I've come across an integral that I'm not sure how to solve (nor am I sure whether it can be solved). I have tried to analyse the integral by the contour method but I'm not really sure where to begin, can anybody give me any pointers?
Integral:
$$\int_{-1}^{1} \frac{(x-a)^2}{\sqrt{(1-x)}\ (1-ax)^4} dx$$
where $0<a<1$ is a constant.
Thanks!
Note that $$ \int_{-1}^{1} \frac{(x-a)^2}{\sqrt{1-x}\ (1-ax)^4} dx =\int_{-1}^{1} \frac{(x-a)^2\sqrt{1-x}}{(1-x)(1-ax)^4} dx\\ =\int_{-1}^{1}\left(\frac{\frac{a}{(a-1)^2}}{ax-1}+ \frac{\frac{a}{a-1}}{(ax-1)^2}+ \frac{\frac{a^2-1}{a}}{(ax-1)^3} +\frac{\frac{(a+1)^2(a+1)}{a}}{(ax-1)^4}+ \frac{\frac{1}{(a-1)^2}}{1-x}\right)\sqrt{1-x} dx\\ $$ Now solve each one by substitution $t=\sqrt{1-x}$: $$ \int\frac{\sqrt{1-x}}{(1-ax)^n}dx= \int\frac{t}{(1-a(1-t^2))^n}\cdot-2tdt= \int\frac{-2t^2}{(1-a+at^2)^n}dt $$ it is not easy but possible...