# Problem of the Week #283 - Dec 11, 2018

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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

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#### Euge

##### MHB Global Moderator
Staff member
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

##### MHB Global Moderator
Staff member
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 $$\int_{-1}^1 \left(\frac{1-x}{1+x}\right)^a\frac{dx}{(x-b)^2} = 2\pi a\csc(\pi a)\, (b-1)^{a-1}(b+1)^{-a-1}$$

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