10354138, mrtaurho, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Your integral seems divergent.
$endgroup$
– Awe Kumar Jha
Dec 20 '18 at 10:39
1
$begingroup$
Come on, people! It is convergent and is equal to $mathrm{B}(a,1-a)=dfrac{pi}{sin api}$.
$endgroup$
– metamorphy
Dec 20 '18 at 12:01
1
$begingroup$
Did you find $mathrm{B}(x,y)=displaystyleint_0^inftyfrac{t^{x-1},dt}{(1+t)^{x+y}}$ there? Please read the article before asking further questions - I wouldn't want to reproduce the whole of it here ;)
$endgroup$
– metamorphy
Dec 20 '18 at 12:18
1
$begingroup$
@John You could solve this integral using a rectangular contour which goes from -R to R on the real axis and which is of height $2pi i$. Thus you have a pole at $x=pi i$. If you edit your question so that it fits the rules of the site, I can maybe answer your question, because I had to solve this exact same integral today also using contour integration.
$endgroup$
– Poujh
Dec 20 '18 at 20:45
1
$begingroup$
@Poujh That's honestly much easier than the approach I took. So thanks for helping me learn today.
$endgroup$
– Dylan
Dec 21 '18 at 3:56
|
show 11 more comments
$begingroup$
I need to solve the following integral.
$$int _{-infty }^{infty }::dfrac{e^{ax}}{e^x+1}dx$$
where $0<a<1$.
integration improper-integrals contour-integration
$endgroup$
I need to solve the following integral.
$$int _{-infty }^{infty }::dfrac{e^{ax}}{e^x+1}dx$$
where $0<a<1$.
integration improper-integrals contour-integration
integration improper-integrals contour-integration
edited Dec 20 '18 at 11:02
Brahadeesh
6,50642364
6,50642364
asked Dec 20 '18 at 10:31
JohnJohn
8110
8110
closed as off-topic by Brahadeesh, Shaun, user10354138, mrtaurho, Paul Frost Dec 20 '18 at 18:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, Shaun, user10354138, mrtaurho, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Brahadeesh, Shaun, user10354138, mrtaurho, Paul Frost Dec 20 '18 at 18:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, Shaun, user10354138, mrtaurho, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Your integral seems divergent.
$endgroup$
– Awe Kumar Jha
Dec 20 '18 at 10:39
1
$begingroup$
Come on, people! It is convergent and is equal to $mathrm{B}(a,1-a)=dfrac{pi}{sin api}$.
$endgroup$
– metamorphy
Dec 20 '18 at 12:01
1
$begingroup$
Did you find $mathrm{B}(x,y)=displaystyleint_0^inftyfrac{t^{x-1},dt}{(1+t)^{x+y}}$ there? Please read the article before asking further questions - I wouldn't want to reproduce the whole of it here ;)
$endgroup$
– metamorphy
Dec 20 '18 at 12:18
1
$begingroup$
@John You could solve this integral using a rectangular contour which goes from -R to R on the real axis and which is of height $2pi i$. Thus you have a pole at $x=pi i$. If you edit your question so that it fits the rules of the site, I can maybe answer your question, because I had to solve this exact same integral today also using contour integration.
$endgroup$
– Poujh
Dec 20 '18 at 20:45
1
$begingroup$
@Poujh That's honestly much easier than the approach I took. So thanks for helping me learn today.
$endgroup$
– Dylan
Dec 21 '18 at 3:56
|
show 11 more comments
2
$begingroup$
Your integral seems divergent.
$endgroup$
– Awe Kumar Jha
Dec 20 '18 at 10:39
1
$begingroup$
Come on, people! It is convergent and is equal to $mathrm{B}(a,1-a)=dfrac{pi}{sin api}$.
$endgroup$
– metamorphy
Dec 20 '18 at 12:01
1
$begingroup$
Did you find $mathrm{B}(x,y)=displaystyleint_0^inftyfrac{t^{x-1},dt}{(1+t)^{x+y}}$ there? Please read the article before asking further questions - I wouldn't want to reproduce the whole of it here ;)
$endgroup$
– metamorphy
Dec 20 '18 at 12:18
1
$begingroup$
@John You could solve this integral using a rectangular contour which goes from -R to R on the real axis and which is of height $2pi i$. Thus you have a pole at $x=pi i$. If you edit your question so that it fits the rules of the site, I can maybe answer your question, because I had to solve this exact same integral today also using contour integration.
$endgroup$
– Poujh
Dec 20 '18 at 20:45
1
$begingroup$
@Poujh That's honestly much easier than the approach I took. So thanks for helping me learn today.
$endgroup$
– Dylan
Dec 21 '18 at 3:56
2
2
$begingroup$
Your integral seems divergent.
$endgroup$
– Awe Kumar Jha
Dec 20 '18 at 10:39
$begingroup$
Your integral seems divergent.
$endgroup$
– Awe Kumar Jha
Dec 20 '18 at 10:39
1
1
$begingroup$
Come on, people! It is convergent and is equal to $mathrm{B}(a,1-a)=dfrac{pi}{sin api}$.
$endgroup$
– metamorphy
Dec 20 '18 at 12:01
$begingroup$
Come on, people! It is convergent and is equal to $mathrm{B}(a,1-a)=dfrac{pi}{sin api}$.
$endgroup$
– metamorphy
Dec 20 '18 at 12:01
1
1
$begingroup$
Did you find $mathrm{B}(x,y)=displaystyleint_0^inftyfrac{t^{x-1},dt}{(1+t)^{x+y}}$ there? Please read the article before asking further questions - I wouldn't want to reproduce the whole of it here ;)
$endgroup$
– metamorphy
Dec 20 '18 at 12:18
$begingroup$
Did you find $mathrm{B}(x,y)=displaystyleint_0^inftyfrac{t^{x-1},dt}{(1+t)^{x+y}}$ there? Please read the article before asking further questions - I wouldn't want to reproduce the whole of it here ;)
$endgroup$
– metamorphy
Dec 20 '18 at 12:18
1
1
$begingroup$
@John You could solve this integral using a rectangular contour which goes from -R to R on the real axis and which is of height $2pi i$. Thus you have a pole at $x=pi i$. If you edit your question so that it fits the rules of the site, I can maybe answer your question, because I had to solve this exact same integral today also using contour integration.
$endgroup$
– Poujh
Dec 20 '18 at 20:45
$begingroup$
@John You could solve this integral using a rectangular contour which goes from -R to R on the real axis and which is of height $2pi i$. Thus you have a pole at $x=pi i$. If you edit your question so that it fits the rules of the site, I can maybe answer your question, because I had to solve this exact same integral today also using contour integration.
$endgroup$
– Poujh
Dec 20 '18 at 20:45
1
1
$begingroup$
@Poujh That's honestly much easier than the approach I took. So thanks for helping me learn today.
$endgroup$
– Dylan
Dec 21 '18 at 3:56
$begingroup$
@Poujh That's honestly much easier than the approach I took. So thanks for helping me learn today.
$endgroup$
– Dylan
Dec 21 '18 at 3:56
|
show 11 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The substitution $e^x mapsto x$ gives
$$ I = int_{-infty}^infty frac{e^{ax}}{e^x+1}dx = int_0^infty frac{1}{x^{1-a}(1+x)}dx $$
where $0 < 1-a < 1$
We can use complex integration here. Due to the pole at $z=-1$, we pick a branch cut on the positive real axis such that $0 le arg(z) < 2pi$. The contour is a standard "keyhole contour" consisting of
$C_1$: left semicircle centered at $0$ with radius $epsilon$, going clockwise from $-epsilon i$ to $epsilon i$
$C_2$: straight line from $epsilon i$ to $R + epsilon i$
$C_3$: circle centered at $0$ with radius $R$, going counterclockwise from $R+iepsilon$ to $R - epsilon i$
$C_4$: straight line from $R - epsilon i$ to $-epsilon i$
In the limits $epsilon to 0$ and $R to infty$, the two straight line integrals converge to
$$ int_{C_2} f(z) dz to int_0^infty frac{1}{x^{1-a}e^{i0(1-a)}(1+x)}dx = I $$
$$ int_{C_4} f(z) dz to -int_0^infty frac{1}{x^{1-a}e^{i2pi(1-a)}(1+x)}dx = -e^{i2pi a}I $$
The remaining integrals should go to $0$
$$ leftvertint_{C_1} f(z)dzrightvert le frac{pi epsilon}{|z|^{1-a}|1+z|} le frac{pi epsilon^a}{1-epsilon} to 0 $$
$$ leftvertint_{C_3} f(z) dz rightvert le frac{2pi R}{|z|^{1-a}|1+z|} le frac{2pi R^a}{R-1} to 0 $$
Therefore
$$ (1-e^{i2pi a})I = 2pi ioperatorname*{Res}_{z=-1} f(z) = -2pi i e^{ipi a} $$
So finally
$$ I = frac{2pi ie^{ipi a}}{e^{2pi a}-1} = frac{2pi i}{e^{ipi a}-e^{-ipi a}} = frac{pi}{sin(api)} $$
Edit: Here's a simpler method, credit to @Poujh
$$ I = int_{-infty}^infty frac{e^{ax}}{e^x+1}dx = int_{-infty}^infty f(x) dx $$
Take a rectangular contour:
$C_1$: Straight line from $-R$ to $R$
$C_2$: Straight line from $R$ to $R + i2pi$
$C_3$: Straight line from $R + i2pi$ to $-R + i2pi$
$C_4$: Straight line from $-R + i2pi$ to $-R$
In the limit of $Rtoinfty$, we have
$$ leftvertint_{C_2} frac{e^{az}}{e^z+1} dzrightvert = leftvertint_0^{2pi} frac{e^{a(R+iy)}}{e^{R+iy}+1} dyrightvert le 2pi leftvertfrac{e^{aR}}{e^R-1}rightvert to 0 $$
$$ leftvertint_{C_4} frac{e^{az}}{e^z+1} dzrightvert = leftvertint_0^{2pi} frac{e^{a(-R+iy)}}{e^{-R+iy}+1} rightvert le 2pi leftvert frac{e^{-aR}}{1-e^{-R}} rightvert to 0 $$
For the horizontal lines
$$ int_{C_1} frac{e^{az}}{e^z+1} dz to int_{-infty}^infty frac{e^{ax}}{e^x+1} dx = I $$
$$ int_{C_3} frac{e^{az}}{e^z+1} dz to -int_{-infty}^infty frac{e^{ax}e^{i2pi a}}{e^x+1} = -e^{i2pi a} I $$
We end up with the same result
$$ (1-e^{i2pi a})I = 2pi ioperatorname*{Res}_{z=ipi} frac{e^{az}}{e^z+1} $$
where the residue is computed using the limit
$$ lim_{zto ipi} e^{az}frac{z-ipi}{e^z+1} = -e^{ipi a} $$
$endgroup$
$begingroup$
What about the other poles in this second method? Of course $exp(z)+1$ has zeros at $(2n+1)pi i$.
$endgroup$
– John
Dec 23 '18 at 12:22
1
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@John The other poles are outside the contour
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– Dylan
Dec 23 '18 at 13:16
1
$begingroup$
yes I got it and the residue is $- exp(i pi a)$, you can edit it.
$endgroup$
– John
Dec 23 '18 at 13:23
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The substitution $e^x mapsto x$ gives
$$ I = int_{-infty}^infty frac{e^{ax}}{e^x+1}dx = int_0^infty frac{1}{x^{1-a}(1+x)}dx $$
where $0 < 1-a < 1$
We can use complex integration here. Due to the pole at $z=-1$, we pick a branch cut on the positive real axis such that $0 le arg(z) < 2pi$. The contour is a standard "keyhole contour" consisting of
$C_1$: left semicircle centered at $0$ with radius $epsilon$, going clockwise from $-epsilon i$ to $epsilon i$
$C_2$: straight line from $epsilon i$ to $R + epsilon i$
$C_3$: circle centered at $0$ with radius $R$, going counterclockwise from $R+iepsilon$ to $R - epsilon i$
$C_4$: straight line from $R - epsilon i$ to $-epsilon i$
In the limits $epsilon to 0$ and $R to infty$, the two straight line integrals converge to
$$ int_{C_2} f(z) dz to int_0^infty frac{1}{x^{1-a}e^{i0(1-a)}(1+x)}dx = I $$
$$ int_{C_4} f(z) dz to -int_0^infty frac{1}{x^{1-a}e^{i2pi(1-a)}(1+x)}dx = -e^{i2pi a}I $$
The remaining integrals should go to $0$
$$ leftvertint_{C_1} f(z)dzrightvert le frac{pi epsilon}{|z|^{1-a}|1+z|} le frac{pi epsilon^a}{1-epsilon} to 0 $$
$$ leftvertint_{C_3} f(z) dz rightvert le frac{2pi R}{|z|^{1-a}|1+z|} le frac{2pi R^a}{R-1} to 0 $$
Therefore
$$ (1-e^{i2pi a})I = 2pi ioperatorname*{Res}_{z=-1} f(z) = -2pi i e^{ipi a} $$
So finally
$$ I = frac{2pi ie^{ipi a}}{e^{2pi a}-1} = frac{2pi i}{e^{ipi a}-e^{-ipi a}} = frac{pi}{sin(api)} $$
Edit: Here's a simpler method, credit to @Poujh
$$ I = int_{-infty}^infty frac{e^{ax}}{e^x+1}dx = int_{-infty}^infty f(x) dx $$
Take a rectangular contour:
$C_1$: Straight line from $-R$ to $R$
$C_2$: Straight line from $R$ to $R + i2pi$
$C_3$: Straight line from $R + i2pi$ to $-R + i2pi$
$C_4$: Straight line from $-R + i2pi$ to $-R$
In the limit of $Rtoinfty$, we have
$$ leftvertint_{C_2} frac{e^{az}}{e^z+1} dzrightvert = leftvertint_0^{2pi} frac{e^{a(R+iy)}}{e^{R+iy}+1} dyrightvert le 2pi leftvertfrac{e^{aR}}{e^R-1}rightvert to 0 $$
$$ leftvertint_{C_4} frac{e^{az}}{e^z+1} dzrightvert = leftvertint_0^{2pi} frac{e^{a(-R+iy)}}{e^{-R+iy}+1} rightvert le 2pi leftvert frac{e^{-aR}}{1-e^{-R}} rightvert to 0 $$
For the horizontal lines
$$ int_{C_1} frac{e^{az}}{e^z+1} dz to int_{-infty}^infty frac{e^{ax}}{e^x+1} dx = I $$
$$ int_{C_3} frac{e^{az}}{e^z+1} dz to -int_{-infty}^infty frac{e^{ax}e^{i2pi a}}{e^x+1} = -e^{i2pi a} I $$
We end up with the same result
$$ (1-e^{i2pi a})I = 2pi ioperatorname*{Res}_{z=ipi} frac{e^{az}}{e^z+1} $$
where the residue is computed using the limit
$$ lim_{zto ipi} e^{az}frac{z-ipi}{e^z+1} = -e^{ipi a} $$
$endgroup$
$begingroup$
What about the other poles in this second method? Of course $exp(z)+1$ has zeros at $(2n+1)pi i$.
$endgroup$
– John
Dec 23 '18 at 12:22
1
$begingroup$
@John The other poles are outside the contour
$endgroup$
– Dylan
Dec 23 '18 at 13:16
1
$begingroup$
yes I got it and the residue is $- exp(i pi a)$, you can edit it.
$endgroup$
– John
Dec 23 '18 at 13:23
add a comment |
$begingroup$
The substitution $e^x mapsto x$ gives
$$ I = int_{-infty}^infty frac{e^{ax}}{e^x+1}dx = int_0^infty frac{1}{x^{1-a}(1+x)}dx $$
where $0 < 1-a < 1$
We can use complex integration here. Due to the pole at $z=-1$, we pick a branch cut on the positive real axis such that $0 le arg(z) < 2pi$. The contour is a standard "keyhole contour" consisting of
$C_1$: left semicircle centered at $0$ with radius $epsilon$, going clockwise from $-epsilon i$ to $epsilon i$
$C_2$: straight line from $epsilon i$ to $R + epsilon i$
$C_3$: circle centered at $0$ with radius $R$, going counterclockwise from $R+iepsilon$ to $R - epsilon i$
$C_4$: straight line from $R - epsilon i$ to $-epsilon i$
In the limits $epsilon to 0$ and $R to infty$, the two straight line integrals converge to
$$ int_{C_2} f(z) dz to int_0^infty frac{1}{x^{1-a}e^{i0(1-a)}(1+x)}dx = I $$
$$ int_{C_4} f(z) dz to -int_0^infty frac{1}{x^{1-a}e^{i2pi(1-a)}(1+x)}dx = -e^{i2pi a}I $$
The remaining integrals should go to $0$
$$ leftvertint_{C_1} f(z)dzrightvert le frac{pi epsilon}{|z|^{1-a}|1+z|} le frac{pi epsilon^a}{1-epsilon} to 0 $$
$$ leftvertint_{C_3} f(z) dz rightvert le frac{2pi R}{|z|^{1-a}|1+z|} le frac{2pi R^a}{R-1} to 0 $$
Therefore
$$ (1-e^{i2pi a})I = 2pi ioperatorname*{Res}_{z=-1} f(z) = -2pi i e^{ipi a} $$
So finally
$$ I = frac{2pi ie^{ipi a}}{e^{2pi a}-1} = frac{2pi i}{e^{ipi a}-e^{-ipi a}} = frac{pi}{sin(api)} $$
Edit: Here's a simpler method, credit to @Poujh
$$ I = int_{-infty}^infty frac{e^{ax}}{e^x+1}dx = int_{-infty}^infty f(x) dx $$
Take a rectangular contour:
$C_1$: Straight line from $-R$ to $R$
$C_2$: Straight line from $R$ to $R + i2pi$
$C_3$: Straight line from $R + i2pi$ to $-R + i2pi$
$C_4$: Straight line from $-R + i2pi$ to $-R$
In the limit of $Rtoinfty$, we have
$$ leftvertint_{C_2} frac{e^{az}}{e^z+1} dzrightvert = leftvertint_0^{2pi} frac{e^{a(R+iy)}}{e^{R+iy}+1} dyrightvert le 2pi leftvertfrac{e^{aR}}{e^R-1}rightvert to 0 $$
$$ leftvertint_{C_4} frac{e^{az}}{e^z+1} dzrightvert = leftvertint_0^{2pi} frac{e^{a(-R+iy)}}{e^{-R+iy}+1} rightvert le 2pi leftvert frac{e^{-aR}}{1-e^{-R}} rightvert to 0 $$
For the horizontal lines
$$ int_{C_1} frac{e^{az}}{e^z+1} dz to int_{-infty}^infty frac{e^{ax}}{e^x+1} dx = I $$
$$ int_{C_3} frac{e^{az}}{e^z+1} dz to -int_{-infty}^infty frac{e^{ax}e^{i2pi a}}{e^x+1} = -e^{i2pi a} I $$
We end up with the same result
$$ (1-e^{i2pi a})I = 2pi ioperatorname*{Res}_{z=ipi} frac{e^{az}}{e^z+1} $$
where the residue is computed using the limit
$$ lim_{zto ipi} e^{az}frac{z-ipi}{e^z+1} = -e^{ipi a} $$
$endgroup$
$begingroup$
What about the other poles in this second method? Of course $exp(z)+1$ has zeros at $(2n+1)pi i$.
$endgroup$
– John
Dec 23 '18 at 12:22
1
$begingroup$
@John The other poles are outside the contour
$endgroup$
– Dylan
Dec 23 '18 at 13:16
1
$begingroup$
yes I got it and the residue is $- exp(i pi a)$, you can edit it.
$endgroup$
– John
Dec 23 '18 at 13:23
add a comment |
$begingroup$
The substitution $e^x mapsto x$ gives
$$ I = int_{-infty}^infty frac{e^{ax}}{e^x+1}dx = int_0^infty frac{1}{x^{1-a}(1+x)}dx $$
where $0 < 1-a < 1$
We can use complex integration here. Due to the pole at $z=-1$, we pick a branch cut on the positive real axis such that $0 le arg(z) < 2pi$. The contour is a standard "keyhole contour" consisting of
$C_1$: left semicircle centered at $0$ with radius $epsilon$, going clockwise from $-epsilon i$ to $epsilon i$
$C_2$: straight line from $epsilon i$ to $R + epsilon i$
$C_3$: circle centered at $0$ with radius $R$, going counterclockwise from $R+iepsilon$ to $R - epsilon i$
$C_4$: straight line from $R - epsilon i$ to $-epsilon i$
In the limits $epsilon to 0$ and $R to infty$, the two straight line integrals converge to
$$ int_{C_2} f(z) dz to int_0^infty frac{1}{x^{1-a}e^{i0(1-a)}(1+x)}dx = I $$
$$ int_{C_4} f(z) dz to -int_0^infty frac{1}{x^{1-a}e^{i2pi(1-a)}(1+x)}dx = -e^{i2pi a}I $$
The remaining integrals should go to $0$
$$ leftvertint_{C_1} f(z)dzrightvert le frac{pi epsilon}{|z|^{1-a}|1+z|} le frac{pi epsilon^a}{1-epsilon} to 0 $$
$$ leftvertint_{C_3} f(z) dz rightvert le frac{2pi R}{|z|^{1-a}|1+z|} le frac{2pi R^a}{R-1} to 0 $$
Therefore
$$ (1-e^{i2pi a})I = 2pi ioperatorname*{Res}_{z=-1} f(z) = -2pi i e^{ipi a} $$
So finally
$$ I = frac{2pi ie^{ipi a}}{e^{2pi a}-1} = frac{2pi i}{e^{ipi a}-e^{-ipi a}} = frac{pi}{sin(api)} $$
Edit: Here's a simpler method, credit to @Poujh
$$ I = int_{-infty}^infty frac{e^{ax}}{e^x+1}dx = int_{-infty}^infty f(x) dx $$
Take a rectangular contour:
$C_1$: Straight line from $-R$ to $R$
$C_2$: Straight line from $R$ to $R + i2pi$
$C_3$: Straight line from $R + i2pi$ to $-R + i2pi$
$C_4$: Straight line from $-R + i2pi$ to $-R$
In the limit of $Rtoinfty$, we have
$$ leftvertint_{C_2} frac{e^{az}}{e^z+1} dzrightvert = leftvertint_0^{2pi} frac{e^{a(R+iy)}}{e^{R+iy}+1} dyrightvert le 2pi leftvertfrac{e^{aR}}{e^R-1}rightvert to 0 $$
$$ leftvertint_{C_4} frac{e^{az}}{e^z+1} dzrightvert = leftvertint_0^{2pi} frac{e^{a(-R+iy)}}{e^{-R+iy}+1} rightvert le 2pi leftvert frac{e^{-aR}}{1-e^{-R}} rightvert to 0 $$
For the horizontal lines
$$ int_{C_1} frac{e^{az}}{e^z+1} dz to int_{-infty}^infty frac{e^{ax}}{e^x+1} dx = I $$
$$ int_{C_3} frac{e^{az}}{e^z+1} dz to -int_{-infty}^infty frac{e^{ax}e^{i2pi a}}{e^x+1} = -e^{i2pi a} I $$
We end up with the same result
$$ (1-e^{i2pi a})I = 2pi ioperatorname*{Res}_{z=ipi} frac{e^{az}}{e^z+1} $$
where the residue is computed using the limit
$$ lim_{zto ipi} e^{az}frac{z-ipi}{e^z+1} = -e^{ipi a} $$
$endgroup$
The substitution $e^x mapsto x$ gives
$$ I = int_{-infty}^infty frac{e^{ax}}{e^x+1}dx = int_0^infty frac{1}{x^{1-a}(1+x)}dx $$
where $0 < 1-a < 1$
We can use complex integration here. Due to the pole at $z=-1$, we pick a branch cut on the positive real axis such that $0 le arg(z) < 2pi$. The contour is a standard "keyhole contour" consisting of
$C_1$: left semicircle centered at $0$ with radius $epsilon$, going clockwise from $-epsilon i$ to $epsilon i$
$C_2$: straight line from $epsilon i$ to $R + epsilon i$
$C_3$: circle centered at $0$ with radius $R$, going counterclockwise from $R+iepsilon$ to $R - epsilon i$
$C_4$: straight line from $R - epsilon i$ to $-epsilon i$
In the limits $epsilon to 0$ and $R to infty$, the two straight line integrals converge to
$$ int_{C_2} f(z) dz to int_0^infty frac{1}{x^{1-a}e^{i0(1-a)}(1+x)}dx = I $$
$$ int_{C_4} f(z) dz to -int_0^infty frac{1}{x^{1-a}e^{i2pi(1-a)}(1+x)}dx = -e^{i2pi a}I $$
The remaining integrals should go to $0$
$$ leftvertint_{C_1} f(z)dzrightvert le frac{pi epsilon}{|z|^{1-a}|1+z|} le frac{pi epsilon^a}{1-epsilon} to 0 $$
$$ leftvertint_{C_3} f(z) dz rightvert le frac{2pi R}{|z|^{1-a}|1+z|} le frac{2pi R^a}{R-1} to 0 $$
Therefore
$$ (1-e^{i2pi a})I = 2pi ioperatorname*{Res}_{z=-1} f(z) = -2pi i e^{ipi a} $$
So finally
$$ I = frac{2pi ie^{ipi a}}{e^{2pi a}-1} = frac{2pi i}{e^{ipi a}-e^{-ipi a}} = frac{pi}{sin(api)} $$
Edit: Here's a simpler method, credit to @Poujh
$$ I = int_{-infty}^infty frac{e^{ax}}{e^x+1}dx = int_{-infty}^infty f(x) dx $$
Take a rectangular contour:
$C_1$: Straight line from $-R$ to $R$
$C_2$: Straight line from $R$ to $R + i2pi$
$C_3$: Straight line from $R + i2pi$ to $-R + i2pi$
$C_4$: Straight line from $-R + i2pi$ to $-R$
In the limit of $Rtoinfty$, we have
$$ leftvertint_{C_2} frac{e^{az}}{e^z+1} dzrightvert = leftvertint_0^{2pi} frac{e^{a(R+iy)}}{e^{R+iy}+1} dyrightvert le 2pi leftvertfrac{e^{aR}}{e^R-1}rightvert to 0 $$
$$ leftvertint_{C_4} frac{e^{az}}{e^z+1} dzrightvert = leftvertint_0^{2pi} frac{e^{a(-R+iy)}}{e^{-R+iy}+1} rightvert le 2pi leftvert frac{e^{-aR}}{1-e^{-R}} rightvert to 0 $$
For the horizontal lines
$$ int_{C_1} frac{e^{az}}{e^z+1} dz to int_{-infty}^infty frac{e^{ax}}{e^x+1} dx = I $$
$$ int_{C_3} frac{e^{az}}{e^z+1} dz to -int_{-infty}^infty frac{e^{ax}e^{i2pi a}}{e^x+1} = -e^{i2pi a} I $$
We end up with the same result
$$ (1-e^{i2pi a})I = 2pi ioperatorname*{Res}_{z=ipi} frac{e^{az}}{e^z+1} $$
where the residue is computed using the limit
$$ lim_{zto ipi} e^{az}frac{z-ipi}{e^z+1} = -e^{ipi a} $$
edited Dec 23 '18 at 16:46
answered Dec 20 '18 at 17:22
DylanDylan
14.2k31127
14.2k31127
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What about the other poles in this second method? Of course $exp(z)+1$ has zeros at $(2n+1)pi i$.
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– John
Dec 23 '18 at 12:22
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@John The other poles are outside the contour
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– Dylan
Dec 23 '18 at 13:16
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yes I got it and the residue is $- exp(i pi a)$, you can edit it.
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– John
Dec 23 '18 at 13:23
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What about the other poles in this second method? Of course $exp(z)+1$ has zeros at $(2n+1)pi i$.
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– John
Dec 23 '18 at 12:22
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@John The other poles are outside the contour
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– Dylan
Dec 23 '18 at 13:16
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yes I got it and the residue is $- exp(i pi a)$, you can edit it.
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– John
Dec 23 '18 at 13:23
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What about the other poles in this second method? Of course $exp(z)+1$ has zeros at $(2n+1)pi i$.
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– John
Dec 23 '18 at 12:22
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What about the other poles in this second method? Of course $exp(z)+1$ has zeros at $(2n+1)pi i$.
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– John
Dec 23 '18 at 12:22
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@John The other poles are outside the contour
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– Dylan
Dec 23 '18 at 13:16
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@John The other poles are outside the contour
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– Dylan
Dec 23 '18 at 13:16
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yes I got it and the residue is $- exp(i pi a)$, you can edit it.
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– John
Dec 23 '18 at 13:23
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yes I got it and the residue is $- exp(i pi a)$, you can edit it.
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– John
Dec 23 '18 at 13:23
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Your integral seems divergent.
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– Awe Kumar Jha
Dec 20 '18 at 10:39
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Come on, people! It is convergent and is equal to $mathrm{B}(a,1-a)=dfrac{pi}{sin api}$.
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– metamorphy
Dec 20 '18 at 12:01
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Did you find $mathrm{B}(x,y)=displaystyleint_0^inftyfrac{t^{x-1},dt}{(1+t)^{x+y}}$ there? Please read the article before asking further questions - I wouldn't want to reproduce the whole of it here ;)
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– metamorphy
Dec 20 '18 at 12:18
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@John You could solve this integral using a rectangular contour which goes from -R to R on the real axis and which is of height $2pi i$. Thus you have a pole at $x=pi i$. If you edit your question so that it fits the rules of the site, I can maybe answer your question, because I had to solve this exact same integral today also using contour integration.
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– Poujh
Dec 20 '18 at 20:45
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@Poujh That's honestly much easier than the approach I took. So thanks for helping me learn today.
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– Dylan
Dec 21 '18 at 3:56