Closed form for certain trigonometric integral

Question

\begin{align}&\mbox{Is there a closed form for} \\[2mm]&\int_0^{\pi/2} \sin^{2}\left(\, nx\,\right)\sin\left(\, mx\,\right)\cot\left(\, x\,\right) \,{\rm d}x\ \quad\mbox{where}\quad m, n\ \mbox{are positive integers ?.} \end{align}

I tried to turn product of sines into a sum but it didn't really get much easier. WolframAlpha doesn't help either.

I'll be grateful for your ideas how to evaluate it.

Answer 1

You could try writing the sines and cosines in terms of complex exponentials.

Say $x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \cdots + x y^{n-2} + y^{n-1} $; hence you can write

$\begin{equation*} \frac{\sin(mx)}{\sin(x)} = \frac{e^{imx}-e^{-imx}}{e^{ix} - e^{-ix}} = e^{i(m-1)x} + e^{i(m-2)x} e^{-ix} + \cdots + e^{ix} e^{-i(m-2)x} + e^{-i(m-1)x} = \sum_{j=0}^{m-1} e^{i(m-1-j)x}e^{-ijx} = \sum_{j=0}^{m-1}e^{i(m-1-2j)x} \end{equation*}$

hence, you have

$\begin{equation*} \sin^2(nx) \sin(mx) \cot(x) = \sin^2(nx) \cos(x) \frac{\sin(mx)}{\sin(x)} = \frac{2-e^{2inx}-e^{-2inx}}{4} \frac{e^{ix} + e^{-ix}}{2} \Big(\sum_{j=0}^{m-1}e^{i(m-1-2j)x} \Big) \end{equation*}$

Just combine the exponentials, and use the fact

$\begin{equation*} \int_0^{\frac{\pi}{2}} e^{i(m-n)x} dx = \begin{cases} \frac{\pi}{2} \text{ if } m = n \\ \frac{(i)^{m-n}-1}{m-n} \text{ if } m \neq n \end{cases} \end{equation*}$

Answer 2

You could convert your trig into complex exponents. $$\sin(mx) = \frac{e^{mxi}-e^{-mxi}}{2i} \\ \sin^2(nx) = \frac{e^{2nxi}+e^{-2nxi}-2}{-4} \\ \cot(x) = \frac{i(e^{2xi}+1)}{e^{2xi}-1} $$ which means $$\int_0^{\pi/2} \sin^2nx\sin mx\cot x \,dx \\ = \int_0^{\pi/2} \frac{e^{2nxi}+e^{-2nxi}-2}{-4}\frac{e^{mxi}-e^{-mxi}}{2i}\frac{i(e^{2xi}+1)}{e^{2xi}-1} \,dx \\ = \frac{-1}{8} \int_0^{\pi/2} (e^{2nxi}+e^{-2nxi}-2)(e^{mxi}-e^{-mxi})\frac{e^{2xi}+1}{(e^{2xi}-1)}dx$$ If you go this route you'll still have to tangle with the algebra monster, but once that's over you should have a mostly easy integral.

Answer 3

Here's an approach I'd probably take, but I can't promise it'll lead anywhere.

Fist look at $\sin nx \sin mx$ and express it as a linear combination of $ \cos (n+m)x$ and $\cos(n-m)x$. Now do the same for $\sin nx \cos x$, which uses up the numerator of the cotangent, and express it as a sum or difference of sines.

Then expand out by the distributive law into four separate integrals, each having the form $$ \frac{\cos kx \sin px}{\sin x}. $$

The numerator can again be expressed as a combination of $\sin(k+p)x$ and $\sin(k-p)x$, so for each of the four you get two integrals of the form $$ \int_0^{\pi/2} \frac{\sin rx}{\sin x} dx $$

I'm not certain that these are any easier, but they look that way to me. :)