We took this idea from Simon Plouffe see here


Can anyone prove this identiy?

We found this identity via a sum calculator by varying Simon Plouffe identities


We can calculate the sum $$\sum_{n = 1}^{\infty}\frac{1}{n(e^{nx} + 1)}$$ as follows.

Let $q = e^{-x}$ and then we have \begin{align} F(q) &= \sum_{n = 1}^{\infty}\frac{1}{n(e^{nx} + 1)}\notag\\ &= \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 + q^{n})}\notag\\ &= \sum_{n = 1}^{\infty}\frac{1}{n}\sum_{k = 1}^{\infty}(-1)^{k - 1}q^{kn}\notag\\ &= \sum_{k = 1}^{\infty}(-1)^{k - 1}\sum_{n = 1}^{\infty}\frac{q^{kn}}{n}\notag\\ &= \sum_{k = 1}^{\infty}(-1)^{k}\log(1 - q^{k})\notag\\ &= \sum_{k \text{ even}}\log(1 - q^{k}) - \sum_{k \text{ odd}}\log(1 - q^{k})\notag\\ &= 2\sum_{k \text{ even}}\log(1 - q^{k}) - \sum_{k = 1}^{\infty}\log(1 - q^{k})\notag\\ &= 2\sum_{k = 1}^{\infty}\log(1 - q^{2k}) - \sum_{k = 1}^{\infty}\log(1 - q^{k})\notag\\ &= 2f(q^{2}) - f(q)\notag \end{align} where we have from the linked answer (the $f(q)$ of this answer is same as $-a(q)$ of the linked answer, also note that the linked answer proves another formula of Plouffe for $\pi$) $$f(q) = \sum_{k = 1}^{\infty}\log(1 - q^{k}) = \frac{\log k}{12} + \frac{\log k'}{3} + \frac{\log 2}{3} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{24K}\tag{1}$$ and $$f(q^{2}) = \frac{\log(kk')}{6} + \frac{\log 2}{6} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{12K}\tag{2}$$ and $$f(q^{4}) = \frac{1}{12}\log(k^{4}k') - \frac{\log 2}{6} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{6K}\tag{3}$$ From the above equations we get $$F(q) = \frac{\log k}{4} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{8K}\tag{4}$$ and $$F(q^{2}) = \frac{\log k}{2} - \frac{\log 2}{2} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{4K}\tag{5}$$ The sum in your question is $$8(F(q) - F(q^{2}))$$ where $q = e^{-\pi}$ and this is clearly equal to $$8\left(-\frac{\log k}{4} + \frac{\log 2}{2} - \frac{\pi K'}{8K}\right)$$ and we note that $k = 1/\sqrt{2}$ and $K' = K$ for $q = e^{-\pi}$. Hence the desired sum is $$8((5/8)\log 2 - (\pi/8)) = 5\log 2 - \pi$$

Most of the sums which involve the expressions of type $e^{n\pi}$ I try to express them as function of $q = e^{-\pi}$ and hope that the result series / product in $q$ can be related to the elliptic integrals $K$ and modulus $k$. If this is possible (like here) then the problem of dealing with infinite series/product is transformed into dealing with finite expressions consisting of $k, K, \pi$ and simple algebraic manipulations on them. This may not work always (or may not be convenient every time), but appears simpler to me when it works compared to general techniques for summing infinite series.

  • $\begingroup$ Thank you that was beautiful, nice proof @Paramanand Singh $\endgroup$ – user339807 May 24 '16 at 16:55
  • $\begingroup$ Good to see that you were able to solve it. (+1). $\endgroup$ – Marko Riedel May 24 '16 at 17:19
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    $\begingroup$ @Marko Riedel: Thanks a lot! I was checking your solution as to why it did not work to cancel all the terms and get the final sum. From your answer the evaluation of $S(\pi) - S(2\pi)$ crucially depends on $T(\pi) - T(2\pi)$. The function $T(x)$ is same as $-f(q)$ of my solution if $q = e^{-x}$. When we change $x$ to $\pi^{2}/x$ it replaces $q$ by another number $q_{1}$ such that the elliptic modulus $k$ and $k'$ are interchanged. To get $T(\pi) - T(2\pi)$ we need a relation between $T(x)$ and $T(2\pi^{2}/x)$ which is same as a relation between $f(q)$ and $f(q_{1}^{2})$ . Contd.. $\endgroup$ – Paramanand Singh May 25 '16 at 2:39
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    $\begingroup$ @MarkoRiedel: The formulas for $f(q)$ and $f(q_{1}^{2})$ (obtained by interchanging $k, k'$ in formula for $f(q^{2})$) show that $f(q) - f(q_{1}^{2})$ contains terms with $K'/K = x/\pi$ and also terms with $k, k'$ which are not possible to express algebraically in terms of $x$ and that's why it is difficult to get the value of $T(x) - T(2\pi^{2}/x)$ in terms of $x$. But for special value of $x = \pi$ we see that $k = k' = 1/\sqrt{2}$ and hence those terms containing $k, k'$ are also evaluated easily and in fact $T(\pi) - T(2\pi)$ has a simple closed form in terms of $\log 2$ and $\pi$. $\endgroup$ – Paramanand Singh May 25 '16 at 2:43
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    $\begingroup$ @MarkoRiedel: BTW it can be easily seen that there is an algebraic relation between $T(x)$ and $T(4\pi^{2}/x)$. From the formulas for $f(q)$ and $f(q_{1}^{4})$ we see that on subtraction only the terms with $k'/K = x/\pi$ remain and terms with $k, k'$ cancel nicely. You mention the same functional relation between $T(x)$ and $T(4\pi^{2}/x)$ in your answer. $\endgroup$ – Paramanand Singh May 25 '16 at 2:53

We present a closely related companion identity, as follows:

$$- \frac{1}{2} \log 2 + \frac{1}{8} \pi = \sum_{n\ge 1} \frac{1}{n} \frac{1}{\exp(\pi n) + 1} - \sum_{n\ge 1} \frac{1}{n} \frac{1}{\exp(2\pi n) + 1} \\ - 3 \sum_{n\ge 1} \frac{1}{n} \frac{1}{\exp(4\pi n) - 1} + 3 \sum_{n\ge 1} \frac{1}{n} \frac{1}{\exp(2\pi n) - 1}.$$

These sums may be evaluated using harmonic summation techniques.

Introduce the sum $$S(x) = \sum_{n\ge 1} \frac{1}{n} \frac{1}{e^{nx}+1}$$ with $x\ge 0.$

We start by trying to compute $S(\pi)-S(2\pi)$ as asked by the OP and obtain the quoted identity when that strategy does not yield the desired result.

The sum term is harmonic and may be evaluated by inverting its Mellin transform.

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 $$\lambda_k = \frac{1}{k}, \quad \mu_k = k \quad \text{and} \quad g(x) = \frac{1}{e^x+1}.$$

We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \frac{1}{e^{x}+1} x^{s-1} dx = \int_0^\infty \frac{e^{-x}}{1+e^{-x}} x^{s-1} dx \\ = \int_0^\infty \sum_{q\ge 1} (-1)^{q+1} e^{-q x} x^{s-1} dx = \sum_{q\ge 1} (-1)^{q+1} \int_0^\infty e^{-q x} x^{s-1} dx \\= \Gamma(s) \sum_{q\ge 1} \frac{(-1)^{q+1}}{q^s} = \left(1-2\frac{1}{2^s}\right)\Gamma(s) \zeta(s).$$

It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x)$ is given by

$$Q(s) = \left(1-\frac{1}{2^{s-1}}\right) \Gamma(s) \zeta(s) \zeta(s+1) \\ \text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{k} \frac{1}{k^s} = \zeta(s+1)$$ for $\Re(s) > 1.$

The Mellin inversion integral here is $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero.

The two zeta function terms cancel the poles of the gamma function term, the factor in front the pole of the first zeta function term and we are left with just

$$\begin{align} \mathrm{Res}(Q(s)/x^s; s=0) & = - \frac{1}{2} \log 2 + \frac{1}{2} \log\pi - \frac{1}{2} \log x \quad\text{and}\\ \mathrm{Res}(Q(s)/x^s; s=-1) & = \frac{x}{8} \end{align}$$

This shows that $$S(x) = - \frac{1}{2} \log 2 + \frac{1}{2} \log\pi -\frac{1}{2} \log x + \frac{x}{8} + \frac{1}{2\pi i} \int_{-3/2-i\infty}^{-3/2+i\infty} Q(s)/x^s ds.$$

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 $Q(s)$

$$\left(1-\frac{1}{2^{s-1}}\right) \frac{1}{\sqrt\pi} 2^{s-1} \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{s+1}{2}\right) \zeta(s) \zeta(s+1)$$

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 $Q(s)$ $$\left(1-\frac{1}{2^{s-1}}\right) \frac{1}{\sqrt\pi} 2^{s-1} \pi^{s-1/2} \Gamma\left(\frac{s+1}{2}\right) \Gamma\left(\frac{1-s}{2}\right) \zeta(1-s)\zeta(s+1) \\ = \left(1-\frac{1}{2^{s-1}}\right) \frac{1}{\sqrt\pi} 2^{s-1} \pi^{s-1/2} \frac{\pi}{\sin(\pi(s+1)/2)} \zeta(1-s)\zeta(s+1) \\ = \left(1-\frac{1}{2^{s-1}}\right) 2^{s-1} \frac{\pi^s}{\sin(\pi(s+1)/2)} \zeta(1-s)\zeta(s+1).$$

Now put $s=-u$ in the remainder integral to get

$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} \left(1-2^{u+1}\right) 2^{-u-1} \\ \times \frac{\pi^{-u}}{\sin(\pi(-u+1)/2)} \zeta(1+u)\zeta(1-u) x^u du \\ = \frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} \left(1-2^{u+1}\right) 2^{-u-1} \\ \times \frac{\pi^{u}}{\sin(\pi(-u+1)/2)} \zeta(1+u)\zeta(1-u) (x/\pi^2)^u du.$$

Now $$\sin(\pi(-u+1)/2) = \sin(\pi(-u-1)/2+\pi) \\ = - \sin(\pi(-u-1)/2) = \sin(\pi(u+1)/2)$$

and furthermore $$(1-2^{u+1}) 2^{-u-1} = \left(\frac{1}{2^{u+1}} - 1\right) = \left(\frac{4}{2^{u+1}} - 1\right) - \frac{3}{2^{u+1}}.$$

We have shown that $$S(x) = - \frac{1}{2} \log 2 + \frac{1}{2} \log\pi - \frac{1}{2}\log x + \frac{x}{8} - 2 S(2\pi^2/x) \\ - 3 \frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} \Gamma(u) \zeta(u) \zeta(u+1) (x/\pi^2/4)^u du$$

or alternatively

$$S(x) = - \frac{1}{2} \log 2 + \frac{1}{2} \log\pi - \frac{1}{2}\log x + \frac{x}{8} - 2 S(2\pi^2/x) - 3 T(4\pi^2/x)$$

where $$T(x) = \sum_{n\ge 1} \frac{1}{n} \frac{1}{\exp(nx)-1}$$

with functional equation $$T(x) = - \frac{1}{24} x - \frac{1}{2}\log 2 - \frac{1}{2}\log\pi + \frac{1}{2}\log x + \frac{\pi^2}{6x} + T(4\pi^2/x).$$

This implies that (evaluate at $\pi$ and $4\pi$) $$T(\pi)-T(4\pi) = -\frac{1}{2}\log 2 + \frac{1}{8}\pi.$$

We seek $S(\pi)-S(2\pi)$ and obtain

$$S(\pi)-S(2\pi) = -\frac{1}{2} \log \pi + \frac{1}{2} \log \pi + \frac{1}{2} \log 2 + \frac{\pi}{8} - \frac{\pi}{4} \\ - 2 (S(2\pi)-S(\pi)) - 3 (T(4\pi) -T(2\pi)).$$

This says that

$$S(\pi)-S(2\pi) = -\frac{1}{2}\log 2 + \frac{\pi}{8} + 3(T(4\pi)-T(2\pi)).$$

which is the claim.

Addendum. Working directly with the term

$$\left(\frac{1}{2^{u+1}} - 1\right)$$

we obtain the alternate equation

$$S(x) = - \frac{1}{2} \log 2 + \frac{1}{2} \log\pi - \frac{1}{2}\log x + \frac{x}{8} - 2 T(2\pi^2/x) + T(4\pi^2/x) \\ = \frac{1}{6} x + \log\pi - \log x - \frac{\pi^2}{6x} + T(x)- 2 T(2\pi^2/x)$$

which yields

$$S(\pi)-S(2\pi) = \log 2 - \frac{1}{4}\pi + 3(T(\pi)-T(2\pi)).$$

  • $\begingroup$ Wow this is a lengthy proof. I never knew this kind of question required such knowledge. Thank you @Marko Riedel. $\endgroup$ – user339807 May 23 '16 at 13:11

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