Not $\zeta$ again!

$\int_0^\infty \log\left(\frac{e^x +1}{e^x -1}\right) dx = \int_0^\infty \log\left(\frac{1 +e^{-x}}{1 -e^{-x}}\right)dx = \int_0^\infty [\log(1 +e^{-x}) - \log(1 -e^{-x})] dx$

Since $e^{-x} <1$ , we can expand the logarithms as a power series around 1. This gives, $\int_0^\infty \left[ \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} e^{-nx} + \sum_{n=1}^\infty \frac{1}{n} e^{-nx} \right] dx$ .

Now, using $\int_0^\infty e^{-nx} dx = \frac{1}{n}$ , we get $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} + \sum_{n=1}^\infty \frac{1}{n^2} = 2 \sum_{n \, odd} \frac{1}{n^2}$

That is, $2 \left[ \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \dots \right]$

Which equals, $2 \left[ ( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots) - ( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots) \right] =2\left( \zeta(2) -\frac{1}{4} \zeta(2) \right) =\frac{3}{2} \zeta(2)$