How to compute $\int_0^s \frac{r\, \mathrm{d}r}{\sqrt{s^2-r^2}\sqrt{t^2-r^2}}$ if we know that $s<t$?

Question

In a integral equation problem, I came across the following integral $$ \phi(s,t) = \int_0^s \frac{r\, \mathrm{d}r}{\sqrt{s^2-r^2}\sqrt{t^2-r^2}} \, . $$

We can remark that the integral is given by $$ \phi(s,t) = \frac{1}{2} \ln \left( \frac{t+s}{t-s} \right) \, . $$

But is there a way to prove that analytically, provided that $s < t$?

Thanks

R

Answer 1

Use the change of variable $\tau = r^2$ and the following indefinite integral $$ \int \frac{d\tau}{\sqrt{\tau^2-a\tau+b}} =\ln\left(\frac{a}{2}-\tau-\sqrt{\tau^2-a\tau+b}\right). $$ Then \begin{align} \int_0^s \frac{r\, \mathrm{d}r}{\sqrt{s^2-r^2}\sqrt{t^2-r^2}} &=\frac{1}{2}\int_0^{s^2}\frac{d\tau}{\sqrt{\tau^2-(s^2+t^2)\tau+s^2t^2}} \cr &=\left.\frac{1}{2}\ln\left(\frac{s^2+t^2}{2}-\tau-\sqrt{\tau^2-(s^2+t^2)\tau+s^2t^2}\right) \right|_{\tau=0}^{s^2} \cr &= \frac{1}{2}\ln\left( \frac{t+s}{t-s}\right). \end{align}