Prove $\int_{0}^{\infty} \frac{x^{p-1}}{1+x} dx = \frac{\pi}{\sin (p \pi)}$ without using residue theorem or Beta function I need to prove that
$$\int \limits_{0}^{\infty} \frac{x^{p-1}}{1+x} dx = \frac{\pi}{\sin (p \pi)}$$
for $0<p<1$.
I know how to do it using residue theorem and using the Beta function integral definitions. Also, I can transform the integral to a summation; but that is cheating.
So, is there any simpler way to evaluate the integral using the integral definition of gamma function, or something lighter than the residue theorem?
 A: I'm not sure if this helps since in the end, I've reduced the problem to evaluating an infinite series (in which standard residue theorems are still required to evaluate the series).
This is done via a really simple decomposition of the integral:
\begin{equation}
\begin{aligned}
&\int^{\infty}_0 \frac{x^{p-1}}{1+x} dx \\
= &\int^{1}_0 \frac{x^{p-1}}{1+x} dx + \int^{\infty}_1 \frac{x^{p-1}}{1+x} dx \\
\end{aligned}
\end{equation}
and we consider the infinite series
\begin{equation}
\frac{1}{1+x} = \sum_{n=0}^\infty (-1)^n x^n
\end{equation}
for $|x| < 1$ and
\begin{equation}
\frac{1}{1+x} = \frac{1}{x}\frac{1}{1+1/x} = \sum_{n=0}^\infty (-1)^n x^{-n-1}
\end{equation}
for $|x| > 1$.
The first integral can be evaluated as:
\begin{equation}
\begin{aligned}
&\int^{1}_0 \frac{x^{p-1}}{1+x} dx \\
= & \int^{1}_0 x^{p-1} \sum_{n=0}^\infty (-1)^n x^n dx \\
= & \sum_{n=0}^\infty (-1)^n \int^{1}_0 x^{p-1+n} dx \\
= & \sum_{n=0}^\infty \frac{(-1)^n}{n+p}. \\
\end{aligned}
\end{equation}
Here, you can easily justify the interchange between the integral and infinite series using Fubini's Theorem.
Similarly, we have:
\begin{equation}
\begin{aligned}
&\int^{\infty}_1 \frac{x^{p-1}}{1+x} dx \\
= & \sum_{n=0}^\infty \frac{(-1)^n}{n+1-p}. \\
\end{aligned}
\end{equation}
The two series can be combined to give:
\begin{equation}
\begin{aligned}
&\int^{\infty}_0 \frac{x^{p-1}}{1+x} dx \\
= &\frac{1}{p} + \sum_{n=1}^\infty \frac{(-1)^n}{n+p} - \sum_{n=1}^\infty \frac{(-1)^n}{n-p}\\
= &\frac{1}{p} - (2p) \sum_{n=1}^\infty \frac{(-1)^n}{n^2-p^2}.\\
\end{aligned}
\end{equation}
At this point, you can look up a table of infinite series or use Mathematica to see that
\begin{equation}
\sum_{n=1}^\infty \frac{(-1)^n}{n^2-p^2} = -\frac{\pi}{2p}\left(\frac{1}{p\pi} + \frac{1}{\sin(p\pi)} \right).
\end{equation}
Substitute the above equation right in to obtain the required expression.
Remark: The way to evaluate the infinite series is usually done via Residue Theorem, so you can't really escape from that, although you could argue that you can search up a table of infinite series to obtain the results that you want.
