Hello am looking for a solution to proving this. $$ I:=\int_0^\infty \log(1+x^2)\frac{\cosh \pi x +\pi x\sinh \pi x}{\cosh^2 \pi x}\frac{dx}{x^2}=4-\pi. $$ This one is related to Integral $\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx=2-\frac{4}{\pi}$ that the community together seemed to solve!
I tried writing $$ I=\int_0^\infty \log(1+x^2)\frac{dx}{x^2\cosh \pi x}+\int_0^\infty \log(1+x^2)\frac{\pi x\tanh \pi x}{\cosh \pi x }\frac{dx}{x^2} $$ but is not so clear now that I have two integrals to solve. I wasn't sure how integrating by parts would give me a clearer integral as the terms do not clean up here like the last one. I am not sure how else

Introducing something like $I(\alpha), I'(\alpha)$ in this situation did help but not much after this: $$ I(\alpha)=\int_0^\infty \log(1+\alpha x^2) \frac{\cosh \pi x +\pi x\sinh \pi x}{\cosh^2 \pi x}\frac{dx}{x^2}, \frac{dI}{d\alpha}= $$ $$ \int_0^\infty \frac{dx}{1+\alpha x^2}\frac{\cosh \pi x +\pi x\sinh \pi x}{\cosh^2 \pi x}=\int_0^\infty \frac{dx}{(1+\alpha x^2)\cosh \pi x}+\pi\int_0^\infty \frac{dx}{1+\alpha x^2}\frac{x\tanh \pi x}{\cosh \pi x} $$ Thank you

  • $\begingroup$ Have you tried contour integration? In the linked problem it works out fine so why not try it here as well? $\endgroup$ – Alex R. May 1 '14 at 23:43
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    $\begingroup$ Please make titles objective and informative. $\endgroup$ – Pedro Tamaroff May 2 '14 at 1:07
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    $\begingroup$ Whenever you see a square in the denominator, ask yourself whether it might not be an application of $\bigg(\dfrac1{f(x)}\bigg)'=-\dfrac{f'(x)}{f^2(x)}$ $\endgroup$ – Lucian May 2 '14 at 7:15
  • $\begingroup$ @Lucian: did you see my answer? $\endgroup$ – Ron Gordon May 2 '14 at 13:49
  • $\begingroup$ @Lucian Very helpful. Thanks! $\endgroup$ – Jeff Faraci May 2 '14 at 15:31

This integral is doable using Fourier methods. First note that

$$\frac{\cosh{\pi x} + \pi x \sinh{\pi x}}{x^2 \cosh^2{\pi x}} = -\frac{d}{dx} \frac1{x \cosh{\pi x}}$$

Then integrate by parts to get that

$$I = \int_{-\infty}^{\infty} dx \frac{\operatorname*{sech}{\pi x}}{1+x^2}$$

This may be evaluated using Parseval's equality, because the Fourier transform of each factor is well-known. Thus

$$I = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, \pi \, e^{-|k|} \operatorname*{sech}{(k/2)} = 2 \int_0^{\infty} dk \frac{e^{-k}}{e^{k/2}+e^{-k/2}}$$

(For a derivation of the FT of the sech term, see this answer.) This integral is easily evaluated by expanding the denominator:

$$I = 2 \sum_{n=0}^{\infty} (-1)^n \int_0^{\infty} dk \, e^{-(n+3/2) k} = 4 \sum_{n=0}^{\infty} \frac{(-1)^n}{2 n+3} = 4-\pi$$


It occurred to me that there's a much more straightforward way to evaluate the integral obtained through Parseval:

$$I = 2 \int_0^{\infty} dk \frac{e^{-k}}{e^{k/2}+e^{-k/2}} = 4 \int_0^1 du \frac{u}{u+1/u} \\ = 4 \int_0^1 du \left (1-\frac1{1+u^2} \right ) = 4 \left (1-\frac{\pi}{4} \right )=4-\pi$$

  • $\begingroup$ By Parseval's equality do you mean $\int f \hat{g} = \int \hat{f} g$? $\endgroup$ – Random Variable May 2 '14 at 1:25
  • $\begingroup$ @RandomVariable: No, I mean $$\int dx \, f(x) g(x) = \frac1{2 \pi} \int dk\, F(k) G(k) $$ $\endgroup$ – Ron Gordon May 2 '14 at 1:30
  • $\begingroup$ Am I mistaken to think that both equalities are essentially equivalent? $\endgroup$ – Random Variable May 2 '14 at 2:16
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    $\begingroup$ I found an answer in which sos440 used the first equality to evaluate a variation of this integral. And it would seem like it follows from expressing $\hat{g}$ as an integral and then changing the order of integration. But strangely I can't find it stated anywhere else. math.stackexchange.com/questions/411058/… $\endgroup$ – Random Variable May 2 '14 at 13:23
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    $\begingroup$ I didn't want to bother sos440 since he's been really busy as of late. So I asked Daniel Fischer. He said it's valid in most situations, even when changing the order of integration is not justified. $\endgroup$ – Random Variable May 2 '14 at 17:05

I will present an alternative solution here.

Denote, for $a$ between $-\pi$ and $\pi$ inclusive, $$F(a)=\int_{-\infty}^{\infty} {\cosh(ax)\over {\cosh(\pi x)}}{1\over {1+x^2}}dx$$ then by differentiation under the integral sign, which is easily justified by uniform convergence, we have $$F''(a)+F'(a)=\int_{-\infty}^{\infty} {\cosh(ax)\over {\cosh(\pi x)}}dx = \sec {a\over 2}$$ The above can be evaluated by standard complex analysis techniques.

Note that $F(\pi)=\pi$ and $F'(0)=0$

These initial value conditions, along with the differential equation above, can be solved with variation of parameters, giving $$F(a)=4\cos{a\over 2}-\pi\cos a-2\sin a\ln\tan({a\over 4}+{\pi\over 4})$$

This gives $$F(0)=\int_{-\infty}^{\infty} {1\over {\cosh(\pi x)}}{1\over {1+x^2}}dx=4-\pi$$ which is the same as the original question according to Ron Gordon's answer.


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