contour integral with rational and cosh

Here is a fun looking integral.

$$\int_{0}^{\infty}\frac{1}{(4x^{2}+{\pi}^{2})\cosh(x)}dx=\frac{\ln(2)}{2\pi}$$.

I rewrote it as $\frac{2e^{z}}{(4z^{2}+{pi}^{2})(e^{2z}+1)}$

It would appear there is a pole of order 2 at $\frac{\pi i}{2}$.

This is due to it being a zero of cosh and the rational part.

I think the residue at $\frac{\pi i}{2}$ is $\frac{-i}{4{\pi}^{2}}$. I found this by looking at the Laurent expansion.

Thus, $2\pi i(\frac{-i}{4\pi^{2}})=\frac{1}{2\pi}$

I thought about using a rectangular contour with vertices $-R,R,R+\pi i, -R+\pi i$that

encloses $\frac{\pi i}{2}$ because there are an infinite number of poles due to the poles

of cosh being $\frac{(2n+1)\pi i}{2}$.

So, the only pole enclosed by C would be $\frac{\pi i}{2}$.

Now, how can I deal with this?.

For the top horizontal , I tried $$\frac{2e^{x+\pi i}}{(4(x+\pi i)^{2}+{\pi}^{2})(e^{2(x+\pi i)}+1)}=\frac{-2e^{x}}{(4(x+\pi i)^{2}+{\pi}^{2})(e^{2x}+1)}$$

But, this looks kind of messy with that square in the denominator.

Does anyone have any ideas on how to arrive at that ln(2)?. Thanks much.

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I think if you use the standard semicircular contour you will get a series like $\sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+1)}$ for the sum of the residues and the sum of this series has something to do with $\ln(2)$. –  kiwi May 12 '12 at 23:24
Thanks. Yes, I got to thinking that a rectangular may not be the best idea. Good idea with the series. –  Cody May 13 '12 at 0:42
May I ask, Kiwi, how you arrived at that series?. I played around with it and managed to set up $$\sum_{n=0}^{\infty}(-1)^{n}e^{x(2n+1)}=\frac{e^{x}}{e^{2x}+1}=sech(x)$$. But, then I got hung up. –  Cody May 13 '12 at 10:08

Making the substitution $$x=\frac{\pi t}{2}$$

results in $$\frac{1}{2\pi}\int_{0}^{\infty}\frac{1}{(t^{2}+1)\operatorname{cosh}(\pi t/2)}dt$$

But, $$\frac{1}{2\pi}\int_{0}^{\infty}\frac{1}{(t^{2}+1)\operatorname{cosh}(at)}dt=\sum_{n=1}^{\infty}\frac{(-1)^{k-1}}{2a+(2k-1)\pi}$$

This is derived from the poles at $$\frac{(2k-1)\pi i}{2a}$$

Resulting in a residue of $$\frac{(-1)^{n-1}\cdot4ia}{4a^{2}-\pi^{2}(2k-1)^{2}}$$

The residue theorem and partial fractions then produces the above result

Let $a=\frac{\pi}{2}$ and the result is $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n\pi}$$

$$=\frac{\ln(2)}{2\pi}$$

I found this in Gradshetyn and Ryzhik's hyperbolic integral paper.

I still think there should be an easier way perhaps using a rectangle. Quite a challenging problem for a Schaum's Outline book.

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What is Gradshetyn and Ryzhik’s hyperbolic integral paper? –  cm007 Oct 30 '12 at 17:00