I would like to prove that $\displaystyle\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{n}{e^{n\pi}+1}=\frac1{24}$.

I found a solution by myself 10 hours after I posted it, here it is:

$$f(x)=\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{nx^n}{1+x^n},\quad\quad g(x)=\displaystyle\sum_{n=1}^{\infty}\frac{nx^n}{1-x^n},$$

then I must prove that $f(e^{-\pi})=\frac1{24}$. It was not hard to find the relation between $f(x)$ and $g(x)$, namely $f(x)=g(x)-4g(x^2)+4g(x^4)$.

Note that $g(x)$ is a Lambert series, so by expanding the Taylor series for the denominators and reversing the two sums, I get


where $\sigma$ is the divisor function $\sigma(n)=\sum_{d\mid n}d$.

I then define for complex $\tau$ the function $$G_2(\tau)=\frac{\pi^2}3\Bigl(1-24\sum_{n=1}^{\infty}\sigma(n)e^{2\pi in\tau}\Bigr)$$ so that $$f(e^{-\pi})=g(e^{-\pi})-4g(e^{-2\pi})+4g(e^{-4\pi})=\frac1{24}+\frac{-G_2(\frac i2)+4G_2(i)-4G_2(2i)}{8\pi^2}.$$

But it is proven in Apostol "Modular forms and Dirichlet Series", page 69-71 that $G_2\bigl(-\frac1{\tau}\bigr)=\tau^2G_2(\tau)-2\pi i\tau$, which gives $\begin{cases}G_2(i)=-G_2(i)+2\pi\\ G_2(\frac i2)=-4G_2(2i)+4\pi\end{cases}\quad$. This is exactly was needed to get the desired result.

Hitoshigoto oshimai !

I find that sum fascinating. $e,\pi$ all together to finally get a rational. This is why mathematics is beautiful!

Thanks to everyone who contributed.

Paramanand Singh
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    See the article on the [odd divisor function](http://mathworld.wolfram.com/OddDivisorFunction.html) on MathWorld. According to this article, your function $f(x)$ is $1/24$ multiplied by the sum of the fourth powers of two [Jacobi theta functions](http://mathworld.wolfram.com/JacobiThetaFunctions.html). – Jim Belk May 12 '13 at 07:09
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    @danodare: If you found an answer by yourself, you are welcome to [answer your own question](http://meta.math.stackexchange.com/questions/1401/what-to-do-if-you-figure-out-the-answer-to-your-own-question). – JavaMan May 12 '13 at 22:30
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    Whence did this question come? – robjohn Aug 21 '13 at 21:34
  • That is a great solution! Wow. – Akiva Weinberger Aug 31 '14 at 17:39
  • **are you sure** ?than the sum is equal to $$\frac{1}{24}$$ when x=1 look –  Dec 29 '15 at 19:44

7 Answers7


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 $f(x)$.

Mhenni Benghorbal
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  • @Potato: Thank you very much for the comment. I really appreciate it. – Mhenni Benghorbal May 12 '13 at 07:30
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    I have seen you use similar techniques many times before. Where did you learn them? You are quite good with integrals. – Potato May 12 '13 at 07:32
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    @Potato: These techniques for summing infinite series are known in the literature as applications of integral transforms such as Mellin and Fourier transform. I have been learning them over the years. They are very effective techniques. See [here](http://math.stackexchange.com/questions/208317/show-sum-n-0-infty-frac1a2n2-frac1a-pi-coth-a-pi2a2) for a Fourier transform technique for summing a series. – Mhenni Benghorbal May 12 '13 at 07:48
  • Are the singularities at the negative integers removable since $\zeta(s+1)$ has zeros at the odd negative integers (excluding $-1$) and $\zeta(s)$ has zeroes at the even negative integers? And how do you calculate that limit to find the residue? It seems daunting. – Random Variable May 12 '13 at 14:07
  • @RandomVariable: I added more material. – Mhenni Benghorbal May 12 '13 at 22:05
  • @RonGordon: Thank you very much for the comment. I really appreciate it. – Mhenni Benghorbal May 12 '13 at 23:30
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    Is it just me or is there an error in the signs of $$\sum_{k\ge 0} \frac{1}{(2k+1)^s} = \zeta(s) (1 - 2^{-s})$$ This error is repeated twice and canceled by the square, so that you still get the right answer. – Marko Riedel May 13 '13 at 00:06
  • This same issue also surfaces in your note. – Marko Riedel May 13 '13 at 00:15
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    This article is very nice work and if it should turn out that it does contain an error it definitely should be fixed so it can serve as a future reference to the method. – Marko Riedel May 13 '13 at 01:13
  • @MarkoRiedel: Thanks for the comment. You are right. I'll be working on it. – Mhenni Benghorbal May 13 '13 at 03:40
  • @MarkoRiedel: It is corrected. – Mhenni Benghorbal May 13 '13 at 03:51
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    @MhenniBenghorbal I added a missing piece to your proof. Enjoy! We now have a complete example computation that we may link to in the future. – Marko Riedel May 13 '13 at 23:36
  • @MarkoRiedel: Good job. Thank you for doing this. – Mhenni Benghorbal May 13 '13 at 23:52
  • @Downvoter: What's the downvote for? – Mhenni Benghorbal Dec 21 '13 at 23:29
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    +1. Somehow, but I don't know why, the identity $\large{{1 \over {\rm e}^{n\pi} + 1} + {1 \over {\rm e}^{-n\pi} + 1} = 1}$ is involved. It is amusing that the result $\large{-\,{1 \over 2}\,\left(-\,{1 \over 12}\right)}$ is related to $\large{\zeta\left(-1\right) = -\,{1 \over 12}}$. It reminds me the 'bizarre' discussion about the sum of the natural numbers. Nice job. – Felix Marin Jul 11 '14 at 08:08
  • @FelixMarin: Thank you for your comment. I realy appreciate it. – Mhenni Benghorbal Jul 12 '14 at 05:47
  • And I've thought that Mellin's transform is just an ugly unuseful brother of that of Laplace and Fourier - This has completely blew my mind !!!! – Machinato Jul 21 '16 at 20:10
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    Two questions about computation of the contour integral : 1) gamma function itself has a residue at s=-1 why its residue s not counted? 2) To work with residues the integral must to be closed somewhere, how is it shown integral vanishes near s=-infty? – Machinato Aug 04 '16 at 09:07

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

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The calculation of the Mellin transform of $f(x)$ is not present in the above answer, so I will show it here.

$$\mathfrak{M}\left(\frac{1}{e^{\pi x}+1};s \right) = \int_0^\infty \frac{1}{e^{\pi x}+1} x^{s-1} dx = \int_0^\infty \frac{1}{e^{\pi x}} \frac{1}{1+e^{-\pi x}} x^{s-1} dx \\= \int_0^\infty \frac{1}{e^{\pi x}} \sum_{q\ge 0} (-1)^q e^{-\pi q x} x^{s-1} dx = \int_0^\infty \sum_{q\ge 0} (-1)^q e^{-\pi (q+1) x} x^{s-1} dx \\ = \Gamma(s) \sum_{q\ge 0} (-1)^q \frac{1}{\pi^s (q+1)^s} = \frac{1}{\pi^s} \Gamma(s) (\zeta(s) - 2 \times 2^{-s} \times \zeta(s)) = \frac{1}{\pi^s} \Gamma(s) (1 - 2\times 2^{-s}) \zeta(s).$$ It now follows from the definition of the Mellin transform that $$\mathfrak{M}\left(\frac{x}{e^{\pi x}+1};s \right) = \frac{1}{\pi^{s+1}} \Gamma(s+1) (1 - 2^{-s}) \zeta(s+1).$$

Marko Riedel
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    There is more material [here](http://math.stackexchange.com/questions/331404/how-to-prove-this-identity-pi-sum-limits-k-infty-infty-left-frac-sin) and [here](http://math.stackexchange.com/questions/297900/compute-sum-k-1-inftye-pi-k2-left-pi-k2-frac14-right/) and [here](http://math.stackexchange.com/questions/339966/closed-form-sum-of-sum-infty-n-1-frac13n-1). – Marko Riedel May 12 '13 at 23:31
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    One more thing -- another interesting example can be found [here](http://math.stackexchange.com/questions/331116/missing-term-in-series-expansion). – Marko Riedel May 12 '13 at 23:39
  • I apologize -- changed it. – Marko Riedel May 12 '13 at 23:53
  • doesn't "above answer" not always the above for everyone? – RE60K Dec 28 '14 at 05:52
  • are you sure that the result 1/24 is correct look the post[link](http://math.stackexchange.com/questions/1592565/proof-of-frac1e-pi1-frac3e3-pi1-frac5e5-pi1-ldots) –  Dec 29 '15 at 20:24

We can link the functions $g(x),g(x^{2}),g(x^{4})$ with elliptic integrals $K,E$ and modulus $k$. Also we switch to variable $q$ instead of $x$. The value of $q$ here is $q = e^{-\pi}$ so that $k = 1/\sqrt{2}$.

From this post we can see that \begin{align} g(q) &= \frac{1 - P(q)}{24} = \frac{1}{24}\left\{1 - \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6E}{K} + k^{2} - 5\right)\right\}\tag{1}\\ g(q^{2}) &= \frac{1 - P(q^{2})}{24} = \frac{1}{24}\left\{1 - \left(\frac{2K}{\pi}\right)^{2}\left(\frac{3E}{K} + k^{2} - 2\right)\right\}\tag{2}\\ g(q^{4}) &= \frac{1 - P(q^{4})}{24} = \frac{1}{24}\left\{1 - \left(\frac{2K}{\pi}\right)^{2}\left(\frac{3E}{2K} + \frac{k^{2} - 2}{4}\right)\right\}\tag{3} \end{align} Using the above equations and the fact that $k = 1/\sqrt{2}$ we get $$f(q) = g(q) - 4g(q^{2}) + 4g(q^{4}) = \frac{1}{24}$$ Note: The formula for $P(q^{4})$ is not given in the linked post, but it can be obtained via the same technique as mentioned in the post, namely finding an expression for $\eta(q^{4})$ in terms of $K, k$ and then applying logarithmic differentiation.

Based on suggestion from robjohn (via comment) I add some details on the theory of elliptic integrals and theta functions so that the identities $(1), (2), (3)$ get some context.

Let us then assume that $k$ is a real number (which is called modulus) with $0 < k < 1$ and let $k' = \sqrt{1 - k^{2}}$ then we define elliptic integrals $$K(k) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - k^{2}\sin^{2}x}}, E(k) = \int_{0}^{\pi/2}\sqrt{1 - k^{2}\sin^{2}x}\,dx\tag{4}$$ The integrals $K(k), E(k), K(k'), E(k')$ are usually denoted by $K, E, K', E'$ when the argument $k, k'$ is available from context. These integrals satisfy the Legendre's Identity $$KE' + K'E - KK' = \frac{\pi}{2}\tag{5}$$ We have the derivatives $$\frac{dK}{dk} = \frac{E - k'^{2}K}{kk'^{2}},\,\,\frac{dE}{dk} = \frac{E - K}{k}\tag{6}$$ It is almost a piece of magic (created by mathemagician Jacobi) that it is possible to obtain the values of $k, k'$ from the values of $K, K'$ using another variable called nome defined by $$q = e^{-\pi K'/K}\tag{7}$$ The desired relations are given by theta functions $\vartheta_{2}(q), \vartheta_{3}(q), \vartheta_{4}(q)$ as follows $$k = \frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)},\, k' = \frac{\vartheta_{4}^{2}(q)}{\vartheta_{3}^{2}(q)},\, \frac{2K}{\pi} = \vartheta_{3}^{2}(q)\tag{8}$$ where \begin{align} \vartheta_{2}(q) &= \sum_{n = -\infty}^{\infty}q^{(n + 1/2)^{2}} = 2(q^{1/4} + q^{9/4} + q^{25/4} + \cdots )\notag\\ &= 2q^{1/4}\prod_{n = 1}^{\infty}(1 - q^{2n})(1 + q^{2n})^{2}\tag{9a}\\ \vartheta_{3}(q) &= \sum_{n = -\infty}^{\infty}q^{n^{2}} = 1 + 2q + 2q^{4} + 2q^{9} + \cdots\notag\\ &= \prod_{n = 1}^{\infty}(1 - q^{2n})(1 + q^{2n - 1})^{2}\tag{9b}\\ \vartheta_{4}(q) &= \sum_{n = -\infty}^{\infty}(-1)^{n}q^{n^{2}} = 1 - 2q + 2q^{4} - 2q^{9} + \cdots\notag\\ &= \prod_{n = 1}^{\infty}(1 - q^{2n})(1 - q^{2n - 1})^{2}\tag{9c} \end{align} Using the relations $(5), (6), (7)$ it is possible to show that $$\frac{dq}{dk} = \frac{\pi^{2}q}{2kk'^{2}K^{2}}\tag{10}$$ We next introduce the function $\eta(q)$ via $$\eta(q) = q^{1/24}\prod_{n = 1}^{\infty}(1 - q^{n})\tag{11}$$ Logarithmic differentiation of the above relation gives rise to the function $P(q)$ i.e. $$P(q) = 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{n}} = 24q\frac{d}{dq}\log\eta(q)\tag{12}$$ Using equations $(8), (9a), (9b), (9c)$ (especially the product representations of theta functions) it is possible to express $\eta(q)$ in terms of $k, k', K$ and we have \begin{align} \eta(q) &= 2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{13a}\\ \eta(q^{2}) &= 2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{13b}\\ \eta(q^{4}) &= 2^{-2/3}\sqrt{\frac{2K}{\pi}}k^{1/3}k'^{1/12}\tag{13c} \end{align} I obtained the expression of $\eta(q^{2})$ directly using the product representation of theta functions as $$2\{\eta(q^{2})\}^{3} = \vartheta_{2}(q)\vartheta_{3}(q)\vartheta_{4}(q)$$ and then equation $(13b)$ follows by using equation $(8)$. The expressions for $\eta(q), \eta(q^{4})$ were obtained by applying Landen's Transformation on these expressions (thus if we replace $k$ by $2\sqrt{k}/(1 + k)$ and $K$ by $(1 + k)K$ the variable $q^{2}$ is replaced by $q$ and if we replace $k$ by $(1 - k')/(1 + k')$ and $K$ by $(1 + k')K/2$ the variable $q$ is replaced by $q^{2}$). The identities $(1), (2)$ and $(3)$ of my answer are obtained by logarithmic differentiation of the above expressions for $\eta(q), \eta(q^{2}), \eta(q^{4})$.

Paramanand Singh
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  • (+1) However, it would be nice to bring the information from the linked post here. – robjohn Mar 13 '16 at 11:05
  • @robjohn: I will post some details here but won't be possible to give entire information with full proof. – Paramanand Singh Mar 13 '16 at 14:11
  • @robjohn: I have added some details regarding the identities linking elliptic integrals with theta functions. The eta function is also introduced in order to get the function $P(q)$. I have given links to my blog posts for more details and proofs. I hope this is useful. – Paramanand Singh Mar 14 '16 at 05:15
  • Epic answer, which deserves a lot more upvotes (+1k) – tired May 16 '17 at 19:49
  • @tired: thanks for encouraging words. Since I am not so much familiar with the powerful techniques of complex analysis, I try to deal with such identities via Ramanujan's techniques. – Paramanand Singh May 16 '17 at 22:57
  • @ParamanandSingh to be honest, i admire your skills in the realm of theta functions and elliptic integrals/functions and hope to learn it better in the next year or so. Concerning the power of contour of integration you might be interested in my recent answer to this old question here, which can, i suppose, also be tackeled by the tools you are familiar with: https://math.stackexchange.com/questions/272909/prove-that-sum-k-1-infty-large-frack-texte2-pi-k-1-frac124/272955?noredirect=1#comment4699890_272955 – tired May 17 '17 at 21:43
  • @tired: I gave an answer to that question based on Ramanujan's ideas. You may have a look. – Paramanand Singh May 17 '17 at 21:57

Actually the above is not quite complete, the missing piece is the proof that we can drop the contribution from the pole at $s=-1,$ which is $x/24.$ To verify this we have to show that $$\int_{-i\infty}^{i\infty} \frac{1}{\pi^{s+1}} \Gamma(s+1) (1-2^{-s})^2 \zeta(s+1)\zeta(s) ds = 0.$$ Now from the functional equation of the Riemann Zeta function we see that this integral is equal to $$-\int_{-i\infty}^{i\infty} \frac{\zeta(-s)}{\sin(1/2s\pi)} (2^s-1) (1-2^{-s}) \zeta(s) ds$$ Actually doing the accounting we find that the kernel $$ g(s) = \frac{\zeta(-s)}{\sin(1/2s\pi)} (2^s-1) (1-2^{-s}) \zeta(s) $$ of this integral has the property that $g(s) = - g(-s)$ on the imaginary axis, so the integral is zero.

To see this consider what effect negation has on the individual terms. $$\zeta(-s)\zeta(s) \to \zeta(s)\zeta(-s),$$ $$(2^s-1)(1-2^{-s}) \to (2^{-s}-1)(1-2^s) = (2^s-1)(1-2^{-s}),$$ $$\sin(1/2 s\pi) \to \sin(1/2 (-s)\pi) = -\sin(1/2 s\pi).$$ The first two terms are even and the last one is odd, QED.

Note that we have taken advantage of the fact that $x=1$ ... for other values of $x$ this trick will not go through. Also relevant is that negation (rotation by 180 degrees about the origin) takes the imaginary axis to itself (this is not the case when we are integrating along some other line parallel to the imaginary axis in the right half plane).

Marko Riedel
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  • So the closed contour is a tall rectangle? – Random Variable May 15 '13 at 20:16
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    Yes, exactly, the two vertical sides are the lines $\Re(s)=3/2$ and $\Re(s) = 0.$ – Marko Riedel May 15 '13 at 22:17
  • Let me rephrase my questions. 1) Can the right side of the rectangle be any vertical line to the right of the line $\Re(s) =1$? 2) There is a simple pole at the origin (albeit with residue 0). Does the contour technically need to be indented? 3) Does the integral go to zero along the top and bottom of the rectangle since $|\Gamma(s)|$ decays quickly as $\Im(s)$ increases? – Random Variable May 16 '13 at 00:10
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    1) Owing to the convergence of $\sum_{k\ge 0} \frac{1}{(2k+1)^s}$ in the half plane $\Re(s)>1$ and the fact that there are no additional poles the Mellin inversion integral can indeed be along any vertical line in that half plane. 2) The simple pole is cancelled by the $(1-2^{-s})$ term, no indentation necessary. 3) This is correct, the decrease is exponential. Finally, let me refer you to one of the experts on this one -- the paper "Mellin Transform and Its Applications" by Szpankowski on academia.edu contains many examples and is highly readable. (For some reason SE won't let me add a link.) – Marko Riedel May 16 '13 at 01:21
  • Thanks. If you have time, I started a related thread earlier today. http://math.stackexchange.com/questions/392706/evaluating-sum-n-1-infty-fracne2-pi-n-1-using-the-inverse-melli?lq=1 – Random Variable May 16 '13 at 01:37

$$S=\sum_{n=1}^{\infty }\frac{2n-1}{1+e^{\pi (2n-1)}},$$ first we know that $$\frac{1}{1+e^x}=\sum_{k=1}^{\infty }(-1)^{k-1}e^{-kx}$$ therefore $$S=\sum_{n=1}^{\infty }(2n-1)\sum_{k=1}^{\infty }(-1)^{k-1}e^{-k(2n-1)\pi }=\sum_{k=1}^{\infty }(-1)^{k-1}\sum_{n=1}^{\infty }(2n-1)e^{-k(2n-1)\pi }$$

but we know $$ \sum_{n=1}^{\infty }e^{-(2n-1)\pi x}=\frac{1}{2sinh(\pi x)},$$ for all $x>0$ but $$\frac{1}{2sinh(\pi x)}=(\frac{i}{2\pi })(\frac{\pi }{sin(i\pi x)})$$ and $$\int_{0}^{\infty }\frac{t^{-x}}{1+t}dt=\frac{\pi }{sin(\pi x)}$$ therefore $$\int_{0}^{\infty }\frac{t^{-x}}{1+t}dt=\int_{0}^{1}\frac{t^{-x}}{1+t}dt+\int_{0}^{1}\frac{t^{x-1}}{1+t}dt=\int_{0}^{1}\frac{t^{-x}+t^{x-1}}{1+t}dt\\ \\ \frac{\pi }{sin(\pi x)}=\int_{0}^{1}\frac{t^{-x}+t^{x-1}}{1+t}dt=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{1}{k-x}+\frac{1}{k-1+x})\\ \therefore from\ \frac{\pi }{sin(i\pi x)}=\int_{0}^{\infty }\frac{t^{-ix}+t^{ix-1}}{1+t}dt=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{1}{k-ix}+\frac{1}{k-1+ix})\\ \\ =\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{k+ix}{x^2+k^2}+\frac{k-1-ix}{(k-1^2)+x^2})=\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{k}{k^2+x^2}+\frac{k-1}{x^2+(k-1)^2})\\ \\ +i\sum_{k=1}^{\infty }(-1)^{k-1}(\frac{x}{x^2+k^2}-\frac{x}{x^2+(k-1)^2})=i(\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}-\sum_{k=1}^{\infty }(-1)^{k}\frac{x}{x^2+k^2})\\ \\ \\ =i[\frac{-1}{x}+\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}+\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}]=i[\frac{-1}{x}+2\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{x^2+k^2}]\\ \\\therefore \frac{\pi }{sinh(\pi x)}=\frac{1}{x}-2\sum_{k=1}^{\infty }(-1)^{k-1}\frac{x}{k^2+x^2}=\int_{0}^{1 }\frac{t^{-ix}+t^{ix-1}}{1+t}dt$$

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Ahmed Hejazi
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    Please do not write everything in one big formula. Break it up, and write text outside of formulas. – quid Aug 18 '19 at 13:54

Let's integrate the function $$f(z) = \frac{z e^{iz}}{\cosh (z) + \cos(z)} \, $$ around the contour $[-\sqrt{2} \pi N, \sqrt{2} \pi N] \cup \sqrt{2} \pi Ne^{i[0, \pi]}$, where $N$ is a positive integer.

The integral vanishes on the semicircle as $N \to \infty$ because $|f(z)|$ decays exponentially fast to zero along the entire semicircle.

We therefore have

$\begin{align} \int_{-\infty}^{\infty}f(x) \, \mathrm dx &= 2 \pi i \left(\sum_{n=0}^{\infty}\operatorname{Res} \left[f(z), \frac{(2n+1) \pi (1+i)}{2} \right] + \sum_{n=0}^{\infty}\operatorname{Res} \left[f(z), \frac{(2n+1) \pi (-1+i)}{2} \right]\right) \\ &= 2 \pi i \left(\sum_{n=0}^{\infty} \lim_{z \to \frac{(2n+1) \pi (1+i)}{2} }\frac{ze^{iz}}{\sinh(z) - \sin(z)} + \sum_{n=0}^{\infty} \lim_{z \to \frac{(2n+1) \pi (-1+i)}{2} }\frac{ze^{iz}}{\sinh(z) - \sin(z)} \right) \\ &= 2 \pi i \left(\frac{\pi}{2}\sum_{n=0}^{\infty} \frac{(2n+1)e^{-(2n+1) \pi /2}}{\cosh \left(\frac{(2n+1) \pi }{2} \right)} + \frac{\pi}{2}\sum_{n=0}^{\infty} \frac{(2n+1)e^{-(2n+1) \pi /2}}{\cosh \left(\frac{(2n+1) \pi }{2} \right)}\right) \\ &= 2 \pi^{2} i \sum_{n=0}^{\infty} \frac{(2n+1) e^{- (2n+1) \pi /2}}{\cosh \left(\frac{(2n+1)\pi }{2} \right)} \\ &= 4 \pi^{2}i \sum_{n=0}^{\infty} \frac{2n+1}{e^{(2n+1)\pi }+1}. \end{align}$

Equating the imaginary parts on both sides of the equation, we get $$ \begin{align} \sum_{n=0}^{\infty} \frac{2n+1}{e^{(2n+1) \pi }+1} &= \frac{1}{4 \pi^{2}}\int_{-\infty}^{\infty} \frac{x \sin(x)}{\cosh (x) + \cos(x)} \, \mathrm dx \\ &= \frac{1}{2 \pi^{2}}\int_{0}^{\infty} \frac{x \sin(x)}{\cosh (x) + \cos(x)} \, \mathrm dx. \end{align}$$

But we know that $$ \begin{align} \int_{0}^{\infty} \frac{x \sin(x)}{\cosh (x) + \cos(x)} \, \mathrm dx &= 2 \, \Im \int_{0}^{\infty} x \sum_{n=1}^{\infty} (-1)^{n-1} e^{(-1+i)nx} \, \mathrm dx \\ &= 2\, \Im \sum_{n=1}^{\infty} (-1)^{n-1} \int_{0}^{\infty} x e^{(-1+i)nx} \, \mathrm dx \\ &= 2\, \Im \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{(1-i)^{2}n^{2}} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}} \\ &= \frac{\pi^{2}}{12} . \end{align} $$

Therefore, $$\sum_{n=0}^{\infty} \frac{2n+1}{e^{(2n+1) \pi }+1} = \frac{1}{2 \pi^{2}} \left(\frac{\pi^{2}}{12} \right) = \frac{1}{24}.$$

The same approach also shows that $$\sum_{n=0}^{\infty} \frac{(2n+1)^{5}}{e^{(2n+1)\pi}+1} = - \frac{2}{\pi^{6}} \int_{0}^{\infty} \frac{x^{5} \sin(x)}{\cosh(x) + \cos(x)} \, \mathrm dx = \frac{60}{\pi^{6}} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{6}} = \frac{31}{504}.$$

And in general, for a nonnegative integer $k$, we have $$ \begin{align} \sum_{n=0}^{\infty} \frac{(2n+1)^{4k+1}}{e^{(2n+1)\pi}+1} &= \frac{(-1)^{k}2^{2k-1}}{\pi^{4k+2}} \int_{0}^{\infty} \frac{x^{4k+1} \sin(x)}{\cosh(x) + \cos(x)} \, \mathrm dx \\ &= \frac{(-1)^{k}2^{2k-1}}{\pi^{4k+2}} \frac{1}{2^{2k}} (-1)^{k} (4k+1)! \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{4k+2}} \\ &\overset{(1)}{=} \frac{B_{4k+2}}{2 (4k+2)} \left(2^{4k+1}-1\right), \end{align}$$ where $B_{k}$ is the $k$th Bernoulli number.

$(1)$ https://en.wikipedia.org/wiki/Dirichlet_eta_function#Particular_values

Random Variable
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