How to Solve Integral with Parameters a and b in POTW #283?

[Euge]

Here is this week's POTW:

-----
Evaluate the integral
$$\int_{-1}^1 \left(\frac{1-x}{1+x}\right)^{\!\!a} \frac{dx}{(x - b)^2}$$
where $0 < a < 1$ and $b > 1$.

-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

Hi all,

Due to the Christmas holiday, solutions to the graduate and university POTW will be posted next week. As for this graduate problem, you might want to consider a contour integral involving a dogbone-like contour.

 

[Euge]

No one answered this week's problem. You can read my solution below.

Spoiler

Consider the contour integral $\int_C (z - 1)^a (z + 1)^{-a} (z - b)^{-2}\, dz$ where $C$ is a dogbone contour around $-1$ and $1$. Since $(z - b)^{-2}$ as a pole of order $2$ at $z = b$ (which is outside the contour) and is $O(\lvert z\rvert^{-2})$ as $\lvert z\rvert \to \infty$, then since the integrand is analytic at infinity, it follows from the residue theorem that

$$(e^{-i\pi a} - e^{i\pi a}) \int_{-1}^1 (1 - x)^a(1 + x)^{-a} (x - b)^{-2}\, dx = -2\pi i\operatorname{Res}\limits_{z = b} (z - 1)^a(z + 1)^{-a}(z - b)^{-2}$$

The residue at $z = b$ is $2a\,(b - 1)^{a - 1}(b + 1)^{-a-1}$ and $e^{-i\pi a} - e^{i\pi a} = -2i\sin(\pi \mu)$. Therefore $$-2i\sin \pi a \int_{-1}^1 (1 - x)^a(1+x)^{-a}(x - b)^{-2}\, dx = -4\pi i a\, (b-1)^{a-1}(b+1)^{-a-1}$$ or $$\int_{-1}^1 \left(\frac{1-x}{1+x}\right)^a\frac{dx}{(x-b)^2} = 2\pi a\csc(\pi a)\, (b-1)^{a-1}(b+1)^{-a-1}$$