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The following formula is well known due to theory of elliptic functions: $\sum _{n=1}^{\infty } \text{sech}^2(n \pi )=-\frac{1}{2}+\frac{1}{2 \pi }+\frac{\Gamma \left(\frac{1}{4}\right)^4}{16 \pi ^3}$.

Question $1$: How to evaluate the following $$T_k=\sum _{n=1}^{\infty }\text{sech}^{2k}(n \pi)$$ For instance Mathematica gives $T_2=\sum _{n=1}^{\infty } \text{sech}^4(\pi n)=-\frac{1}{2}+\frac{1}{3 \pi }+\frac{\Gamma \left(\frac{1}{4}\right)^8}{192 \pi ^6}+\frac{\Gamma \left(\frac{1}{4}\right)^4}{24 \pi ^3}$ but I don't know how it's generated (it cannot evaluate $T_3$, etc).

Note that series of class $S_k=\sum _{n=1}^{\infty }\text{csch}^{2k}(n \pi)$ is evaluated in this post, by manipulating the normalized Eisenstein series $G_{2k}(i)=\sum_{m,n\in\mathbb R, mn\not=0}\ \frac{1}{(m+ni)^{2k}}$. For $T_k$ the corresponding series should be $\sum_{m,n\in\mathbb R, mn\not=0}\ \frac{1}{(m+\frac12+ni)^{2k}}$, but so far I can't see how to compute it for arbitrary $k$.

Question $2$: By Jacobi theta identity one have $\sum _{n=1}^{\infty } \text{sech}(\pi n)=\frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \pi ^{3/2}}-\frac{1}{2}$ (Mathematica also 'remembers' this result), while no closed-form of $\sum _{n=1}^{\infty } \text{csch}(\pi n)$ seems to exist. So, what causes the difference between two cases? More generally, is it possible to evaluate the odd-weight class $$\tilde T_k=\sum _{n=1}^{\infty }\text{sech}^{2k-1}(n \pi)$$

Any help will be appreciated.

Infiniticism
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1 Answers1

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This is how I see the theory behind those things, which means there could very well be a simpler and more concrete way to solve it in this particular case.

  • The elliptic way

We can evaluate $$\sum_{m,n}\ \frac{1}{(m+\frac12+ni)^{2k}},\qquad \sum_{(m,n)\ne (0,0)}\ \frac{1}{(m+ni)^{2k}}$$ in term of $\wp_i^{(k-2)}(1/2)$ and $(\wp_i^{(k-2)}-z^{-1})(0)$ where $\wp_i(z) = \sum_{m,n}\frac{1}{(z+m+ni)^2}-\frac{1_{(m,n)\ne (0,0)}}{(m+in)^2}$,

which reduces to show that $\wp'(z)^2-4\wp(z)^3-g_2(i) \wp(z)=0$ (bounded entire function) then find $g_2(i)$ from $1=\int_0^1 dz=\int_0^1 \frac{d\wp(z)}{\sqrt{4\wp(z)^3+g_2(i)\wp(z)}}=\int_C \frac{dx}{\sqrt{4x^3+g_2(i)x}}$ $=A g_2(i)\int_0^1 \frac{du}{\sqrt{u^3-1}}=2A g_2(i)B(1/2,3/4)=C g_2(i)\Gamma(1/4)^2$.

This way we'll obtain some recursive formulas for the derivatives of $\wp_i$ in term of $\wp_i,\wp_i'$ which will give for the values at $0,1/2$ a rational expression in term of $g_2(i)$.

  • The modular way

For $f$ a level $N$ modular form with integer coefficients then the coefficients of $F_z(X)=\prod_{\gamma\in SL_2(Z)/\Gamma(N)}(X-f|_k\gamma)$ are level $1$ weight $\le m$ modular forms thus in $\sum_{4a+6b=m} \Bbb{Q} E_4^aE_6^b$ so that $f(i)$ is algebraic over $E_4(i)$ (as $E_6(i)=0$), and the complex multiplication stuffs show that in $F_i(X)$ we can replace $SL_2(Z)/\Gamma(N)$ by an abelian group (the class group of $\Bbb{Z}[Ni]$) so that $f(i)$ is radical over $E_4(i)$ whose $\Gamma(1/4)$ expression is found from the elliptic way.

reuns
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