Does the integral $\displaystyle \int\limits_{0}^{1} \frac{\log{x}}{1-x^{2}}$ converge where $\displaystyle x \in [0,1]$.
I don't undestand what needs further demostration...
... may be this step?...
$\displaystyle \int_{0}^{1} x^{n}\cdot \ln x \cdot dx = -\frac {1}{(n+1)^{2}}$ (1)
... or may be this?...
$\displaystyle \int_{0}^{1} \frac{\ln x}{1-x^{2}}\cdot dx= \sum_{n=0}^{\infty} \int_{0}^{1} x^{2n}\cdot \ln x \cdot dx = $
$\displaystyle = - \sum_{n=0}^ {\infty} \frac{1}{(2n+1)^{2}} = - \frac{\pi^{2}}{8}$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
$\displaystyle \int_{1}^{\infty }{\frac{\ln x}{x^{2}-1}\,dx}=\int_{1}^{2}{\frac{\ln x}{x^{2}-1}\,dx}+\int_{2}^{\infty }{\frac{\ln x}{x^{2}-1}\,dx}.$
the first integral converges by limit comparison test with $\displaystyle \int_1^2\frac{\ln x}{x-1}\,dx,$ now put $\displaystyle x\mapsto x+1$ in this integral to get $\displaystyle \int_0^1\frac{\ln(x+1)}{x}\,dx$ which exists since the integrand has limit as $\displaystyle x\to0.$ (1)
for $\displaystyle x\ge2$ we have $\displaystyle \frac{\ln x}{x^{2}-1}\le \frac{2\ln x}{x^{2}},$ and $\displaystyle \int_{2}^{\infty }{\frac{\ln x}{x^{2}}\,dx}<\infty .$ (2)
in your integral put $\displaystyle x\mapsto\frac1x$ and get $\displaystyle -\int_1^\infty\frac{\ln x}{x^2-1}\,dx,$ then by (1) and (2) the conclusion follows.