I'm trying to solve the following indefinite integral: $$I = \int \ln\left ( \frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}} \right ) \frac{dx}{1-x^2}$$
The integral is a general case which comes from a physics problem of potential in a capacitor with a rod inside.
I tried to figure out any plausible substitution or a transform by introducing a special function, but failed to do that.
- An approach I tried was by introducing a function: $h(z) = \mathrm{ln}|z| +i\phi$ for $0 < \phi < 2\pi$.
- An approach with a special function seemed applicable to the integral, such as the Dilogarithm.
Update-1: Applying numerical integration from $0$ to $1$, I obtained (via Matlab, for self-check):
fun = @(x) log((1+sqrt(1-x.^2))./(1-sqrt(1-x.^2))).*(1./(1-x.^2))
q = integral(fun,0,1)
>> 4.9348
Which is $\frac{1}{2} \pi^2$.
Start by using the substitution $x=\frac{2z}{1+z^2}$ then:$$I = \int \ln\left ( \frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}} \right ) \frac{dx}{1-x^2}=-4\int \frac{\ln z}{1-z^2}dz=2\int \frac{\ln z}{z-1}dz-2\int \frac{\ln z}{z+1}dz$$ $$=-2\operatorname{Li}_2(1-z)-2\operatorname{Li}_2(-z)-2\ln z \ln(1+z)+C,\quad z=\frac{1-\sqrt{1-x^2}}{x},x< |1|$$
The same substitution yields:
$$\int_0^1 \ln\left ( \frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}} \right ) \frac{dx}{1-x^2}=-4\int_0^1 \frac{\ln z}{1-z^2}dz$$$$=-4\sum_{n=0}^\infty \int_0^1 z^{2n}\ln zdz=4\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{\pi^2}{2} $$