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

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.

let $$x=\sin t$$,

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

$$I'(\alpha) = \int_0^{\pi/2}\frac{-2\alpha\sin^2 t}{1-\alpha^2 \sin^2 t}dt =\frac2{\alpha}\int_0^{\pi/2}\left(1-\frac{1}{1-\alpha^2 \sin^2 t}\right)dt = \frac{\pi}{\alpha} - J$$

where

$$J= \frac{2}{\alpha} \int_0^{\pi/2}\frac{dt }{1-\alpha^2 \sin^2 t} = \frac{2}{\alpha} \int_0^{\pi/2}\frac{d(\tan t)}{1+(1-\alpha^2) \tan^2 t} =\frac{\pi}{\alpha\sqrt{1-\alpha^2}}$$

Thus,

$$I(\alpha) = \int_0^{\alpha} I'(s)ds= \pi \int_0^{\alpha}\left( \frac1{s} -\frac{1}{s\sqrt{1-s^2}}\right)ds =\pi \ln\left(1+\sqrt{1-s^2}\right)_0^{\alpha}$$ $$=-\pi\ln\left(\frac2{1+\sqrt{1-\alpha^2}}\right)$$

• Thanks for your answer. Other than the general steps of finding $\frac{df}{d\alpha}$, are there general principles by which you rearrange functions to compute the derivatives? The computations seem rather ad hoc otherwise. – Mnifldz Nov 22 '19 at 7:06
• @Mnifldz - The steps presented in the answer is intended to illustrate the Feynman technique. It may seem ad hoc otherwise, but only because certain non-essential steps are skipped. A smooth presentation would have been quite lengthy. For example, the last integral $\int_0^{\alpha}\frac{1}{s\sqrt{1-s^2}}ds$ could be carried out quiet naturally with a few more steps. – Quanto Nov 22 '19 at 15:18
• That's fine. I understand how all of the computations were made, no confusion there. I was just wondering, in addition to the technique itself, if there were underlying methods to carry out the integrals themselves. – Mnifldz Nov 22 '19 at 15:46