Integral for $\Gamma(z) \Gamma(1-z)$ In gamma function How do you evaluatate
$\Gamma(z)\Gamma(1-z) = \int_{0}^\infty \frac{v^{z-1}}{v+1} dv$
which method could be fit to solve the integral?
The result should be
$ \Gamma(z)\Gamma(1-z) = \int_{0}^\infty \frac{v^{z-1}}{v+1} =\frac{\pi}{\sin \pi z}$
 A: On the one hand, the substitution $v=\tan^2 t$ converts the integral to $$2\int_0^{\pi/2}\sin^{2z-1}t\cos^{1-2z}t\,dt=\operatorname{B}(z,\,1-z)=\Gamma(z)\Gamma(1-z).$$(Everything you need to know about Beta and Gamma functions to follow that is found here.) On the other hand, the reflection formula is proven as thus. I'll make this answer more self-contained later, when it's clear which question the OP had in mind.
A: $$\Gamma(s)=\int_0^\infty x^{s-1}e^{-x}dx$$
$$\frac{\pi}{2} e^{-|x|}=\int_0^\infty\frac{\cos(xz)}{z^2+1}dz$$ 
The second integral can be calculated through the use of complex residues.
$$\Gamma(s)=\frac{2}{\pi}\int_0^\infty x^{s-1}\int_0^\infty\frac{\cos(xz)}{z^2+1}dzdx$$
By Fubini's Theorem: $$\Gamma(s)=\frac{2}{\pi}\int_0^\infty\frac{1}{z^2+1}\int_0^\infty x^{s-1}\cos(zx)dxdz$$
The inner integral is the Mellin Transform of $\Re(e^{-izx})$ which is equal to $\Gamma(s)\Re(iz)^{-s}=\Gamma(s)\cos(\frac{\pi s}{2})z^{-s}$
So $$\Gamma(s)=\frac{2}{\pi}\Gamma(s)\cos(\frac{\pi s}{2})\int_0^\infty\frac{z^{-s}}{z^2+1}dz$$
$$\int_0^\infty\frac{z^{-s}}{z^2+1}dz=\frac{\pi}{2}\sec(\frac{\pi s}{2})$$
Let $u=z^2$ $dz=\frac{du}{2\sqrt{u}}$
$$\frac{1}{2}\int_0^\infty\frac{u^{\frac{-s-1}{2}}}{u+1}du=\frac{\pi}{2}\sec(\frac{\pi s}{2})$$
Letting $s\to (1-2s)$ we arrive at the answer
$$\int_0^\infty \frac{u^{s-1}}{u+1}du=\pi\csc(\pi s)$$ 
To achieve the integral in the form of the $\Gamma(s)$ function, all that is required is either Euler's Reflection Formula, or a substitution and the definition of the Beta function as explained in @J.G.'s answer.
