integration question $\int_{-1}^{1}\left(\frac{1-x}{1+x}\right)^a\frac{dx}{(x-b)^2}$ How to integrate this integral $\displaystyle\int_{-1}^{1}\bigg(\frac{1-x}{1+x}\bigg)^a\frac{dx}{(x-b)^2}$ where $0<a<1$ and $b>1$
Answer. I put this into online integral calculator, but it said it cann't do this integration.
 A: First, substitute $t = (1-x)/(1+x)$.
$$\begin{aligned} I = \int_{-1}^{1}\left(\frac{1-x}{1+x}\right)^{a}\frac{\mathrm{d}x}{(x-b)^{2}} &= \frac{1}{2}\int_{0}^{\infty}t^{a}\left(\frac{x+1}{x-b}\right)^{2}\mathrm{d}t \\
&= 2\int_{0}^{\infty}\frac{t^{a}\,\mathrm{d}t}{(1-t-b(1+t))^{2}} \\
&= 2\int_{0}^{\infty}\frac{t^{a}\,\mathrm{d}t}{((1-b) - (1+b)t)^{2}} \\
&= 2\int_{0}^{\infty}\frac{t^{a}\,\mathrm{d}t}{((b+1)t + (b-1))^{2}}\end{aligned}$$
Let $b_{\pm} = b\pm 1$. Then after $u = b_{+}t/b_{-}$,
$$\begin{aligned} I = 2\int_{0}^{\infty}\frac{t^{a}\,\mathrm{d}t}{(b_{+}t + b_{-})^{2}} &= \frac{2}{b_{-}^{2}}\left(\frac{b_{-}}{b_{+}}\right)^{a+1}\int_{0}^{\infty}\frac{u^{a}\,\mathrm{d}u}{(u+1)^{2}}.\end{aligned}$$
Using the beta function
$$ \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = \int_{0}^{\infty}\frac{z^{y-1}\,\mathrm{d}z}{(z+1)^{x+y}},$$
we identify $x=1-a$ and $y=1+a$. Then
$$ I = \frac{2}{b_{-}^{2}}\left(\frac{b_{-}}{b_{+}}\right)^{a+1}\frac{\Gamma(1-a)\Gamma(1+a)}{\Gamma(2)}.$$
Using the reflection identity $\Gamma(z)\Gamma(1-z) = \pi/\sin\pi z$ and recursion $\Gamma(1+a) = a\Gamma(a)$, we have
$$ \frac{\Gamma(1-a)\Gamma(1+a)}{\Gamma(2)} = \Gamma(1-a)a\Gamma(a) = \frac{\pi a}{\sin\pi a}.$$
The answer is then
$$ I = \frac{2}{(b-1)^{2}}\left(\frac{b-1}{b+1}\right)^{a}\frac{b-1}{b+1}\frac{\pi a}{\sin\pi a} = \boxed{\frac{2}{b^{2}-1}\left(\frac{b-1}{b+1}\right)^{a}\frac{\pi a}{\sin\pi a}.}$$
