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I would like to know a nice approach for problems of this type

$I_n = \int_{-\pi}^{\pi} \frac{\sin (nx)}{\sin x} dx$

Then find $I_n$

My Approach:- Trying to write $I_n$ as a difference of two integrals, and then summing, in a telescopic fashion.

But, I can't seem to find the required difference.

There are several questions like this and i'd also appreciate some sort of common approach.

Gregory Grant
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Bayern ROX
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3 Answers3

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First show that

\begin{eqnarray}\frac{\sin nx}{\sin x}=\frac{\sin(n-2)x}{\sin x}+2\cos(n-1)x\end{eqnarray}

From there we have the reduction formula \begin{eqnarray}I_n=I_{n-2}\end{eqnarray} I'll leave the rest to you.

Alex Fok
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Consider $I_{n+2}-I_n$, and use the formula for $sinA-sinB$

David Quinn
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Another way would be using residue theorem. Let $z= e^{ix} ,$ so $ dz = ie^{ix} dx, \sin nx = (z^n - z^{-n})/2i, \sin x = (z-z^{-1}) /2i$. So the integral becomes $$ \int _{|z|=1} \frac{z^{n} -z^{-n}}{z-z^{-1}}\frac{dz}{iz} \\ = \int _{|z|=1} \frac{z^{2n} -1}{z^{n}(z^2-1)}\frac{dz}{i} \\ = \int _{|z|=1} \frac{1}{z^{n}} \left(\sum_{k=0}^{n-1} z^{2k} \right)\frac{dz}{i} .$$ Therefore the result is $2\pi$ if $n $ is odd, and $0$ if $n$ is even.

anumosh
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