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In need a help in computing the sum $$\sum_{i=0}^{n-1}\frac{1}{a+\sin^2(\frac{ij\pi}{n})},$$ where $a$ is rational and $j$ is a non-negative integer. The case $a=0, j=1$ is solved here sum of reciprocals squared sines. It seems that the answer 0 in the same post gives a solution for $j=1$ and any $a$. I have tried to generalize the computation in answer 7, but this was beyond my skills.

Alex
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1 Answers1

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First we solver the easier problem: will find $$S_0=\sum_{k=0}^{n-1}\frac1{a+\sin^2\frac{\pi k}n}, \,a>0$$ Denoting for a while $1+2a=\cosh(2\pi b)$ $$S_0=\sum_{k=0}^n\frac1{a+\sin^2\frac{\pi k}n}-\frac1a=2\sum_{k=0}^n\frac1{\cosh(2\pi b)-\cos\frac{2\pi k}n}-\frac1a=2S_1-\frac1a$$ To evaluate $\displaystyle S_1$ we integrate the function $f(z)=\frac{\pi \cot\pi z}{\cosh(2\pi b)-\cos\frac{2\pi z}n}$ along the rectangular contour $C$ in the complex plane: $$-iR\to-iR+n\to iR+n\to iR\to -iR$$ with added arches around the poles: $z=0; \,n;\,\pm ibn;\,n\pm ibn$, leaving all the poles inside the contour (going counter-clockwise).

As the integrand is periodic with the period $\pi$, the integrals (in the PV sense) $-iR+n\to iR+n$ and $iR\to -iR$ cancel each other. Also, the integrals $-iR\to-iR+n$ and $iR+n\to iR$ exponentially tend to zero - as $R\to\infty$

Therefore, we are left with the integral along the added smal arches: $$\oint_C\frac{\pi\cot\pi z}{\cosh(2\pi b)-\cos\frac{2\pi z}n}dz=\pi i\left(\underset{z=0; \,n}{\operatorname{Res}}\frac{\pi\cot\pi z}{\cosh(2\pi b)-\cos\frac{2\pi z}n}+\underset{\binom{z=\pm ibn}{z=n\pm ibn}}{\operatorname{Res}}\frac{\pi\cot\pi z}{\cosh(2\pi b)-\cos\frac{2\pi z}n}\right)$$ On the other hand, the residues in the points $z=0,1,...,n$ give us $S_1$; therefore $$\oint_C\frac{\pi\cot\pi z}{\cosh(2\pi b)-\cos\frac{2\pi z}n}dz=2\pi i\left(S_1+\underset{\binom{z=\pm ibn}{z=n\pm inb}}{\operatorname{Res}}\frac{\pi\cot\pi z}{\cosh(2\pi b)-\cos\frac{2\pi z}n}\right)$$ The residue evaluation is straightforward; using also $\cosh(2\pi b)=2a+1; \,\sinh(2\pi b)=2\sqrt{a(1+a)}$ $$S_1=n\frac{\coth(\pi bn)}{\sinh(2\pi b)}+\frac1{2a}$$ and the desired sum $$S_0=2S_1-\frac1a=2n\frac{\coth(\pi bn)}{\sinh(2\pi b)}$$ Using $e^{2\pi b}=1+2a+2\sqrt{a(1+a)}$ $$\boxed{\,\,S_0=\sum_{k=0}^{n-1}\frac1{a+\sin^2\frac{\pi k}n}=\frac n{\sqrt{a(1+a)}}\frac{\big(1+2a+2\sqrt{a(1+a)}\,\big)^n+1}{\big(1+2a+2\sqrt{a(1+a)}\,\big)^n-1}\,\,}$$ We can also consider the sum $$\hat{S}(a)=\sum_{k=1}^{n-1}\frac1{a+\sin^2\frac{\pi k}n}=\frac n{\sqrt{a(1+a)}}\frac{\big(1+2a+2\sqrt{a(1+a)}\,\big)^n+1}{\big(1+2a+2\sqrt{a(1+a)}\,\big)^n-1}-\frac1a$$ Leading $a\to 0$ we get the well-known expression: $$\hat{S}(a=0)=\sum_{k=1}^{n-1}\frac1{\sin^2\frac{\pi k}n}=\frac{n^2-1}3$$ Acting exactly in the same way we can find, for example $$S_j=\sum_{k=0}^{n-1}\frac{1}{a+\sin^2(\frac{\pi j k}n)}, \,j=2, 3, ...$$ The only difference is that we get additional poles inside the contour (at $z=\frac nj(l\pm ib);\,l=1,\,2,\,...\,j-1$). Generally, we can get different cases (double poles, for example). In the most simple case ($\frac nj\neq\text{integer}$), when all poles are simple, $$S_j=\frac n{j\sqrt{a(1+a)}}\left(\coth\frac{\pi n b}j+\sum_{l=1}^{j-1}\frac{\sinh\frac{2\pi bn}j}{\cosh\frac{2\pi bn}j-\cos(\frac{2\pi n}jl)}\right)$$ where $2\pi b=\ln\big(1+2a+2\sqrt{a(1+a)}\big)$

Svyatoslav
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