Integrating integer powers of a sine function via a hypergeometric function

I recently needed to integrate the fifth power of a sine function from zero to $\pi$ and did this via a hypergeometric function, which I found quite interesting. I will make a quick note of it here. I came across the following entry in a table of integrals:

$\int \sin^n(ax) dx = -\frac{1}{a} \cos(ax) \ {_2F_1 \big[\frac{1}{2}, \frac{1-n}{2}, \frac{3}{2}, \cos^2(ax)\big]} \qquad \qquad \qquad \qquad \qquad (1)$

The function $_2F_1(a, b, c, z)$ is a special function known as the ordinary hypergeometric function. The required definite integral for $n = 5$ , $a = 1$ and limits $0$ to $\pi$ is then

$\int_0^{\pi} dx \sin^5(x) = \bigg[ -\cos(x) \ {_2F_1 \big[\frac{1}{2}, -2, \frac{3}{2}, \cos^2(x)\big]} \bigg]_0^{\pi} = 2 {_2F_1 \big[\frac{1}{2}, -2, \frac{3}{2}, 1 \big]} \qquad \qquad \qquad \qquad \qquad (2)$

What interested me is that there is a summation theorem for the case $_2F_1(a, b, c, 1)$ which yields a simple ratio of products of the gamma function when $Re(c) > Re(a+b)$ :

$_2F_1(a, b, c, 1) = \frac{\Gamma(c) \Gamma(c - a - b)}{\Gamma(c-a) \Gamma(c-b)} \qquad \qquad \qquad \qquad \qquad (3)$

Since this condition happens to be satisfied in our case, we thus have

${_2F_1 \big[\frac{1}{2}, -2, \frac{3}{2}, 1 \big]} = \frac{\Gamma \big(\frac{3}{2}\big) \Gamma \big(\frac{3}{2} - \frac{1}{2} + 2 \big)}{\Gamma \big(\frac{3}{2}-\frac{1}{2}\big) \Gamma \big(\frac{3}{2} + 2 \big)} = \frac{\big(\frac{\sqrt{\pi}}{2}\big)(2!)}{(1)\big(\frac{15\sqrt{\pi}}{8}\big)} = \frac{8}{15}$

Therefore

$\int_0^{\pi} dx \sin^5(x) = \frac{16}{15}$

Integrating integer powers of a sine function via a hypergeometric function

Published by Dr Christian P. H. Salas

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