Evaluate $\int_1^\infty \cosh^{-1}(x) \ln(x^2-1) \exp \left(- \frac{x}{T} \right) dx $ 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.
 A: Hint:
$\int_1^\infty\cosh^{-1}x\ln(x^2-1)e^{-\frac{x}{T}}~dx$
$=\int_0^\infty\cosh^{-1}\cosh x\ln((\cosh x)^2-1)e^{-\frac{\cosh x}{T}}~d(\cosh x)$
$=\int_0^\infty xe^{-\frac{\cosh x}{T}}\sinh x\ln\sinh^2x~dx$
$=2\int_0^\infty xe^{-\frac{\cosh x}{T}}\sinh x\ln\sinh x~dx$
According to http://people.math.sfu.ca/~cbm/aands/page_376.htm,
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 $\int_0^\infty e^{-z\cosh x}\sinh^vx~dx=\dfrac{2^\frac{v}{2}~\Gamma\left(\dfrac{v+1}{2}\right)K_\frac{v}{2}(z)}{\sqrt\pi z^\frac{v}{2}}$ ,
$\int_0^\infty e^{-z\cosh x}\sinh^vx\ln\sinh x~dx=\dfrac{\partial}{\partial v}\left(\dfrac{2^\frac{v}{2}~\Gamma\left(\dfrac{v+1}{2}\right)K_\frac{v}{2}(z)}{\sqrt\pi z^\frac{v}{2}}\right)$
$\int_0^\infty e^{-z\cosh x}\sinh x\ln\sinh x~dx=\dfrac{\partial}{\partial v}\left(\dfrac{2^\frac{v}{2}~\Gamma\left(\dfrac{v+1}{2}\right)K_\frac{v}{2}(z)}{\sqrt\pi z^\frac{v}{2}}\right)_{v=1}$
