Definite integral $\int_0^\infty \frac{1}{x^a(1+x)}dx$ I would like to know why the identity
$\int_0^\infty \frac{1}{x^a(1+x)}dx = \frac{\pi}{\sin(a \pi)}$
holds. I found it in a reference work without any proof. Euler's reflection formula may help, but I do not know how.
 A: Note that we have
$$\begin{align}
\int_0^\infty \frac{1}{x^a(1+x)}\,dx&=B(a,1-a)\tag 1\\\\
&=\frac{\Gamma(a)\Gamma(1-a)}{\Gamma(1)}\tag 2\\\\
&=\frac{\pi}{\sin(\pi a)}\tag 3
\end{align}$$
In arriving at $(1)$ we used that fact that $B(x,y)=\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}\,dt$
In going from $(1)$ to $(2)$ we used the relationship $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$.
And in going from $(2)$ to $(3)$ we used the reflection principal for the Gamma function, $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi a)}$.

In THIS ANSWER, I used real analysis methodologies only to develop the relationship between the Beta and Gamma function and prove the reflection principal for the Gamma function.

A: Here I show that:
$$B(m,n)=\int_{0}^{\infty} \frac{x^{m-1}}{(1+x)^{m+n}}$$
Here we have $m-1=-a$ and $m+n=1$. So that $m=1-a$ and $n=a$. So we have,
$$\int_0^\infty \frac{1}{x^a(1+x)}\, dx$$
$$=B(a,1-a)$$
Which by the Beta-Gamma function relationship is,
$$=\frac{\Gamma(a)\Gamma(1-a)}{\Gamma(1)}$$
By the reflection formula this is,
$$=\frac{\pi}{\sin(\pi a)}$$
