Cauchy-Ramanujan Formula $ \displaystyle \sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}} $

Cauchy and Ramanujan both gave the formula:

$$ \sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}} = (2\pi)^{4p+3}\sum_{k=0}^{2p+2} (-1)^{k+1} \frac{B_{2k}}{(2k)!}\frac{B_{4(p+1)-2k}}{(4(p+1)-2k)!} $$

If we lightly manipulate things, we can use a bit of umbral calculus to simplify the formula

\begin{eqnarray} -\frac{[4(p+1)]!}{(2\pi)^{4p+3}}\sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}} &=& \sum_{k=0}^{2p+2} (-1)^{k} \binom{4(p+1)}{2k}B_{2k}B_{4(p+1)-2k} \\ &=& \frac{1}{2}\left[ (B+iB)^{2(p+1)}+(B-iB)^{2(p+1)}\right] \end{eqnarray}

It's still a mess, but at least we have normalization. Prove LHS = RHS

Note that this sum is in fact $$2\sum_{m\ge 1} \frac{\coth m\pi}{m^{4p+3}}.$$

The sum term $$T_p(x) = 2\sum_{m\ge 1} \frac{\coth mx}{m^{4p+3}}$$ is harmonic and may be evaluated by inverting its Mellin transform.

Observe that $$\coth x = \frac{e^x+e^{-x}}{e^x-e^{-x}} = 1 + 2\frac{e^{-x}}{e^x-e^{-x}} = 1 + 2\frac{e^{-2x}}{1-e^{-2x}}.$$

This yields $$T_p(x) = 2\zeta(4p+3) + 4\sum_{m\ge 1} \frac{1}{m^{4p+3}} \frac{e^{-2mx}}{1-e^{-2mx}}.$$

We will now work with $$S(x) = \sum_{m\ge 1} \frac{1}{m^{4p+3}} \frac{e^{-2mx}}{1-e^{-2mx}}$$ and establish a functional equation for $S(x)$ that has $x=\pi$ as a fixed point.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = \frac{1}{k^{4p+3}}, \quad \mu_k = k \quad \text{and} \quad g(x) = \frac{\exp(-2x)}{1-\exp(-2x)}.$$

We need the Mellin transform $g^*(s)$ of $g(x)$ which we compute as follows: $$\int_0^\infty \frac{\exp(-2x)}{1-\exp(-2x)} x^{s-1} dx = \int_0^\infty \sum_{q\ge 1} \exp(-2qx)\; x^{s-1} dx \\ = \sum_{q\ge 1} \int_0^\infty \exp(-2qx)\; x^{s-1} dx = \Gamma(s) \sum_{q\ge 1} \frac{1}{2^s q^s} = \frac{1}{2^s} \Gamma(s) \zeta(s).$$

with fundamental strip $\langle 1,\infty \rangle.$

Hence the Mellin transform $Q(s)$ of $S(x)$ is given by $$ Q(s) = \frac{1}{2^s} \Gamma(s) \zeta(s) \zeta(s+4p+3) \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \zeta(s+4p+3)$$ where $\Re(s+4p+3) > 1$ or $\Re(s) > -4p-2$.

Intersecting the fundamental strip and the half-plane from the zeta function term we find that the Mellin inversion integral for an expansion about zero is $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ which we evaluate in the left half-plane $\Re(s)<3/2.$

The plain zeta function term cancels the poles of the gamma function term at even negative integers $2q\le -2$ and the compound zeta function term the poles at odd negative integers $2q+1\le -4p-5$. We are left with the contributions from $s=1$, $s=0$ and $s=-1$ which are

$$\begin{align} \mathrm{Res}(Q(s)/x^s; s=1) & = \frac{1}{2x}\zeta(4p+4) \\ \mathrm{Res}(Q(s)/x^s; s=0) & = -\frac{1}{2}\zeta(4p+3) \\ \mathrm{Res}(Q(s)/x^s; s=-1) & = \frac{1}{6} x\zeta(4p+2). \end{align}$$

The remaining contributions are $$\sum_{q=1}^{2p+1} \mathrm{Res}(Q(s)/x^s; s=-2q-1) \\ = \sum_{q=1}^{2p+1} 2^{2q+1} x^{2q+1} \frac{(-1)^{2q+1}}{(2q+1)!} \zeta(-2q-1) \zeta(4p+2-2q) \\ = \sum_{q=1}^{2p+1} 2^{2q+1} x^{2q+1} \frac{1}{(2q+1)!} \frac{B_{2q+2}}{2q+2} (-1)^{2p+1-q+1} \frac{B_{4p+2-2q}(2\pi)^{4p+2-2q}}{2(4p+2-2q)!} \\ = \sum_{q=1}^{2p+1} 2^{2q+1} x^{2q+1} \frac{B_{2q+2}}{(2q+2)!} (-1)^{-q} \frac{B_{4p+2-2q}(2\pi)^{4p+2-2q}}{2(4p+2-2q)!} \\ = \sum_{q=2}^{2p+2} 2^{2q-1} x^{2q-1} \frac{B_{2q}}{(2q)!} (-1)^{q+1} \frac{B_{4p+4-2q}(2\pi)^{4p+4-2q}}{2(4p+4-2q)!} \\ = 2^{4p+2} \sum_{q=2}^{2p+2} x^{2q-1} \frac{B_{2q}}{(2q)!} (-1)^{q+1} \frac{B_{4p+4-2q}\pi^{4p+4-2q}}{(4p+4-2q)!} \\ = 2^{4p+2} \sum_{q=2}^{2p+2} x^{2q-1} \pi^{4p+4-2q} \frac{B_{2q}}{(2q)!} (-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}$$

and the one from the pole of the compound zeta function term at $s=-4p-2$ which we'll do in a moment.

Now some algebra shows that setting $q=0$ and $q=1$ in the sum produces precisely the values that we obtained earlier for the poles at $s=1$ and $s=-1$ so we may extend the sum to start at zero, keeping only the residue from the pole at $s=0$ to get $$2^{4p+2} \sum_{q=0}^{2p+2} x^{2q-1} \pi^{4p+4-2q} \frac{B_{2q}}{(2q)!} (-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}.$$

We have thus established that $$S(x) = -\frac{1}{2}\zeta(4p+3) + \mathrm{Res}(Q(s)/x^s; s=-4p-2) \\ + 2^{4p+2} \sum_{q=0}^{2p+2} x^{2q-1} \pi^{4p+4-2q} \frac{B_{2q}}{(2q)!} (-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!} \\ + \frac{1}{2\pi i} \int_{-4p-4-i\infty}^{-4p-4+i\infty} Q(s)/x^s ds.$$

To treat the integral recall the duplication formula of the gamma function: $$\Gamma(s) = \frac{1}{\sqrt\pi} 2^{s-1} \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{s+1}{2}\right).$$

which yields for the integral $$\int_{-4p-4-i\infty}^{-4p-4+i\infty} \frac{1}{2\sqrt\pi} \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{s+1}{2}\right) \zeta(s) \zeta(s+4p+3)/x^s ds.$$

Furthermore observe the following variant of the functional equation of the Riemann zeta function: $$\Gamma\left(\frac{s}{2}\right)\zeta(s) = \pi^{s-1/2} \Gamma\left(\frac{1-s}{2}\right) \zeta(1-s)$$

which gives for the integral $$\int_{-4p-4-i\infty}^{-4p-4+i\infty} \frac{\pi^{s-1}}{2} \Gamma\left(\frac{s+1}{2}\right) \Gamma\left(\frac{1-s}{2}\right) \zeta(1-s) \zeta(s+4p+3)/x^s ds \\ = \int_{-4p-4-i\infty}^{-4p-4+i\infty} \frac{\pi^{s}}{2} \frac{1}{\sin(\pi/2(s+1))} \zeta(1-s) \zeta(s+4p+3)/x^s ds.$$

Now put $s=-4p-2-u$ to get $$\int_{2-i\infty}^{2+i\infty} \frac{\pi^{-4p-2-u}}{2} \frac{1}{\sin(\pi/2(-4p-2-u+1))} \\ \times \zeta(u+4p+3) \zeta(1-u)/x^{-4p-2-u} du \\ = \frac{x^{4p+2}}{\pi^{4p+2}} \int_{2-i\infty}^{2+i\infty} \frac{\pi^{-u}}{2} \frac{1}{\sin(\pi/2(-(u+1)))} \\ \times \zeta(u+4p+3) \zeta(1-u)/x^{-u} du \\ = - \frac{x^{4p+2}}{\pi^{4p+2}} \int_{2-i\infty}^{2+i\infty} \frac{\pi^{-u}}{2} \frac{1}{\sin(\pi/2(u+1))} \\ \times \zeta(u+4p+3) \zeta(1-u)/x^{-u} du.$$

We have established the functional equation $$S(x) = -\frac{1}{2}\zeta(4p+3) + \mathrm{Res}(Q(s)/x^s; s=-4p-2) \\ + 2^{4p+2} \sum_{q=0}^{2p+2} x^{2q-1} \pi^{4p+4-2q} \frac{B_{2q}}{(2q)!} (-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!} -\left(\frac{x}{\pi}\right)^{4p+2} S\left(\frac{\pi^2}{x}\right).$$

Setting $x=\pi$ we have $$S(\pi) = -\frac{1}{2}\zeta(4p+3) + \mathrm{Res}(Q(s)/\pi^s; s=-4p-2) \\ + 2^{4p+2} \pi^{4p+3} \sum_{q=0}^{2p+2} \frac{B_{2q}}{(2q)!} (-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!} -\left(\frac{\pi}{\pi}\right)^{4p+2} S\left(\pi\right)$$ and hence $$S(\pi) = -\frac{1}{4}\zeta(4p+3) + \frac{1}{2} \mathrm{Res}(Q(s)/\pi^s; s=-4p-2) \\ + 2^{4p+1} \pi^{4p+3} \sum_{q=0}^{2p+2} \frac{B_{2q}}{(2q)!} (-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}.$$

To conclude we treat the residue that we have defered until now. Recall the alternate form of $Q(s)/\pi^s,$ $$\frac{\pi^{s}}{2} \frac{1}{\sin(\pi/2(s+1))} \zeta(1-s) \zeta(s+4p+3)/\pi^s \\ = \frac{1}{2} \frac{1}{\sin(\pi/2(s+1))} \zeta(1-s) \zeta(s+4p+3).$$ It follows that the residue at $s=-4p-2$ is $$\frac{1}{2} \frac{1}{\sin(\pi/2(-4p-1))} \zeta(4p+3) = \frac{1}{2} \frac{1}{\sin(-\pi/2)} \zeta(4p+3) = -\frac{1}{2} \zeta(4p+3).$$

This finally yields for $S(\pi)$ $$S(\pi) = -\frac{1}{2}\zeta(4p+3) + 2^{4p+1} \pi^{4p+3} \sum_{q=0}^{2p+2} \frac{B_{2q}}{(2q)!} (-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}.$$

Computing $T_p(\pi)$ we thus obtain $$2\zeta(4p+3) -4\times\frac{1}{2}\zeta(4p+3) + 2^{4p+3} \pi^{4p+3} \sum_{q=0}^{2p+2} \frac{B_{2q}}{(2q)!} (-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!}$$ or $$(2\pi)^{4p+3} \sum_{q=0}^{2p+2} \frac{B_{2q}}{(2q)!} (-1)^{q+1} \frac{B_{4p+4-2q}}{(4p+4-2q)!},$$ QED.

The inspiration for this calculation is from the paper "Mellin Transform and its Applications" by Szpankowski.

There is also a way to obtain this formula using the partial fraction expansion of cotangent: $$ \frac{\pi}{2x} \coth(\pi x) - \frac{1}{2x^2} = \sum_{k=1}^{\infty} \frac{1}{k^2+x^2} $$ Let $m\geq 0$ be an integer, set $x=n$, divide by $n^{4m+2}$ and then sum from $n=1$ to infinity to get $$ \sum_{n=1}^{\infty} \frac{\pi}{2n^{4m+3}} \coth(\pi n) - \sum_{n=1}^{\infty} \frac{1}{2n^{4m+4}} = \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{n^{4m+2}(k^2+n^2)} $$ The sum on the right may be handled by switching the order of summation (justified by absolute convergence) and then rewriting things as follows: $$ \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m}} \frac{1}{n^2(k^2+n^2)} = \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m}k^2} \left(\frac{1}{n^2}-\frac{1}{k^2+n^2} \right) = \zeta(4m+2)\zeta(2) - \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m}k^2(k^2+n^2)} $$ Repeating the procedure results in $$ \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m+2}(k^2+n^2)} = \zeta(4m+2)\zeta(2) - \zeta(4m)\zeta(4) + \cdots + \zeta(2) \zeta(4m+2) - \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{k^{4m+2}(k^2+n^2)}$$ The final sum is the same as the first if we switch the dummy variables, and so we have that $$ 2\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m+2}(k^2+n^2)} = 2 \sum_{k=0}^{m-1} (-1)^k \zeta(4m+2-2k) \zeta(2k+2) +(-1)^m \zeta^2(2m+2) $$ Replacing this in the equality above and cleaning up a bit gives the result $$ \sum_{n=1}^{\infty} \frac{\coth(\pi n)}{n^{4m+3}} = \frac{\zeta(4m+4)}{\pi} + \frac{2}{\pi} \sum_{k=0}^{m-1} (-1)^k \zeta(4m+2-2k)\zeta(2k+2) + \frac{(-1)^m}{\pi} \zeta^2(2m+2) $$ This may be expressed in the form you wrote by using the well known formula for zeta at the even integers.