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 May 14th, 2012, 10:50 AM #1 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 another somewhat challenging contour integral?. Here's another integral I ran across. $\int_{0}^{\infty}\frac{1}{(4x^{2}+\pi^{2})cosh(x)} dx=\frac{ln(2)}{2\pi}$ I thought perhaps a rectangular contour may be in order, but turns out maybe a semi-circular one after all?. I tried rewriting it as $\int_{0}^{\infty}\frac{2e^{x}}{(4x^{2}+\pi^{2})(e^ {2x}+1)}dx$ cosh has an infinite number of poles $x=\frac{(2n-1)\pi i}{2}, \;\ n\in \mathbb{N}$. $\frac{\pi i}{2}$ is a pole of $4x^{2}+\pi^{2}$ as well as for cosh(x). Anyone have a good method for this one?. I am not certain I am doing it correctly due to cosh and the rational part having poles at the same place. Namely, $\frac{\pi i}{2}$. Here is a link to a similar problem: http://129.81.170.14/~vhm/papers_html/final21.pdf The residue at $\frac{(2k-1)\pi i}{2a}$ they get $\frac{(-1)^{k-1}4ia}{4a^{2}-\pi^{2}(2k-1)^{2}}$. I get close, but I do not know from where they get that residue. They do not explain...just show it.
 May 15th, 2012, 10:37 AM #2 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: another somewhat challenging contour integral?. $residue=\lim_{z \to \frac{(2k-1)\pi i}{2a}} \frac{2e^{az}(z-\frac{(2k-1)\pi i}{2a})}{(z^{2}+1)(e^{2az}+1)}$ For k=2 I get f.i. $\lim_{z\to \frac{\pi i (2\ 2-1)}{2 a}} \, \frac{2 \left(z-\frac{\pi i (2\ 2-1)}{2 a}\right) e^{a z}}{\left(z^2+1\right) \left(e^{2 a z}+1\right)}=\frac{4 i a}{4 a^2-9 \pi ^2}$ http://www.wolframalpha.com/input/?i=Re ... a%29%7D%5D
 May 15th, 2012, 11:44 AM #3 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: another somewhat challenging contour integral?. ...
 May 15th, 2012, 01:39 PM #4 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: another somewhat challenging contour integral?. If we make the sub $x=\frac{\pi t}{2}$ in the original integral, we get: $\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{(t^{2}+1)cosh(\frac{\pi t}{2})}dt$ cosh and $t^{2}+1$ both have a pole at x=i and cosh at $(2k+1)i$ For i the residue is $\frac{-i}{2\pi}$ $\frac{1}{2\pi}(\frac{-i}{2\pi})=1$ Multiply by the $\frac{1}{2\pi}$ and we get $\frac{1}{2\pi}$ The other 0's of cosh are at $(2k+1)i$. Where k=0 would correspond to i, the one we just found. So, for the other poles for k=1,2,3,.... the residues are at $\frac{(-1)^{k}i}{2k(k+1)\pi}$ $2\pi i\left(\frac{i(-1)^{k}}{2k(k+1)\pi}\right)=\frac{(-1)^{k-1}}{k(k+1)}$ Taking the sum, beginning at 1, and summing them up we get: $\frac{1}{2\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k(k+1)}=\frac{2ln(2)-1}{2\pi}$ Now, do not forget to add the k=0 for i from before and we get: $\frac{ln(2)}{\pi}-\frac{1}{2\pi}+\frac{1}{2\pi}=\frac{ln(2)}{\pi}$ Since this is an even function, divide by 2 and get $\frac{ln(2)}{2\pi}$

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