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I would be interested in any clue on how to evaluate the following integral $$\int_1^\infty \cosh^{-1}(x) \ln(x^2-1) \exp \left(- \frac{x}{T} \right) dx $$

I have tried integration by parts but it seems to lead only to other integrals of the same form, with additional powers of $x$ in the integrand.

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This integral seems to be composed by random functions :-) Does it come from anything special? – Siminore Jun 29 '12 at 13:58
@Siminore If you pick up a random function it won't be smooth like that. This theorem is due to Banach :) – Norbert Jun 29 '12 at 14:10
It arises in the context of Faraday polarization transfer. – chris Jun 29 '12 at 14:19
@chris The function seems to grow like $x^{3/2}$ for small $T$ ($0<T<50$). – Pedro Tamaroff Jun 30 '12 at 23:22
up vote 2 down vote accepted



$=\int_0^\infty\cosh^{-1}\cosh x\ln((\cosh x)^2-1)e^{-\frac{\cosh x}{T}}~d(\cosh x)$

$=\int_0^\infty xe^{-\frac{\cosh x}{T}}\sinh x\ln\sinh^2x~dx$

$=2\int_0^\infty xe^{-\frac{\cosh x}{T}}\sinh x\ln\sinh x~dx$

According to,

Consider $\int_0^\infty e^{-z\cosh x}\cosh(vx)~dx=K_v(z)$ ,

$\int_0^\infty xe^{-z\cosh x}\sinh(vx)~dx=\dfrac{\partial K_v(z)}{\partial v}$

$\int_0^\infty xe^{-z\cosh x}\sinh x~dx=\dfrac{\partial K_v(z)}{\partial v}(v=1)$

Consider $\int_0^\infty e^{-z\cosh x}\sinh^vx~dx=\dfrac{2^\frac{v}{2}~\Gamma\left(\dfrac{v+1}{2}\right)K_\frac{v}{2}(z)}{\sqrt\pi z^\frac{v}{2}}$ ,

$\int_0^\infty e^{-z\cosh x}\sinh^vx\ln\sinh x~dx=\dfrac{\partial}{\partial v}\left(\dfrac{2^\frac{v}{2}~\Gamma\left(\dfrac{v+1}{2}\right)K_\frac{v}{2}(z)}{\sqrt\pi z^\frac{v}{2}}\right)$

$\int_0^\infty e^{-z\cosh x}\sinh x\ln\sinh x~dx=\dfrac{\partial}{\partial v}\left(\dfrac{2^\frac{v}{2}~\Gamma\left(\dfrac{v+1}{2}\right)K_\frac{v}{2}(z)}{\sqrt\pi z^\frac{v}{2}}\right)_{v=1}$

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Thanks! Very impressive! FYI this computation was needed for this paper for which we gave up presenting the corresponding anisotropic closed form. – chris Sep 17 '13 at 19:16

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