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let $n, m$ be integers such that $m$ is odd integer , My simple Guess about evaluation of this :$\int_{0}^{\pi}(\frac{\sin n x}{\sin x})^m$ is to get the following , for $n$ is even integer the integrand is $0$ and for $n$ is odd the integrand is give something like : $k\pi$ with k is such form which i can't get it . My question here is What is the closed form of this:$$\int_{0}^{\pi}\left(\frac{\sin n x}{\sin x}\right)^m$$ for $m$ is odd ?

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    $\begingroup$ If $m$ is even, then the integrand is nonnegative, so the integral could only be $0$ if the integrand were everywhere $0$, which is obviously not the case as long as $n\neq 0$. $\endgroup$ – Fimpellizieri Aug 16 '18 at 22:03
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    $\begingroup$ You could look at the following link : en.wikipedia.org/wiki/Dirichlet_kernel for $m=1$ and $n$ odd. $\endgroup$ – tmaths Aug 16 '18 at 22:11
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    $\begingroup$ Your attempts? Once you exploit $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and the binomial theorem, it is not difficult to figure out what is the value of the wanted integral. $\endgroup$ – Jack D'Aurizio Aug 16 '18 at 22:13
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For $m,n \in \mathbb{N}$ let $$ I(m,n) = \int \limits_0^\pi \left(\frac{\sin(n x)}{\sin(x)}\right)^m \, \mathrm{d} x \, . $$ If $m$ is odd and $n$ is even, the substitution $x = \pi - y$ yields $$ I(m,n) = \int \limits_0^\pi \left(\frac{(-1)^{n+1} \sin(n y)}{(-1)^2 \sin(y)}\right)^m \, \mathrm{d} y = (-1)^m I(m,n) = - I(m,n) \, ,$$ so $I(m,n) = 0$ .

In any other case $(n-1)m$ is even and the integral does not vanish. Then we can use the partial sums of the geometric series to find \begin{align} I(m,n) &= \int \limits_0^\pi \left(\frac{\mathrm{e}^{\mathrm{i} n x} - \mathrm{e}^{-\mathrm{i} n x}}{\mathrm{e}^{\mathrm{i} x} - \mathrm{e}^{-\mathrm{i} x}}\right)^m \, \mathrm{d} x = \int \limits_0^\pi \mathrm{e}^{\mathrm{i} (n-1) m x}\left(\frac{1 - \mathrm{e}^{-2 \mathrm{i} n x}}{1 - \mathrm{e}^{-2\mathrm{i} x}}\right)^m \, \mathrm{d} x \\ &= \int \limits_0^\pi \mathrm{e}^{\mathrm{i} (n-1) m x}\left(\sum \limits_{k=0}^{n-1} \mathrm{e}^{-2\mathrm{i} k x}\right)^m \, \mathrm{d} x \\ &= \sum \limits_{k_1=0}^{n-1} \cdots \sum \limits_{k_m=0}^{n-1} \, \int \limits_0^\pi \exp\left[2 \mathrm{i}\left(\frac{(n-1)m}{2} - \sum \limits_{i=1}^m k_i\right) x\right] \, \mathrm{d} x \, . \end{align} Now note that the remaining integral is non-zero (and equal to $\pi$) if and only if $$ \frac{(n-1)m}{2} - \sum \limits_{i=1}^m k_i = 0 \, \Longleftrightarrow \, \sum \limits_{i=1}^m k_i = \frac{(n-1)m}{2}$$ holds. Therefore we have $$ I(m,n) = N\left(\frac{(n-1)m}{2} , m , n-1\right) \pi \, ,$$ where $N\left(\frac{(n-1)m}{2} , m , n-1\right)$ is the number of $m$-tuples $(k_1 , \,\dots \, , k_m) \in \{0 \, \dots \, , n-1\}$ whose elements sum to $\frac{(n-1)m}{2}$ .

$N(b,c,d)$ can also be interpreted as the number of ways to fill $c$ containers with $b$ balls such that there are at most $d$ balls in each container. These numbers are discussed in this question.

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  • $\begingroup$ +1 for nice solution $\endgroup$ – user547564 Aug 18 '18 at 18:34

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