Calculating the improper integral $\int_{0}^{+\infty}\frac{\log \mid 1-x^2 \mid}{x^2} dx$

Question

I tried to calculate the improper integral

$$\int_{0}^{+\infty}\frac{\log \mid 1-x^2 \mid}{x^2} dx$$ using integration by parts but the function is not continuous in $x=0$ and $x=1$ and this makes trouble for finding the convergence value.

Thanks to any help for calculating the value of this integral.

Answer 1

$$I=\int_0^{\infty} \frac{\log|1-x^2|}{x^2} dx \\ \overset{x=\frac 1t}{=} \int_0^{\infty}\log\left|\frac{t^2-1}{t^2}\right| dt\\ =\int_0^1 \log\left(\frac{1-t^2}{t^2}\right)dt +\int_1^{\infty} \log\left(\frac{t^2-1}{t^2}\right)dt = I_1 + I_2$$

Now,

$$I_1\overset{\text{IBP}}=\left[t\log\left(\frac{1-t^2}{t^2}\right) \right]_0^1 -\int_0^1 \frac{2}{t^2-1}dt \\ =\lim_{t\to 1} \ \log\left(\frac{1-t^2}{t^2}\right) -\left[\log\left(\frac{1-t}{1+t}\right)\right]_0^1 \\ =\lim_{t\to 1}\bigg[ \log\left(\frac{1-t^2} {t^2}\right) -\log\left(\frac{1-t}{1+t}\right)\bigg] \\ =\log (4) $$

Similarly, $I_2= \log\left({\frac 14}\right)$

And so $$I=I_1 + I_2 = \boxed 0$$