Integration of $\int_0^{+\infty} \frac{\ln(1/x)}{1-x^2}\,dx$ I am trying to integrate the below problem, but not sure if its integrable.
$$\int_0^\infty \frac{\ln(1/x)}{1-x^2}\,dx$$
Had the denominator been $1+x^2$,  I would have used tan(x) substitution.
Thanks.
 A: HINT: $$\int_0^{\infty}\frac{\ln\left(\frac 1x\right)}{1-x^2}\ dx$$
$$=\lim_{a\to \infty}\int_0^{a}\frac{-\ln(x)}{(1+x)(1-x)}\ dx$$
$$=\frac 12\lim_{a\to \infty}\int_0^{a}\ln(x)\left(\frac{1}{x-1}-\frac{1}{x+1}\right)\ dx$$
A: Split the integral into two pieces $[0,1)$ and $[1,\infty)$.
On the interval $[1,\infty)$ you do the change of variable $x=\frac{1}{t}$ 
Thus you will get that 
$$\int_{0}^{\infty}\frac{\log(\frac{1}{x})}{1-x^{2}}\, dx = -2\int_{0}^{1}\frac{\log(x)}{1-x^{2}} \, dx$$
Now recall the geometric series expansion $\frac{1}{1-x^{2}} =\sum_{n\geq 0} x^{2n}$ 
So by The monotone convergence theorem and integration by parts we have
$$-2\int_{0}^{1}\frac{\log(x)}{1-x^{2}} \, dx = -2\sum_{n\geq 0} \int_{0}^{1}\log(x) x^{2n} \, dx = 2\sum_{n\geq 0} \frac{1}{(2n+1)^{2}}$$
The later sum is known to evaluate at $\frac{\pi^{2}}{8}$, Thus $$\int_{0}^{\infty}\frac{\log(\frac{1}{x})}{1-x^{2}}\, dx = \frac{\pi^{2}}{4}$$
A: Substituting $x=e^y$ gives $$\int_0^\infty\frac{\ln x}{x^2-1}\text{d}x=\int_{-\infty}^\infty\frac{y\text{e}^y}{\text{e}^{2y}-1}\text{d}y.$$ The integrand is the even function $\frac{y}{2\sinh y}$, so the integral is $$\int_0^\infty\frac{2y\text{e}^{-y}}{1-\text{e}^{-2y}}\text{d}y=2\sum_{n=0}^\infty \int_0^\infty y\text{e}^{-\left( 2n+1 \right) y}\text{d}y =2\sum_{n=0}^\infty \frac{1}{\left( 2n+1 \right)^2}.$$ Famously, $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ and $\sum_{n=0}^\infty \frac{1}{\left( 2n+1 \right)^2}=\frac{\pi^2}{8}$, so the final result is $\frac{\pi^2}{4}$.
