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I was able to calculate $$\int_0^\infty\frac{\sqrt{e^x-1}}{1-2\cosh x}\,dx=-\frac\pi{\sqrt3}.$$ It turns out the integrand even has an elementary antiderivative (see here).

Now I'm interested in a similar integral $$I=\int_0^\infty\frac{x\,\sqrt{e^x-1}}{1-2\cosh x}\,dx.$$ Numerically, it is $$I\approx4.0336314630136915863337257797951491097354689684433117747419802...$$

Is it possible to find a closed form for this integral?

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  • $\begingroup$ What techniques have you used to evaluate the first integral? $\endgroup$ – science Mar 22 '15 at 23:04
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    $\begingroup$ @science When a program gives you a purpoted closed-form antiderivative, the proof is usually easy. You just guess that the form is likely to be correct, and prove it by direct differentiation. :) Calculation of limits may be an interesting part sometimes. $\endgroup$ – Vladimir Reshetnikov Mar 22 '15 at 23:45
  • $\begingroup$ @VladimirReshetnikov: Yes you are right. $\endgroup$ – science Mar 22 '15 at 23:47
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This is another one of those integrals that looks grim, then turns out to be fine once you work out what's going off and choose the right path. In this case, the trick is simply to change variables to $$ u^2 = e^x - 1, $$ which doesn't change the range of integration; this is the same as $ x = \log{(1+u^2)} $ Then $$ dx = \frac{2u}{1+u^2}, $$ and most significantly, $$ 2\cosh{x} -1 = (1+u^2)-1 + \frac{1}{1+u^2}, $$ and we find that your first integral becomes (rewriting it to be positive) $$ \int_0^{\infty} \frac{\sqrt{e^x-1}}{2\cosh{x}-1} \int_0^{\infty} \frac{2u^2}{u^2(1+u^2)+1} \, du = \int_{-\infty}^{\infty} \frac{u^2}{1+u^2+u^4} \, du $$ The integrand being even, this is straightforward to finish off using the residue theorem, giving the $\pi/\sqrt{3}$.


Of course, what you actually care about is this with another $\log{(1+u^2)}$ in it. The usual trick here is to set $$ I(a) = \int_{-\infty}^{\infty} \frac{u^2\log{(1+a u^2)}}{1+u^2+u^4} \, du. $$ $I(1)$ is what we want, $I(0)$ is clearly $0$, so we can write $I(1) = \int_0^1 I'(a)$. It so happens that $I'(a)$ is much easier to compute, since it is just $$ I'(a) = \int_{-\infty}^{\infty} \frac{u^4}{(1+u^2+u^4)(1+a u^2)} \, du, $$ which another tedious residue computation, or equally uninteresting partial fractions gives as $$ I'(a) = \frac{\pi}{\sqrt{3}} \left( \frac{a-2}{a^2- a+1} + \frac{\sqrt{3}}{\sqrt{a} (a^2- a+1)} \right), $$ and so $$ I(1) = \frac{\pi}{\sqrt{3}} \int_0^1 \left( \frac{a-2}{a^2- a+1} + \frac{\sqrt{3}}{\sqrt{a} (a^2- a+1)} \right) \, da, $$ another integral whose difficulty lies more in patience than in ingenuity, and we eventually obtain the final answer $$ I(1) = \frac{\pi^2}{6}+\frac{\pi}{2 \sqrt{3}} \log{\left(7+4 \sqrt{3}\right)}.$$

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By substituting $x\mapsto\ln(1+x^2)$ we see that the integral is equivalent to $$I=-\int^\infty_{-\infty}\frac{x^2\ln(1+x^2)}{x^4+x^2+1}dx$$ This is equivalent to $2\pi i$ times the residues of $f(z)=-\dfrac{2z^2\ln(1-iz)}{z^4+z^2+1}$ in the upper-half plane at $z=e^{\pi i/3}$ and $z=e^{2\pi i/3}$. The result is $$I=\frac{\pi^2}{6}+\frac{\pi}{\sqrt{3}}\ln(2+\sqrt{3})$$

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    $\begingroup$ Nice answer again! +1. $\endgroup$ – user153012 Mar 23 '15 at 0:18

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