# Integral $\int_0^{\pi/2} \ln \sinh x \, dx$

$$I_1=\int_0^{\pi/2} \ln \sinh x \, dx,\quad I_2=\int_0^{\pi/2} \ln \cosh x \, dx, \quad I_1\neq I_2.$$ I am trying to calculate these integrals. We know the similar looking integrals $$\int_0^{\pi/2} \ln \sin x \, dx = \int_0^{\pi/2} \ln \cos x \, dx = -\frac{\pi\ln 2}{2}\approx -1.088793045....$$ are solved by using the symmetry of the integrands. We cannot use the same symmetry trick here to solve $$I_1, I_2$$ because the hyperbolic functions don't have that symmetry. What can we do?

We have $$\ln(\sinh(x)) = \ln(e^x-e^{-x}) - \ln(2) = x - \ln(2) + \ln(1-e^{-2x}) = x - \ln(2) - \sum_{k=1}^{\infty} \dfrac{e^{-2kx}}{k}$$ Hence, $$\int \ln(\sinh(x)) dx = \dfrac{x^2}2 - x\ln(2) + \sum_{k=1}^{\infty} \dfrac{e^{-2kx}}{2k^2} = \dfrac{x^2}2 - x\ln(2) + \dfrac{\text{Li}_2(e^{-2x})}2+C$$
Similarly, we have $$\ln(\cosh(x)) = \ln(e^x+e^{-x}) - \ln(2) = x - \ln(2) + \ln(1+e^{-2x}) = x - \ln(2) - \sum_{k=1}^{\infty} \dfrac{(-1)^ke^{-2kx}}{k}$$ Hence, $$\int \ln(\cosh(x)) dx = \dfrac{x^2}2 - x\ln(2) + \sum_{k=1}^{\infty} \dfrac{(-1)^ke^{-2kx}}{2k^2}+C$$
$$\int_0^{\pi/2}dx \ln \sinh x$$ by partial integration $$x \ln \sinh x - \int dx x \frac{1}{\sinh x} \cosh x$$ where $\cosh x$ is the term from the inner differentiation. Then underivative of the second term then contributes $$\int \frac{x dx}{\tanh x} = -\frac{x^2}{2}+x \ln[1-e^x] +Li_2(e^x)+x\ln[1+e^x]+Li_2(-e^x)$$ where $Li_2(y)\equiv \int_1^y \frac{\ln t}{1-t}dt$ is (one definition of) the dilogarithm http://en.wikipedia.org/wiki/Dilogarithm .