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

  • Thread starter Thread starter Euge
  • Start date Start date
Euge
Gold Member
MHB
POTW Director
Messages
2,072
Reaction score
245
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!
 
Physics news on Phys.org
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.
 
No one answered this week's problem. You can read my solution below.
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 $$\int_{-1}^1 \left(\frac{1-x}{1+x}\right)^a\frac{dx}{(x-b)^2} = 2\pi a\csc(\pi a)\, (b-1)^{a-1}(b+1)^{-a-1}$$
 
Back
Top