MHB Contour integration with rational function and cosh

[fredoniahead]

Hi All:

I am new to the site, so I thought this would be a good time to post an interesting integral I ran across that I am having a time with. It is a miscellaneous problem in Schaum's Outline of Complex Variables, #86 in ch. 7. I have been self-teaching a little CA when I get time and this one is a little more challenging than others I have encountered.

[math]\int_{0}^{\infty}\frac{1}{(4x^{2}+\pi^{2})\cosh(x)}dx=\frac{\ln(2)}{2\pi}[/math]

Does anyone have a good idea of how to approach this one?.

At first, I thought perhaps a rectangular contour may do the trick, but now I am thinking perhaps a semicircular one may be in order.

It would appear there is a double pole at [math]\frac{\pi i}{2}[/math].

Using the Laurent series, I think the residue is [math]\frac{-i}{4\pi^{2}}[/math]

Which gives [math]2\pi i\left(\frac{-i}{4\pi^{2}}\right)=\frac{1}{2\pi}[/math]

But, how to get that ln(2)?. Maybe use [math]\cos(ix)[/math] in some respect?.

 

[chisigma]

Hi All:

I am new to the site, so I thought this would be a good time to post an interesting integral I ran across that I am having a time with. It is a miscellaneous problem in Schaum's Outline of Complex Variables, #86 in ch. 7. I have been self-teaching a little CA when I get time and this one is a little more challenging than others I have encountered.

[math]\int_{0}^{\infty}\frac{1}{(4x^{2}+\pi^{2})\cosh(x)}dx=\frac{\ln(2)}{2\pi}[/math]

Does anyone have a good idea of how to approach this one?.

At first, I thought perhaps a rectangular contour may do the trick, but now I am thinking perhaps a semicircular one may be in order.

It would appear there is a double pole at [math]\frac{\pi i}{2}[/math].

Using the Laurent series, I think the residue is [math]\frac{-i}{4\pi^{2}}[/math]

Which gives [math]2\pi i\left(\frac{-i}{4\pi^{2}}\right)=\frac{1}{2\pi}[/math]

But, how to get that ln(2)?. Maybe use [math]\cos(ix)[/math] in some respect?.

There are poles also for $\cosh z=0$ , i.e. for $z=i\ (2k+1)\ \frac{\pi}{2}$, not only for $z= \pm i\ \frac {\pi}{2}$...

Kind regards

$\chi$ $\sigma$

 

[fredoniahead]

Thanks, chisigma. Yes, I know. I thought perhaps a contour could be chosen that enclosed a finite number of poles. Namely, [math]\frac{\pi i}{2}[/math].

 

[chisigma]

Thanks, chisigma. Yes, I know. I thought perhaps a contour could be chosen that enclosed a finite number of poles. Namely, [math]\frac{\pi i}{2}[/math].

The integration contour You have to choose is represented in red here...

View attachment 169

... so that, if R tends to infinity, all the poles with imaginary part >0 are involved...

Kind regards

$\chi$ $\sigma$

 

[fredoniahead]

I made a sub $t=\frac{\pi x}{2}$ and arrived at

$$\frac{1}{2\pi}\int_{0}^{\infty}\frac{1}{(t^{2}+1)cosh(\frac{\pi t}{2})}$$

There is an identity that says $$\frac{1}{2\pi}\int_{0}^{\infty}\frac{1}{(t^{2}+1)cosh(at)}dt=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2a+(2n-1)\pi}$$.

It is derived from the poles being at $\frac{(2n-1)\pi i}{2a}$. Then, the residue turns out to be:

$\frac{(-1)^{n-1}4ia}{4a^{2}-\pi^{2}(2n-1)^{2}}$. Now, using the residue theorem and expanding gives the aforementioned series.

Letting $a=\frac{\pi}{2}$ gives $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n\pi}=\frac{1}{2\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\frac{ln(2)}{2\pi}$$

This is in a paper I have on hyperbolic integrals by Gradshteyn and Ryzhik. Though, it still seems to me that a rectangular contour is a possibility.