How to solve the following integral, using Euler functions (Gamma, Beta, and Phi):
$$\int_{0}^{\pi}\frac{\sin^px}{1+\cos x} dx $$
I got this integral from "Collection of problems in mathematical analysis, Kudryavtsev. Book 3.". Under the theme "Euler functions".
There are also functions, which can help:
\begin{align*}
\phi(p) &= \beta(p,1-p)= \frac{\pi}{\sin(p\pi)}\\
\phi'(p) &= – \frac{\pi^2\cos(p\pi)}{\sin^2(p\pi)}\\
\phi''(p) &= \frac{\pi^3}{\sin(p\pi)}\left(\cot^2(p\pi)+\frac1{\sin^2(p\pi)}\right)
\end{align*}
Best Answer
Using $1+\cos(x)=2\cos^2(x/2)$ we see that
$$\begin{align} \int_0^\pi \frac{\sin^p(x)}{1+\cos(x)}\,dx&=\int_0^\pi \frac{\sin^p(x)}{2\cos^2(x/2)}\,dx\\\\ &=\int_0^{\pi/2} \frac{\sin^p(2x)}{\cos^2(x)}\,dx\\\\ &=2^p\int_0^{\pi/2} \sin^p(x)\cos^{p-2}(x)\,dx\\\\ &=2^{p-1}\text{B}\left(\frac{p+1}{2},\frac{p-1}{2}\right)\\\\ &=2^{p-1}\text{B}\left(\frac{p-1}{2}+1,\frac{p-1}{2}\right)\tag1\\\\ &=\text{B}\left(\frac12,\frac{p-1}{2}\right)\tag2 \end{align}$$
where in going from $(1)$ to $(2)$ we used the identities $\text{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$, $\Gamma(x+1)=x\Gamma(x)$, and
$$\frac1{\Gamma(2x)}=2^{1-2x}\frac{\Gamma(1/2)}{\Gamma(x)\Gamma(x+1/2)}=2^{1-2x}\frac{\text{B}\left(\frac12,x\right)}{ \Gamma^2(x)}$$