I stumbled on the following rather appealing trigonometric definite integral, \begin{equation} \int_0^y \left(\frac{\sin x}{\sin (y-x)}\right)^a \mathrm{d}x = \pi \frac{\sin(ya)}{\sin(\pi a)} \end{equation} for $a\in (-1,1)$ and $y\in[0,\pi]$. Does anyone know a reference or a simple proof?


1 Answer 1


Just after submitting the question I realised there is a rather simple Mellin transform solution to this problem, which I record below. I'd still be interested in a reference or generalizations of this identity.

Making the substitution $z = \sin(y-x)/\sin x = \sin y \cot x -\cos y$ we get the integral

\begin{align} \int_0^y \left(\frac{\sin x}{\sin (y-x)}\right)^{-a} \mathrm{d}x &= \int_0^\infty \frac{\sin y}{1+z^2+2z \cos y} z^{-a} \mathrm{d}z \\ &= \frac{1}{2i} \int_0^\infty \frac{z^{-a}}{z+e^{-iy}}\mathrm{d}z-\frac{1}{2i} \int_0^\infty \frac{z^{-a}}{z+e^{-iy}}\mathrm{d}z\\ &= \frac{1}{2i}e^{i a y} \frac{\pi}{\sin(\pi a)} -\frac{1}{2i} e^{-i a y} \frac{\pi}{\sin(\pi a)}\\ &= \pi \frac{\sin(ya)}{\sin(\pi a)}. \end{align}


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