When calculating $\int_{0}^{\infty}\frac{x^{a-1}}{1+x}dx（0<a<1）$, one way is to draw the path C1 like the first picture. Then we have
$\int_{C1}\frac{z^{a-1}}{1+z}dz=\int_r^R\frac{x^{a-1}}{1+x}dx+\int_R^r\frac{x^{a-1}e^{i2\pi(a-1)}}{1+x}dx+\int_{|z|=R}+\int_{|z|=r}=Res(-1)$.
Let $R\to\infty$ and $r\to0$, the first and second term on the right side will become:
$[1-e^{2i\pi(a-1)}]\int_0^{\infty}\frac{x^{a-1}}{1+x}dx$
Then the rest of the problem will be easy.
I want to use similar method to solve $\int_{0}^{\infty}\frac{x^{a-1}}{1-x}dx$ so I draw the path C2 like the second picture and have this equation.
$\int_{C2}\frac{z^{a-1}}{1-z}dz=\int_{CA}+\int_{BD}+\int_{|z|=R}+\int_{|z|=r}=Res(1)$
But I am confused when intergrating the function on CA and BD.
When integrating on CA, I let $z=xe^{i\pi}=-x$ so $z=e^{logx+i\pi}$ and $dz=-dx$, then
$\int_{CA}\frac{z^{a-1}}{1-z}dz=\int_{R}^{r}\frac{e^{(a-1)(logx+i\pi)}}{1+x}(-dx)$
Similarly, on BD, let $z=xe^{-i\pi}=-x$ ,then
$\int_{BD}\frac{z^{a-1}}{1-z}dz=\int_{r}^{R}\frac{e^{(a-1)(logx-i\pi)}}{1+x}(-dx)$
and the terms about $1-x$ disappeared. It seems like I'm still solving $\int_{0}^{\infty}\frac{x^{a-1}}{1+x}dx$.
Did I choose the wrong path or have other mistakes? If so, how to draw the suitable and correct path?
