Prove that $\int_0^\infty\frac1{t^x+1}dt = \frac\pi x \csc \frac\pi x$

Question

I'm stuck on this identity:

$$ \int_0^\infty\frac1{t^x+1}dt = \frac\pi x \csc \frac\pi x $$

Could someone show me a proof for this?

What I've tried:

I've thrown a bunch of substitutions and integration by parts at this, but they haven't led me to the answer. I did, however, find these identites: $$ \int_0^\infty\frac1{t^x+1}dt = \int_0^\infty\frac{t^{x-2}}{t^x+1}dt = \int_0^1\left( \frac{1-t}t \right)^{1/x}dt $$ But none of these seem to lead anywhere helpful.

I also tried introducing another variable to turn it into something similar to the Laplace Transform, but I'm not very familiar with methods like that, so they've led nowhere.

Answer 1

You have a Beta function \begin{eqnarray*} B(a,b) = \int_0^1 t^{a-1} (1-t)^{b-1} dt = \frac{ \Gamma(a) \Gamma(b) } {\Gamma(a+b)}. \end{eqnarray*} Now use the well known gamma function identity \begin{eqnarray*} \Gamma(z) \Gamma(1-z) = \frac{ \pi} { \sin( \pi z)}. \end{eqnarray*}