# Nice problem of parametric integral

Evaluate

$$I(a)=\int\limits_0^{\infty}\frac{1}{{\cosh}({\pi}x)(1+ax^2)}dx\;,\;\;\text{with}\;\;a>0$$

For example, if $a=4$,then$I(4)=\frac{1}{2}\log(2)$.

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do you want to evaluate the integral ??? – what'sup Oct 18 '13 at 13:35
"Show that"? Did you mean "evaluate"? What've you done so far, though? – DonAntonio Oct 18 '13 at 13:44
Ooo, these are my favorite kind of integrals. You mind sharing where the question came from? – David H Oct 18 '13 at 13:53
look at this too math.stackexchange.com/questions/411058/… – what'sup Oct 18 '13 at 13:56
This is 3.522.4 from Gradshtein & Ryzhik. – user64494 Oct 18 '13 at 14:10

i think the OP wants to evaluate it

$$I = \int_0^{\infty} \frac{dx}{\cosh(ax)(1+x^2)}$$

$$u = bx \Rightarrow I = \frac{1}{b}\int_0^{\infty} \frac{dx}{\cosh \left(\frac{a}{b}x \right)\left( 1 + \frac{x^2}{b^2} \right)}$$

put $\frac{a}{b} = \pi$

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The integrand is even. Switch the integral to $\left(-\infty,\infty\right)$. Use complex integration $\left(~\mbox{Residue}\ \mbox{Theorem}~\right)$. Poles of $\cosh\left(\pi x\right)$ are at ${\rm i}\left(n + 1/2\right)$. $n \in {\mathbb Z}$. The "other piece" has poles at $z = \pm{\rm i}a^{-1/2}$. That's all we need.

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