How to evaluate $\sum\limits_{k=0}^{n-1} \sin^t(\pi k/2n)$?

Solution 1:

Here is an answer when $t = 2s$ is even.

Let $n, s$ be non-negative integers. Then with $\zeta = \mathrm{e}^{i\pi/n}$ and $\omega = \zeta^2 = \mathrm{e}^{2i\pi/n}$, we have

\begin{align*} 2^{2s} \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n} \right) &= \sum_{k=0}^{n-1} \left( \frac{\zeta^k + \zeta^{n-k}}{i} \right)^{2s} \\ &= (-1)^{-s} \sum_{k=0}^{n-1}\sum_{l=0}^{2s} \binom{2s}{l} \zeta^{(2s-l)k+l(n-k)} \\ &= (-1)^{-s} \sum_{k=0}^{n-1}\sum_{l=0}^{2s} \binom{2s}{l} (-1)^l \omega^{(s-l)k}. \end{align*}

Interchanging the order of summation, we have

\begin{align*} 2^{2s} \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n} \right) &= \sum_{l=0}^{2s} (-1)^{l-s} \binom{2s}{l} \left( \sum_{k=0}^{n-1} (\omega^{s-l})^k \right) \\ &= \sum_{l=0}^{2s} (-1)^{l-s} \binom{2s}{l} \left( n \cdot \mathbf{1}_{\{ l \equiv s \ (\mathrm{mod} \ n)\}} \right) \\ &= n \sum_{j} (-1)^{nj} \binom{2s}{s+nj}, \end{align*}

where the last summation runs over all integers $j$ such that $-\frac{s}{n} \leq j \leq \frac{s}{n}$.

Examples. As special cases, plugging $s = n$ yields

$$ \sum_{k=0}^{n-1} \sin^{2n} \left(\frac{\pi k}{n} \right) = \frac{n}{2^{2n}} \left[ \binom{2n}{n} + (-1)^n 2 \right], $$

and similarly, replacing $n$ by $2n$ and plugging $s = n$ yields

$$ \sum_{k=0}^{n-1} \sin^{2n} \left(\frac{\pi k}{2n} \right) = \frac{n}{2^{2n}} \binom{2n}{n} - \frac{1}{2}. $$

Addendum. Here is an intuition on why we expect the sum to be simplified. If $t = 2s$ is even, then we can expand $\sin^{2s} x$ into a linear combination of $1, \cos 2x, \cos 4x, \cdots, \cos 2sx$. So in our case, we may write

$$ \sin^{2s} \left(\frac{\pi k}{n}\right) = \sum_{j=0}^{s} a_j \cos \left(\frac{2\pi j k}{n}\right). $$

Now, as you sum this over $k = 0, \cdots, n-1$, all the cosine terms will cancel out except when $j$ is a multiple of $n$. This means that

$$ \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n}\right) = n (a_0 + a_n + a_{2n} + \cdots). $$

So it is enough to identify $a_j$ for $j$'s multiple of $n$. This can be done by expanding $\sin^{2s} x$ using complex exponential and the binomial theorem. This is essentially what we did in the computation above.

On the other hand, this trick does not work for odd $t$. Indeed, when $t = 2s+1$ is odd, we can write

$$ \sin^{2s+1} \left(\frac{\pi k}{n}\right) = \sum_{j=0}^{s} a_j \sin \left(\frac{\pi (2j+1)k}{n}\right) $$

for some constants $a_0, \cdots, a_n$. (In fact, $a_j = (-1)^j 2^{-2s} \binom{2s+1}{s-j}$.) Now summing both sides for $k = 0, \cdots, n-1$,

\begin{align*} \sum_{k=0}^{n-1}\sin^{2s+1} \left(\frac{\pi k}{n}\right) &= \sum_{j=0}^{s} a_j \sum_{k=0}^{n-1} \sin \left(\frac{\pi (2j+1)k}{n}\right) \\ &= \frac{1}{2^{2s}} \sum_{j=0}^{s} (-1)^j \binom{2s+1}{s-j} \cot \left(\frac{\pi(2j+1)}{2n}\right). \end{align*}

It is hard for me to believe that this will ever simplify except for some nice $s$.

Addendum 2 (Just for fun). Although no longer simpler than the original sum, one can also prove that for complex $t$ with $\Re(t) > 0$ and for positive integer $n$ the following formula holds:

$$\sum_{k=0}^{n-1} \sin^{t} \left(\frac{\pi k}{n} \right) = \frac{n}{2^{t}} \sum_{j=-\infty}^{\infty} (-1)^{nj} \binom{t}{\frac{t}{2}+nj}. $$

Here, $\binom{n}{k} = \frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}$ is the extended binomial coefficient.

Solution 2:

This integral remind me of Matsubara sum. Let me use contour integral to provide another perspective of the case $t = 2s$.

Define $a_n = \sum_{k = 1}^{n-1} \sin ^{t} \left( \frac{\pi k }{n} \right) $. What OP is looking for is \begin{equation} \sum_{k=1}^{n-1} \sin^{t} \left( \frac{\pi k }{2n} \right) = \frac{1}{2} \left( \sum_{k=1}^{2n-1} \sin^{t} \left( \frac{\pi k }{2n} \right) -1 \right) = \frac{1}{2} ( a_{2n } - 1 ) \end{equation} One can extend the summation to all the roots of $\zeta^{2n} = 1$ if $t = 2s$ \begin{equation} a_n = \frac{1}{2} \sum_{k=1}^{2n} \sin^{2s} \left( \frac{\pi k }{n} \right ) = \frac{1}{2} \sum_{ k = 1}^{2n} f( \zeta_k ) \qquad \zeta_k = \exp( \frac{ik\pi }{n}) \end{equation} where \begin{equation} f(z ) = \left( \frac{z - \frac{1}{z}}{2i} \right)^{2s} \end{equation}

The summation can be regarded as the sum of the residue of the follow contour integral \begin{equation} \sum_{ k = 1}^{2n} f( \zeta_k ) = \frac{2n}{2\pi i }\oint_{C} \frac{z^{2n-1}}{ z^{2n} - 1} f( z) dz \end{equation} where the integrand is specifically designed such that \begin{equation} \text{Res} ( \frac{ 2n z^{2n-1}}{z^{2n} - 1} f(z) , \zeta_k ) = 2n \frac{z^{2n-1}(z - \zeta_k)}{z^{2n} - \zeta^{2n}_k} f(\zeta_k ) \Big|_{z = \zeta_k } = f(\zeta_k ) \end{equation} Here the closed contour should enclose all $\zeta_k$ but avoid the singularities of $f$, which can be taken as two concentric circles as shown in the following figure(red spots are those roots $\xi_k$).

The inner contour gives (minus) the residue at $0$, and the outer contour gives (minus) the residue at $\infty$, hence \begin{equation} \begin{aligned} a_n &= n \left[ -\text{Res}( \frac{z^{2n-1}}{ z^{2n} - 1} f( z) , 0 ) - \text{Res}( \frac{z^{2n-1}}{ z^{2n} - 1} f( z) , \infty ) \right] \\ \end{aligned} \end{equation} Convert the residue of $\infty$ to $0$ \begin{equation} \begin{aligned} a_n =& n \left[ \text{Res}( \frac{z^{2n-1}}{ 1 - z^{2n} } f( z) , 0 ) + \text{Res}( \frac{\frac{1}{z}}{1 - z^{2n} } f( \frac{1}{z}) , 0 ) \right] \\ =& n \left[ \text{Res}( \frac{1}{z}\frac{z^{2n} + 1 }{ 1 - z^{2n} } f(z) , 0 ) \right] \\ =& n \left[ \text{Res}( \frac{1}{z} ( 1 + 2\sum_{k=1}^{\infty} z^{2nj} ) f(z) , 0 ) \right] \\ \end{aligned} \end{equation} Finally, expand $f(z)$ and extract the terms with power $0, -2nj, -4nj$ etc, \begin{equation} \begin{aligned} a_n &= n \frac{(-1)^s}{2^{2s}} \left( (-1)^s {2s \choose s } + 2 \sum_{0 < nj < s } (-1)^{s+ nj} {2s \choose s+nj } \right)\\ &= \frac{n}{2^{2s}} \sum_{ -s < nj < s } (-1)^{nj} {2s \choose s + nj } \\ \end{aligned} \end{equation} This is the same as what @Sangchul Lee obtained.