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May 2nd, 2012, 01:39 PM   #1
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contour integral with sin and e

I am somewhat stuck on this one. Zardoz?.

$\int_{0}^{\infty}\frac{sin(ax)}{e^{x}+1}dx$

I rewrote as $\int_{0}^{\infty}\frac{e^{aiz}}{e^{z}+1}$

The poles are at $z=(2n+1)\pi i$, but the only one inside the contour is $\pi i$

It was suggested to use a rectangular contour with vertices $0, \;\ R, \;\ R+2\pi i, \;\ 2\pi i$

with an indent around $\pi i$.

Unless I made an error, the residue is $\frac{-1}{e^{a\pi}$

What would be a good approach for this one?.

It should work out to $\frac{1}{2a}-\frac{\pi e^{a\pi}}{e^{2a\pi}-1}=\frac{1}{2a}-\frac{\pi}{2sinh(\pi a)}$.

I managed to evaluate this by obtaining a series $\sum_{n=0}^{\infty}\frac{a(-1)^{n}}{a^{2}+n^{2}}$

Then, using $\pi csc(\pi z)$. This has poles at $-ai, \;\ ai$ and the residues were easily found and summed up to arrive at the required result.

The contour method with the rectangular contour is what has me a little befuddled.
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 contour.GIF (2.8 KB, 1184 views)

 May 2nd, 2012, 03:01 PM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: contour integral with sin and e [color=#000000]I will try it some other time, but there is a theorem called fractional residue theorem which states the following $\fbox{\text{If } z_0\text{ is a simple pole of f(z) and }C_{r}\text{ is an arc of the circle }|z-z_0|=r\text{ of angle } a \text{ then }\\\hspace{335pt}\lim_{r\to 0 }\int_{C_r}f(z)\;dz=\pi a \textrm{Res}\left(f(z);z=a\right)}$ .[/color]
May 6th, 2012, 06:52 AM   #3
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Re: contour integral with sin and e

If anyone is interested, I managed to find a solution to [color=red]***[/color]$\int_{0}^{\infty}\frac{sin(ax)}{e^{2\pi x}-1}dx$.

Which can be written as $\int_{C}\frac{e^{iaz}}{e^{2\pi z}-1}dz$

which is similar to the aforementioned integral and can be done in a similar fashion avoiding the poles created by $e^{x}-1=0$.

It is rather involved and uses a contour that has a line segment AND quarter circle. The quarter circles are around the poles we're trying to avoid at 4 and 6. These are poles at 0 and i.

$\int_{C}\frac{e^{iaz}}{e^{2\pi z}-1}dz=I_{1}+I_{2}+I_{3}+I_{4}+I_{5}+I_{6}$

starting with the line segment $[\epsilon,R]$ and going counterclockwise.

As $\epsilon\to 0^{+}, \;\ R\to \infty, \;\ I_{1}\to I$.

For $I_{3}, \;\ z=x+i$ where x varies from R to $\epsilon$.

$I_{3}=\int_{R}^{\epsilon}\frac{e^{ia(x+i)}}{e^{2\p i (x+i)}-1}dx=-e^{-a}\int_{\epsilon}^{R}\frac{e^{iax}}{e^{2\pi x}-1}dx$

As $\epsilon\to 0^{+}, \;\ R\to \infty, \;\ I_{3}\to -e^{-a}I$

For $I_{2}, \;\ z=R+iy$, and y varies from 0 to 1

Thus, $\frac{e^{ia(R+iy)}}{e^{2\pi (R+iy)}-1}\to 0, \;\ as \;\ R\to \infty$.

$I_{5}, \;\ z=iy$ and y varies from $1-\epsilon \;\ to \;\ \epsilon$.

$I_{5}=\int_{1-\epsilon}^{\epsilon}\frac{e^{ia(iy)}}{e^{2\pi (iy)}-1}idy=-i\int_{\epsilon}^{\epsilon}\frac{e^{-ay}}{e^{2\pi iy}-1}dy$

$I_{4}$ is over the quarter circle from $(\epsilon, \;\ \epsilon+i)$ and we get:

$\lim_{\epsilon\to 0}\int\frac{e^{iaz}}{e^{2\pi z}-1}dz=-i\frac{\pi}{2}\frac{e^{ia(i)}}{2\pi e^{2\pi (i)}}=-i\frac{e^{-a}}{4}$.

$I_{6}$ is over the quarter circle from $(0,i\epsilon) \;\ to \;\ (\epsilon,0)$

$\lim_{\epsilon\to 0}\int\frac{e^{iaz}}{e^{2\pi z}-1}dz=i\frac{-\pi}{2}\frac{e^{ia(0)}}{2\pi e^{2\pi (0)}}=\frac{-i}{4}$

Put them together: $\lim_{\epsilon\to 0}\left[\underbrace{\int_{\epsilon}^{\infty}\frac{e^{iax}} {e^{2\pi x}-1}dx}_{\text{I1}}-\underbrace{e^{-a}\int_{\epsilon}^{\infty}\frac{e^{iax}}{e^{2\pi x}-1}dx}_{\text{I3}}-\underbrace{i\int_{\epsilon}^{\epsilon}\frac{e^{-ay}}{e^{2\pi iy}-1}dy}_{\text{I5}}\right]=\overbrace{i\frac{e^{-a}}{4}+\frac{i}{4}}^{\text{I4&I6}}=$. Of course, $I_{2}=0$ so is omitted.

or $\lim_{\epsilon\to 0}\left[\int_{\epsilon}^{\infty}\frac{e^{iax}}{e^{2\pi x}-1}dx-e^{-a}\int_{\epsilon}^{\infty}\frac{e^{iax}}{e^{2\pi x}-1}dx-i\int_{\epsilon}^{1-\epsilon}\frac{e^{-ay}}{e^{2\pi iy}-1}dy\right]=i\left(\frac{e^{-a}}{4}+\frac{1}{4}\right)$

The limit of each improper integral does not exist, but the limit of the sum does.

Take imaginary parts on both sides:

$\lim_{\epsilon\to 0}Im\left[\int_{\epsilon}^{\infty}\frac{e^{iax}}{e^{2\pi x}-1}dx-e^{-a}\int_{\epsilon}^{\infty}\frac{e^{iax}}{e^{2\pi x}-1}dx-i\int_{\epsilon}^{1-\epsilon}\frac{e^{-ay}}{e^{2\pi iy}-1}dy\right]=\frac{e^{-a}+1}{4}$...........[2]

Now,

$Im\left[\int_{\epsilon}^{\infty}\frac{e^{iax}}{e^{2\pi x}-1}dx-e^{-a}\int_{\epsilon}^{\infty}\frac{e^{iax}}{e^{2\pi x}-1}dx-i\int_{\epsilon}^{1-\epsilon}\frac{e^{-ay}}{e^{2\pi iy}-1}dy\right]$

$=\int_{\epsilon}^{\infty}\frac{sin(ax)}{e^{2\pi x}-1}(1-e^{-a})dx+\int_{\epsilon}^{1-\epsilon}Im\left(\frac{(-i)e^{-ay}}{e^{2\pi iy}-1}\right)dy$.............[3]

$Im\left(\frac{(-i)e^{-ay}}{e^{2\pi iy}-1}\right)=-Re\left(\frac{e^{-ay}}{e^{2\pi iy}-1}\right)$

$=-e^{-ay}Re\left(\frac{1}{e^{2\pi iy}-1}\right)=-e^{-ay}\left(\frac{e^{-i\pi y}}{e^{i\pi y}-e^{-i\pi y}}\right)$

$=-e^{-ay}Re\left(\frac{cos(\pi y)-isin(\pi y)}{2isin(\pi y)}\right)=\frac{e^{-ay}}{2}$

Plug this into [3] and use [2], and we get:

$\lim_{\epsilon\to 0}\left[\int_{\epsilon}^{\infty}\frac{sin(ax)}{e^{2\pi x}-1}(1-e^{-a})dx+\int_{\epsilon}^{1-\epsilon}\frac{e^{-ay}}{2}dy\right]=\frac{e^{-a}+1}{4}$

Evaluate the second integral:

$\lim_{\epsilon\to 0}\left[\int_{\epsilon}^{\infty}\frac{sin(ax)}{e^{2\pi x}-1}(1-e^{-a})dx+\frac{e^{-a(1-\epsilon)}-e^{-a\epsilon}}{2(-a)}\right]=\frac{e^{-a}+1}{4}$

$(1-e^{-a})\lim_{\epsilon\to 0}\int_{\epsilon}^{\infty}\frac{sin(ax)}{e^{2\pi x}-1}dx+\frac{1-e^{-a}}{2a}=\frac{e^{-a}+1}{4}$

So, we're getting to the home stretch:

$(1-e^{-a})\lim_{\epsilon\to 0}\int_{\epsilon}^{\infty}\frac{sin(ax)}{e^{2\pi x}-1}dx=\frac{e^{-a}+1}{4}-\frac{1-e^{-a}}{2a}$

$=\frac{1}{1-e^{-a}}\left[\frac{e^{-a}+1}{4}-\frac{1-e^{-a}}{2a}\right]$

$=\frac{-1}{2a}+\frac{1}{4}\frac{e^{a}+1}{e^{a}-1}=\frac{-1}{2a}+\frac{1}{4}coth(a/2)$

WHEW!!!!!

[color=red]***[/color]In Applied Complex Analysis with PDE's by Asnar Akhle
From what I have seen of it, this is a very nice CA text. Good for class or even self-studying.

EDIT: I managed to evaluate $\int_{0}^{\infty}\frac{sin(ax)}{e^{x}+1}dx$ by using the same technique as above. Only I used a rectangle with vertices $0,R,R+2\pi i, 2\pi i$ and avoided the pole at $\pi i$ by drawing a little semicircle around it. $(0,\pi i+\epsilon) \;\ to \;\ (0,\pi i -\epsilon)$
Attached Images
 contour1.GIF (2.6 KB, 1139 views)

May 9th, 2012, 03:05 PM   #4
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Re: contour integral with sin and e

Quote:
 Originally Posted by galactus I am somewhat stuck on this one. Zardoz?. $\int_{0}^{\infty}\frac{sin(ax)}{e^{x}+1}dx$

[color=#000000]The reason why you can't find the answer is because you use the wrong function, try $f(z)=\frac{e^{iza}-e^{-\pi a }}{e^z+1}$, with this function $i\pi$ is a removable singularity. When the problem with the attachments (cannot upload a figure) is resolved I will upload a solution.[/color]

 May 10th, 2012, 06:54 AM #5 Senior Member   Joined: May 2011 Posts: 501 Thanks: 5 Re: contour integral with sin and e Thanks, Z, I would like to see your method. I managed to work it out using an indent around $\pi i$ and a rectangle with vertices at $2\pi i$. I would like to see your method because it sounds easier. I did not think of that function you mentioned with the removable singularity. Cool.
May 14th, 2012, 08:53 AM   #6
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Re: contour integral with sin and e

[color=#000000][attachment=0:3mi8ld2o]gal.png[/attachment:3mi8ld2o]

Hi G. I almost forgot it. We will use the above figure and we will integrate the function $f(z)=\frac{e^{iaz}-e^{-\pi a}}{e^{z}+1}$. Now f is analytic in F and so $\oint_{\partial F}f(z)\;dz=0\Leftrightarrow \int_{\sigma_1}+\int_{\sigma_2}+\int_{\sigma_3}+\i nt_{\sigma_4}+\int_{\sigma_5}+\int_{\sigma_6}f(z)\ ;dz=0$.

$\bullet\;\;\int_{\sigma_1}f(z)\;dz=\int_{0}^{R}\fr ac{e^{iax}-e^{-\pi a}}{e^{x}+1}\;dx$

$\bullet\;\;\int_{\sigma_2}f(z)\;dz=i\int_{0}^{2\pi }\frac{e^{iaR}e^{-ay}-e^{-\pi a }}{e^{R}e^{iy}+1}\;dy\Rightarrow \lim_{R\to\infty}\int_{\sigma_2}f(z)\;dz=\lim_{R\t o\infty}i\int_{0}^{2\pi}\frac{e^{iaR}e^{-ay}-e^{-\pi a }}{e^{R}e^{iy}+1}\;dy=0$, because using the M-L inequality

$|I_2|\leq 2\pi\max_{y\in[0,2\pi]}\left|\frac{e^{iaR}e^{-ay}-e^{-\pi a }}{e^{R}e^{iy}+1}\right|\leq 2\pi \frac{e^{-\pi a }+1}{e^{R}-1}\to 0$

$\bullet\;\;\int_{\sigma_3}f(z)\;dz=\int_{R}^{0}\fr ac{e^{ia(x+2\pi i) }-e^{-\pi a }}{e^{x+2\pi i }+1}\;dx=-\int_{0}^{R}\frac{e^{iax} e^{-2\pi a}}{e^{x}+1}+\int_{0}^{R}\frac{e^{-\pi a }}{e^{x}+1}\;dx$

adding up $\lim_{R\to\infty}\int_{\sigma_1}+\int_{\sigma_{2}} +\int_{\sigma_{3}}f(z)\;dz=\left(1-e^{-2\pi a }\right)\int_{0}^{\infty}\frac{e^{iax}}{e^x+1}\;dx$.

$\bullet\;\; \int_{\sigma_4}f(z)\;dz=i\int_{2\pi}^{\pi+r}\frac{ e^{-ya}-e^{-\pi a }}{e^{iy}+1}\;dy\;\;\overset{r\to0}{\to}\;\;ie^{-a\pi}\int_{\pi}^{0}\frac{e^{-aw}-1}{1-e^{iw}}\;dw=ie^{-a\pi}\int_{0}^{\pi}\frac{1-e^{-aw}}{1-e^{iw}}\;dw$

$\bullet\;\;\int_{\sigma_5}f(z)\;dz \;\;\overset{r\to 0 }{\to}\;\; 0$, because $|I_{5}|\leq \pi r M\to 0$.

$\bullet\;\;I_{6}=i\int_{\pi-r}^{0}\frac{e^{-ya}-e^{-\pi a }}{e^{iy}+1}\;dy\;\;\overset{r\to 0}{\to}\;\;ie^{-a\pi}\int_{0}^{\pi}\frac{1-e^{aw}}{1-e^{-iw}}\;dw$

$\lim_{r\to 0}I_4+I_5+I_6=ie^{-a\pi}\int_{0}^{\pi}\frac{1-e^{-aw}}{1-e^{iw}}+\frac{1-e^{wa}}{1-e^{-iw}}\;dw$

Now,

$\int_{0}^{\infty}\frac{\sin(ax)}{e^{x}+1}\;dx=\Im\ left(\int_{0}^{\infty}\frac{e^{iax}}{e^x+1}\;dx\ri ght)=-\frac{1}{1-e^{-2\pi a}}\Im\left(I_4+I_6\right)_{\small r\to0}=-\frac{e^{-\pi a }}{1-e^{-2\pi a }}\Re\left(\int_{0}^{\pi}\frac{1-e^{aw}}{1-e^{-iw}}+\frac{1-e^{-aw}}{1-e^{iw}}\;dw\right)=$

$-\frac{1}{2\sinh(\pi a )}\int_{0}^{\pi}\frac{1-e^{aw}}{2}+\frac{1-e^{-aw}}{2}\;dw=\ldots=\frac{1}{2a}\left(1-\frac{a\pi}{\sinh(a\pi)}\right)$.[/color]
Attached Images
 gal.png (13.5 KB, 1062 views)

 May 14th, 2012, 09:10 AM #7 Senior Member   Joined: May 2011 Posts: 501 Thanks: 5 Re: contour integral with sin and e Thanks much, Z. That's fantastic. May I ask what you used to generate that nice graph?. Paint?.
 May 14th, 2012, 10:04 AM #8 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: contour integral with sin and e [color=#000000]I always use geogebra, which is a free program. I use the version for linux, I think you will have to download the windows version.[/color]
 May 14th, 2012, 10:47 AM #9 Senior Member   Joined: May 2011 Posts: 501 Thanks: 5 Re: contour integral with sin and e Thanks, I'll check it out.
 May 22nd, 2012, 04:38 AM #10 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: contour integral with sin and e [color=#000000]I received a pm asking me about the last computation of $-\frac{1}{2\sinh(\pi a )}\int_{0}^{\pi}\frac{1-e^{aw}}{2}+\frac{1-e^{-aw}}{2}\;dw$. $\int_{0}^{\pi}\frac{1-e^{aw}}{2}+\frac{1-e^{-aw}}{2}\;dw=\int_{0}^{\pi}\frac{1}{2}+\frac{1}{2}-\frac{e^{aw}}{2}-\frac{e^{-aw}}{2}\;dw=\int_{0}^{\pi}1\;dw-\int_{0}^{\pi}\frac{e^{aw}+e^{-aw}}{2}\;dw=\pi-\int_{0}^{\pi}\cosh(aw)\;dw=\pi-\frac{\sinh(\pi a )}{a}$ and so $-\frac{1}{2\sinh(\pi a )}\left(\pi-\frac{\sinh(\pi a )}{a}\right)=-\frac{\pi}{2\sinh(\pi a )}+\frac{1}{2a}=\frac{1}{2a}\left(1-\frac{\pi a }{\sinh(\pi a )}\right)$.[/color]

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