Feynman Trick Demonstration for $ \int_0^1 \frac{\ln\left(1-\alpha^2x^2 \right)}{\sqrt{1-x^2}}dx $

Question

I've been tracking this post today on math.SE where the OP was asking for a proof of convergence for the integral

$$ \int_0^1 \frac{\ln\left(1-\alpha^2x^2 \right)}{\sqrt{1-x^2}}dx. $$

I want to ask a related but distinct question of how to explicitly compute the value of this integral. Something I had learned of today for the first time was the Feynman trick for evaluating integrals with some parameter $\alpha$ included in it (here is a helpful link). Essentially what this comes down to is interpreting the result of this integral as a function of $\alpha$: $$ f(\alpha) \;\; =\;\; \int_0^1 \frac{\ln\left(1-\alpha^2x^2 \right)}{\sqrt{1-x^2}}dx. $$

Feynman's approach to solving this was to essentially uncover a differential equation for $f$ by first differentiating this expression with respect to $\alpha$:

$$ \frac{df}{d\alpha} \;\; =\;\; \int_0^1\frac{-2\alpha x^2}{\sqrt{1-x^2} \left (1-\alpha^2x^2\right )}dx. $$

Next we fix an initial condition for this differential equation by picking a value for $\alpha$ that makes our original integral easy to compute. For instance, $\alpha=0$ yields: $$ f(0) \;\; =\;\; \int_0^1 \frac{\ln\left(1 \right)}{\sqrt{1-x^2}}dx \;\; =\;\; 0, $$

and then to proceed with finding $\frac{df}{d\alpha}$ via the integral expression determined above. Plugging this into Wolfram Alpha shows us that

$$ \int_0^1 \frac{\ln\left(1-\alpha^2x^2 \right)}{\sqrt{1-x^2}}dx \;\; =\;\; -\pi\ln \left (\frac{2}{1+ \sqrt{1-\alpha^2}} \right ) $$

but I would like some guidance on how to actually compute this integral. I had never heard of this method until today and am quite intrigued by how it works.

Note: I did a quick scan of all Feynman integral posts on here and I didn't see an example where the integrand included a square-root expression in the denominator. If I am mistaken I will gladly remove this post.

Answer 1

let $x=\sin t$,

$$ I(\alpha)=\int_0^1 \frac{\ln\left(1-\alpha^2x^2 \right)}{\sqrt{1-x^2}}dx =\int_0^{\pi/2}\ln(1-\alpha^2 \sin^2 t)dt $$

\begin{align} I'(\alpha) =& \int_0^{\pi/2}\frac{-2\alpha\sin^2 t}{1-\alpha^2 \sin^2 t}dt \\ =&\ \frac2{\alpha}\int_0^{\pi/2}\left(1-\frac{1}{1-\alpha^2 \sin^2 t}\right)dt = \frac{\pi}{\alpha} - \frac{\pi}{\alpha\sqrt{1-\alpha^2}} \end{align}

Thus

$$I(\alpha) = \int_0^{\alpha} I'(s)ds= \pi \int_0^{\alpha}\left( \frac1{s} -\frac{1}{s\sqrt{1-s^2}}\right)ds =\pi \ln\left(1+\sqrt{1-s^2}\right)_0^{\alpha}$$ $$=\pi\ln\frac{1+\sqrt{1-\alpha^2}}2$$