Proof of $\frac{1}{e^{\pi}+1}+\frac{3}{e^{3\pi}+1}+\frac{5}{e^{5\pi}+1}+\ldots=\frac{1}{24}$

Solution 1:

We will use the Mellin transform technique. Recalling the Mellin transform and its inverse

$$ F(s) =\int_0^{\infty} x^{s-1} f(x)dx, \quad\quad f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} F(s)\, ds. $$

Now, let's consider the function

$$ f(x)= \frac{x}{e^{\pi x}+1}. $$

Taking the Mellin transform of $f(x)$, we get

$$ F(s)={\pi }^{-s-1}\Gamma \left( s+1 \right) \left(1- {2}^{-s} \right) \zeta \left( s+1 \right),$$

where $\zeta(s)$ is the zeta function . Representing the function in terms of the inverse Mellin Transform, we have

$$ \frac{x}{e^{\pi x}+1}=\frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma \left( s+1 \right) \left( 1-{2}^{-s} \right) \zeta \left( s+1 \right) x^{-s}ds. $$

Substituting $x=2n+1$ and summing yields

$$\sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{2\pi i}\int_{C}{\pi}^{-s-1}\Gamma \left( s+1 \right)\left(1-{2}^{-s} \right) \zeta\left( s+1 \right) \sum_{n=0}^{\infty}(2n+1)^{-s}ds$$

$$ = \frac{1}{2\pi i}\int_{C}{\pi }^{-s-1}\Gamma \left( s+1 \right) \left(1-{2}^{-s} \right)^2\zeta\left( s+1 \right) \zeta(s)ds.$$

Now, the only contribution of the poles comes from the simple pole $s=1$ of $\zeta(s)$ and the residue equals to $\frac{1}{24}$. So, the sum is given by

$$ \sum_{n=0}^{\infty}\frac{2n+1}{e^{\pi (2n+1)}+1}=\frac{1}{24} $$

Notes: 1)

$$ \sum_{n=0}^{\infty}(2n+1)^{-s}= \left(1- {2}^{-s} \right) \zeta \left( s \right). $$

2) The residue of the simple pole $s=1$, which is the pole of the zeta function, can be calculated as

$$ r = \lim_{s=1}(s-1)({\pi }^{-s-1}\Gamma \left( s+1 \right) \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right) \zeta(s))$$

$$ = \lim_{s\to 1}(s-1)\zeta(s)\lim_{s\to 1} {\pi }^{-s-1}\Gamma \left( s+1 \right) \left({2}^{-s}-1 \right)^2\zeta\left( s+1 \right) = \frac{1}{24}. $$

For calculating the above limit, we used the facts

$$ \lim_{s\to 1}(s-1)\zeta(s)=1, \quad \zeta(2)=\frac{\pi^2}{6}. $$

3) Here is the technique for computing the Mellin transform of $f(x)$.

Solution 2:

Let's start with $$ \sum_{n=0}^\infty x^n=\frac1{1-x}\tag{1} $$ Differentiating $(1)$ and multiplying by $x$, we get $$ \sum_{n=0}^\infty nx^n=\frac{x}{(1-x)^2}\tag{2} $$ Taking the odd part of $(2)$ yields $$ \sum_{n=0}^\infty(2n+1)x^{2n+1}=\frac{x(1+x^2)}{(1-x^2)^2}\tag{3} $$ Using $(3)$, we get $$ \begin{align} \sum_{n=0}^\infty\frac{2n+1}{e^{(2n+1)\pi}+1} &=\sum_{n=0}^\infty\sum_{k=1}^\infty(-1)^{k-1}(2n+1)e^{-(2n+1)k\pi}\\ &=\sum_{k=1}^\infty\sum_{n=0}^\infty(-1)^{k-1}(2n+1)e^{-(2n+1)k\pi}\\ &=\sum_{k=1}^\infty(-1)^{k-1}\frac{e^{-k\pi}\left(1+e^{-2k\pi}\right)}{\left(1-e^{-2k\pi}\right)^2}\\ &=\frac12\sum_{k=1}^\infty(-1)^{k-1}\frac{\cosh(k\pi)}{\sinh^2(k\pi)}\tag{4} \end{align} $$

We can use the formula proven in this answer $$ \pi\cot(\pi z)=\sum_{k\in\mathbb{Z}}\frac1{z+k}\tag{5} $$ to derive $$ \begin{align} \pi\csc(\pi z) &=\pi\cot(\pi z/2)-\pi\cot(\pi z)\\[9pt] &=\sum_{k\in\mathbb{Z}}\frac2{z+2k}-\sum_{k\in\mathbb{Z}}\frac1{z+k}\\ &=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{z+k}\\ \pi^2\frac{\cos(\pi z)}{\sin^2(\pi z)} &=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{(z+k)^2}\tag{6} \end{align} $$ then rotate coordinates with $z\mapsto iz$ to get $$ \pi^2\frac{\cosh(\pi z)}{\sinh^2(\pi z)}=\sum_{j\in\mathbb{Z}}\frac{(-1)^j}{(z+ij)^2}\tag{7} $$

Now plug $(7)$ into $(4)$: $$ \begin{align} \sum_{n=0}^\infty\frac{2n+1}{e^{(2n+1)\pi}+1} &=\frac1{2\pi^2}\sum_{k=1}^\infty\sum_{j\in\mathbb{Z}}(-1)^{j+k-1}\frac1{(k+ij)^2} \\ &=\frac1{2\pi^2}\sum_{k=1}^\infty(-1)^{k-1}\frac1{k^2}\\ &+\frac1{2\pi^2}\sum_{k=1}^\infty\sum_{j=1}^\infty(-1)^{j+k-1}\left(\frac1{(k+ij)^2}+\frac1{(k-ij)^2}\right)\\ &=\frac1{2\pi^2}\frac{\pi^2}{12}\\ &+\frac1{2\pi^2}\sum_{k=1}^\infty\sum_{j=1}^\infty(-1)^{j+k-1}\frac{2(k^2-j^2)}{(k^2+j^2)^2}\\ &=\frac1{24}+0\tag{8} \end{align} $$