Solve complex integral with $\Gamma$-function Let $s\in\mathbb C$ and $r\in\mathbb R$. In the integral
$$\int_{-\infty}^\infty \frac{1}{z^{r+s}\overline{z}^s} dx$$
we have $z=x+iy$ where $y>0$ is fixed. I read that you can explicitly compute the value of the integral using the $\Gamma$-function. How do you do that?
Edit:
Using integration by substitution (substitute $x$ with $yx$) we get
$$\int_{-\infty}^\infty \frac{1}{z^{r+s}\overline{z}^s} dx = y^{1-2s-r} i^{-r} \int_{-\infty}^\infty \frac{1}{(1-ix)^r|1-ix|^{2s}} dx.$$
Maybe that helps.
 A: The $\Gamma$ function enters in the following way. Let $f_a(x)=x^{a-1}e^{-x}$ for $x>0$ and $0$ for $x<0$, with $a>1$. Then the Fourier transform is
$$
\hat{f}_a(\xi)\;=\; \int_0^\infty x^{a-1}e^{-x} \,e^{-2\pi i x\xi}\;dx
\int_0^\infty x^ae^{-x(1+2\pi i\xi)}\;{dx\over x}
\;=\; (1+2\pi i\xi)^{-a}\Gamma(a)
$$
by the analytically continued form of the identity $\int_0^\infty x^ae^{-xt}\;{dx\over x}=t^{-a}\Gamma(a)$ for $t>0$ real. Then notice that the indicated integral is the hermitian inner product of two of these, so we can invoke Plancherel after a small change of variables...
$$
\int_{-\infty}^\infty {1\over (1+ix)^a} {1\over (1-ix)^b}\;dx
\;=\; 2\pi \int_{-\infty}^\infty {1\over (1+2\pi ix)^a}{1\over (1-2\pi ix)^b}\;dx
$$
$$
\;=\; {2\pi\over \Gamma(a)\Gamma(b)}\int_0^\infty x^{a-1}e^{-x}\cdot x^{b-1}e^{-x}\;dx \;=\;{2\pi\over \Gamma(a)\Gamma(b)}\int_0^\infty x^{a+b-1}e^{-2x}\;{dx\over x}
$$
$$
\;=\; \pi\, 2^{2-a-b} {\Gamma(a+b-1)\over \Gamma(a)\Gamma(b)}
$$
modulo typos and such...
