No, I didn't, I thought you referenced it because it contained a proof--I didn't realize it had six. Thank you.
At first i was thinking how to solve $\displaystyle \int_1^{\infty} \frac{\ln(x)}{x^2 - 1 }~dx $ without using the fact $\displaystyle \zeta(2) = \frac{\pi^2}{6} $ , in other words , I shall give another way to prove this fact starting at this point .
However , i failed , all my attempts eventually led to other troublesome integrals so I gave up , until I was asked to solve another integral $\displaystyle \int_0^1 \frac{ \tan^{-1}(x)}{\sqrt{1-x^2 }}~dx $ which brought some ideas to me in solving the integral I was struggling two weeks before .
Very creative! I'm impressed!
Two questions though:
First, how do we know $\displaystyle \displaystyle \int_1^\infty \frac{\ln(x)}{x^2-1}dx = \frac34 \zeta(2)? $
Also for $\displaystyle \displaystyle \int_0^1 \left[ \tan^{-1}\left(t\sqrt{1-y^2}\right) \right]_0^{\infty} \frac{dy}{\sqrt{1-y^2} } $, we have $\displaystyle \displaystyle y\in[0,1] $.
So $\displaystyle \displaystyle \lim_{t\to\infty} \tan^{-1}\left(t\sqrt{1-y^2}\right) = \begin{cases} \frac\pi2 ,\text{ if } y\neq1\\0,\text{ if } y=1 \end{cases} $.
Therefore, what do we do when $\displaystyle \displaystyle y=1? $
Here is my derivation of the first fact, although we need this definition of the zeta function
$\displaystyle \displaystyle \Gamma(s)\zeta(s) =\int_0^\infty\frac{x^{s-1}}{e^x-1}\,dx$
So, in the initial integral, let $\displaystyle u = \ln(x)$. Then $\displaystyle dx = x du = e^{u} du$. $\displaystyle u(\infty) = \infty$ and $\displaystyle u(1) = 0$. This last bit gives you the new limits of integration.
$\displaystyle \displaystyle \int_{1}^{\infty} \frac{\ln(x)}{x^2-1}\,dx = \int_{0}^{\infty}\frac{xe^x}{e^{2x}-1}\,dx$
Where on the RHS, I've let u = x so that i don't have to use u from now on. Now lets focus on the integrand.
$\displaystyle \displaystyle \frac{xe^x}{e^{2x}-1} = \frac{x(e^x+1-1)}{e^{2x}-1} = \frac{x}{e^{x}-1} - \frac{x}{e^{2x}-1}$.
The integral becomes
$\displaystyle \displaystyle \int_{0}^{\infty}\frac{x}{e^{x}-1} \,dx- \int_{0}^{\infty}\frac{x}{e^{2x}-1}\,dx
$
provided both converge. Now on the second piece, let $\displaystyle u = 2x$, $\displaystyle dx = du/2$ and $\displaystyle x = u/2$ to give that the piece is
$\displaystyle \displaystyle \frac{1}{4}\int_{0}^{\infty}\frac{u}{e^u-1}\,du$
Combining both pieces now we get
$\displaystyle \displaystyle \int_{0}^{\infty}\frac{x}{e^{x}-1} \,dx- \int_{0}^{\infty}\frac{x}{e^{2x}-1}\,dx = \frac{3}{4}\int_{0}^{\infty}\frac{x}{e^x-1}\,dx = \frac{3}{4}\Gamma(2)\zeta(2) = \frac{3}{4}\zeta(2)$