The problem I am currently working on requires proving the following identity: $$\intop_\rho^a \dfrac{\arcsin \dfrac{x}{a}}{\sqrt{x^2-\rho^2}}\mathbb{d}x=\dfrac{\pi}{2}\ln \dfrac{a+\sqrt{a^2-\rho^2}}{a}$$
My questions here are:
- How to prove it?
- I guess that this integral cannot be evaluated using only elementary techniques, and it appears that I have to use residue theorem. Is there a way to evaluate it without using residue theorem?
Many thanks in advance!
Differentiating the left side wrt $a$ gives $$\frac{\pi}{2\sqrt{a^2-\rho^2}}-\frac1a\int_{\rho}^a\frac{xdx}{\sqrt{(a^2-x^2)(x^2-\rho^2)}}.$$ The last integral is easily evaluated by change of the variables $t=\rho^2+(a^2-\rho^2)x^2$, which gives $$\int_{\rho}^a\frac{xdx}{\sqrt{(a^2-x^2)(x^2-\rho^2)}}=\frac12\int_0^1\frac{dt}{\sqrt{t(1-t)}}=\frac12\,B\left(\frac12,\frac12\right)=\frac{\pi}{2}.$$
Differentiating the right side gives $$\frac{\pi}{2}\left[\frac{1}{a+\sqrt{a^2-\rho^2}}\left(1+\frac{a}{\sqrt{a^2-\rho^2}}\right)-\frac1a\right]=\frac{\pi}{2}\left(\frac{1}{\sqrt{a^2-\rho^2}}-\frac1a\right)$$
Since the results of differentiation coincide, it remains to check the initial integral for some value of $a$ (the easiest thing is to set $a=\rho$).