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Can anyone show how to evaluate this integral?

$\displaystyle \int^{\infty}_{-\infty} \frac{dx}{(1+4 x^2)\cosh(\pi x)} = \ln2$

user147462
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  • Let $f(z) = \frac{1}{(1+4z^{2}) \cosh \pi z}$ and sum up all the residues in the upper half-plane. It's similar to another integral that was asked about recently. http://math.stackexchange.com/questions/776679/integrate-int-0-infty-log1x2-frac-cosh-frac-pi-x2-sinh2-frac/776971#776971 – Random Variable May 02 '14 at 23:48
  • The result is so simple. There should be some 'trick' which makes obvious the result. – Felix Marin May 15 '14 at 04:44

1 Answers1

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This is a candidate for Parseval's theorem:

$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k) $$

Here,

$$f(x) = \frac1{1+4 x^2} \implies F(k) = \frac{\pi}{2} e^{-\frac12 |k|}$$ $$g(x) = \operatorname{sech}{\pi x} \implies G(k) = \operatorname{sech}{\frac{k}{2}}$$

The integral is then

$$\frac12 \int_0^{\infty} dk \, e^{-k/2} \, \operatorname{sech}{\frac{k}{2}} = 2 \int_0^1 du \frac{u}{1+u^2} = \log{2}$$

Ron Gordon
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