A Contour Integral I

Question

What is the value of the integral \begin{align} \int_{-a}^{c} \sqrt{ \frac{a+x}{c-x} } \ \frac{dx}{(d-x)(x-b)} \end{align}

Answer 1

Hint

This integral simplifies if you first make a change of variable $$ \sqrt{ \frac{a+x}{c-x} }=u^2$$ that is to say $$x=\frac{c u^2-a}{u^2+1}$$ So $$I=\int \sqrt{ \frac{a+x}{c-x} } \ \frac{dx}{(d-x)(x-b)}=-\int \frac{2 u^2 (a+c)~~ du}{\left(u^2 (b-c)+a+b\right) \left(u^2 (d-c)+a+d\right)}$$ Using partial fractions, this leads to $$I=\int \frac{2 (a+b)}{(b-d) \left(u^2 (b-c)+a+b\right)}- \int \frac{2 (a+d)}{(b-d) \left(u^2 (d-c)+a+d\right)}$$ Using $u$ as variable, the bounds are $0$ and $\infty$.

I am sure that you can take from here.

Answer 2

Presumably $a,c>0$, and $b$ and $d$ are not in the interval $[-a,c]$.

Use a branch of $f(z) = \sqrt{(a+z)/(c-z)}$ that is analytic away from that interval. Compare the integral over a "dumbbell contour" that goes around the branch cut with a large circular contour. Note that the limit of $f(z)$ as you approach a point on the branch cut from above is $-$ the limit as you approach the same point from below. Don't forget the poles at $b$ and $d$.