$\sum_{k=1}^\infty \int_{a}^{b}c^k\sin^k(x)\,\mathrm{d}x$=?

Question

Let $a,b,c\in\mathbb{R}$ such that $b\ge a$ and $|c|<1$. Consider the series $$ \sum_{k=1}^\infty \int_{a}^{b} c^k \sin^k(x)\,\mathrm{d}x. $$

The above series clearly converges since $|c|<1$ by assumption and $|\sin^k(x)|\le 1$, for all $k$, but does the limit value of the series admit a "nice" closed-form expression?

Any help is really appreciated. Thank you!

Answer 1

We want to compute $$\int_a^b f(c,x) dx$$ with $f(c,x) = \frac{1}{1-c \sin (x)}$. Let $N$ be the smallest integer such that $a \le (2N+1)\pi$ and $M$ the largest integer such that $(2M-1) \pi \le b$ then : $$\int_a^b f(c,x) dx = \int_a^{(2N + 1)\pi} f(c, x) dx + \sum_{k = N+1}^{M-1} \int_{(2k-1)\pi}^{(2k+1)\pi} f(c,x) dx + \int_{(2M-1)\pi}^b f(c,x)dx$$ Now we can use the substitution $u = \tan (\frac{x}{2})$ and have : $$\int_a^b f(c,x) dx = \int_{\tan (a/2)}^{+\infty} g(u,c) du + (M-N-1) \int_{-\infty}^{+\infty} g(u,c) du + \int_{-\infty}^{\tan(b/2)} g(u,c) du$$ with $g(u,c) = \frac{2}{u^2 -2cu + 1} = 2\frac{\partial}{\partial u} \frac{ \text{Arctan} \left(\frac{u-c}{\sqrt{1-c^2}}\right)}{\sqrt{1-c^2}}$. This ends the calculations.

Answer 2

Too long for a comment.

For general $(a,b)$, the integral $$I_k=\int_{a}^{b} \sin^k(x)\,\mathrm{d}x$$is very complex (involving hypergeometric functions).

If we use $a=0$ and $b=\frac \pi 2$, we have the usual $$I_k=\frac{\sqrt{\pi }}2\frac{ \Gamma \left(\frac{k+1}{2}\right)}{ \Gamma \left(\frac{k+2}{2}\right)}$$ and $$\sum_{k=1}^\infty c^k\,I_k=\frac{\pi(1- \sqrt{1-c^2})+2 \sin ^{-1}(c)}{2 \sqrt{1-c^2}}$$ If $a=0$ and $b=\pi$, just double.

Answer 3

Proceeding naively,

$\begin{array}\\ \sum_{k=1}^\infty \int_{a}^{b} c^k \sin^k(x)dx &=\int_{a}^{b}\sum_{k=1}^\infty c^k \sin^k(x)dx\\ &=\int_{a}^{b}\dfrac{c\sin(x)dx}{1-c\sin(x)}\\ &= -\left(\dfrac{2 \arctan\left(\dfrac{c - \tan(x/2)}{\sqrt{1 - c^2}}\right)}{\sqrt{1 - c^2}} + x\right)\bigg|_a^b\\ \end{array} $

The last step was done by Wolfy since I didn't have the slightest idea how to do it.

Evidently the $u=\tan(x/2)$ as suggested by Youem is the key.