An series related to the eta Dedekind function

Prove that

    \[\sum_{n=1}^{\infty} \frac{1}{\sinh^2 n \pi} = \frac{1}{6} - \frac{1}{2\pi }\]

Solution

Let us consider the function

    \[f(z) = \frac{\cot \pi  z}{\sinh^2 \pi z}\]

and integrate it along a quadratic counterclockwise contour \Gamma_N with vertices \displaystyle \frac{N}{2} \left ( \pm 1 \pm i \right ) where N is a big odd natural number. Hence,

    \begin{align*} \oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z &= \int_{N/2}^{-N/2} f \left ( x + \frac{i N}{2} \right )\, \mathrm{d}x + i \int_{N/2}^{-N/2} f \left ( i y + \frac{N}{2} \right ) \, \mathrm{d}y \\ &\quad \quad \quad + \int_{-N/2}^{N/2} f \left ( x - \frac{i N}{2} \right ) \, \mathrm{d}x + i \int_{-N/2}^{N/2} f \left ( iy + \frac{N}{2} \right ) \, \mathrm{d}y \end{align*}

We note that \displaystyle \lim_{N \rightarrow +\infty} f \left (x \pm \frac{i N}{2} \right ) = \frac{\mp i}{\cosh^2 \pi x} and that \displaystyle \lim_{N \rightarrow +\infty} f \left (i y \pm \frac{N}{2} \right ) = 0.

It’s also easy to see that

    \[\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\cosh^2 \pi x } = \left [ \frac{\tanh \pi x}{\pi} \right ]_{-\infty}^{\infty} = \frac{2}{\pi}\]

Hence, as N \rightarrow +\infty we have that

    \[\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z = - \frac{4 i}{\pi}\]

By Residue theorem we have that

    \[\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z = 2 \pi i \sum \mathfrak{Res} \left ( f; z = k \right ) + \mathfrak{Res} \left ( f; z = ik \right )\]

It is straightforward to show that

    \[\begin{matrix} \mathfrak{Res} \left ( f;z=k \right ) = &\mathfrak{Res} \left ( f;z=ik \right ) = & \dfrac{1}{\pi \sinh^2 \pi k} \\\\ \mathfrak{Res} \left ( f;z=0 \right ) & = & -\dfrac{2}{3\pi} \end{matrix}\]

Hence,

    \[\oint \limits_{\Gamma_N} f(z)\, \mathrm{d}z = 8 i\sum_{k=1}^{\infty}\frac{1}{\sinh^2 n \pi}-\frac{4 i}{3}\]

in the limit N \rightarrow +\infty. The result follows.

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Fourier transformation

Let \xi \in \mathbb{R} and \alpha \in (0, 1). Show that

    \[\int_{- \infty}^{\infty} {e}^{-2 \pi i x \xi} \frac{ \sin \pi \alpha }{ \cosh \pi x + \cos \pi \alpha } \mathrm{d} x = \frac{2 \sinh 2 \pi \alpha \xi }{ \sinh 2 \pi \xi }\]

Solution

We note that

    \begin{align*} \frac{\sin \pi \alpha}{\cosh \pi x + \cos \pi \alpha} &= \frac{1}{1+e^{-\pi(x+i\alpha)}}-\frac{1}{1+e^{-\pi(x - i\alpha)}}\\ &= 2\sum\limits_{n=1}^{\infty} (-1)^{n}e^{-n\pi x}\sin n\pi\alpha \end{align*}

Thus,

    \begin{align*} \int_{-\infty}^{\infty} \frac{e^{-2\pi i \xi x}\sin \pi \alpha}{\cosh \pi x + \cos \pi \alpha}\,\mathrm{d}x &= \int_{0}^{\infty} \frac{2\sin \pi \alpha \cos 2\pi\xi x}{\cosh \pi x + \cos \pi \alpha}\,\mathrm{d}x\\ &= 4\int_0^{\infty}\sum\limits_{n=1}^{\infty}(-1)^n e^{-n\pi x}\cos 2\pi\xi x \sin n\pi \alpha\,\mathrm{d}x\\ &= \frac{4}{\pi}\sum\limits_{n=1}^{\infty}(-1)^n \frac{n\sin n\pi\alpha}{n^2+4\xi^2}\\ &= \frac{2}{\pi}\sum\limits_{n=1}^{\infty} (-1)^n\left(\frac{1}{n+2i\xi}+\frac{1}{n-2i\xi}\right)\sin n\pi \alpha\\ &= \frac{2}{\pi}\sum\limits_{n = -\infty}^{\infty} \frac{(-1)^n\sin n\pi\alpha}{n+2i\xi}\\ &= \frac{2 \sin 2\pi i \alpha\xi}{\sin 2\pi i\xi} \qquad \textrm{ (by residue theorem) }\\ &= \frac{2 \sinh 2\pi \alpha\xi}{\sinh 2\pi \xi} \end{align*}

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Value of a numeric expression

In a \mathrm{AB} \Gamma triangle it holds that

(1)   \begin{equation*} \cos^2 2\mathrm{A} + \cos^2 2 \mathrm{B} + \cos^2 2 \Gamma =1 \end{equation*}

Prove that

    \[\left ( \mathrm{A} - \frac{\pi}{4} \right ) \left ( \mathrm{A} - \frac{3 \pi}{4} \right ) \left ( \mathrm{B} - \frac{\pi}{4} \right ) \left ( \mathrm{B} - \frac{3 \pi}{4} \right ) \left ( \Gamma - \frac{\pi}{4} \right ) \left( \Gamma - \frac{3 \pi}{4} \right ) = 0\]

Triangle relation

Prove that in any triangle \mathrm{AB} \Gamma it holds that

    \[\left ( \frac{\beta}{\gamma} + \frac{\gamma}{\beta} \right ) \cos \mathrm{A} + \left ( \frac{\gamma}{\alpha} + \frac{\alpha}{\gamma} \right ) \cos \mathrm{B} + \left ( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \right ) \cos \Gamma = 3\]

Convergence of series

Let \mathrm{gpf} denote the greatest prime factor of n. For example \mathrm{gpf}(17) = 17 , \mathrm{gpf} (18) =3. Define \mathrm{gpf}(1)=1. Examine if the sum

    \[\mathcal{S} = \sum_{n=1}^{\infty} \frac{1}{n \; \mathrm{gpf}(n)}\]

converges.

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