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For a research work I ended up needing to give a proof of Zeta's trivial zeros, in order to do so I tried using the Abel-Plana formula.

$\zeta(s)=\frac{2^{s-1}}{s-1}-2^{s}\int_{0}^{\infty} \frac{\sin(s\cdot \arctan(t))}{(e^{t\pi}+1)(1+t^2)^{\frac{s}{2}}} \,dt \implies \frac{2^{-2k-1}}{-2k-1}=-2^{-2k}\int_{0}^{\infty} \frac{\sin[2k\cdot \arctan(t)]}{(e^{t\pi}+1)(1+t^2)^{-k}} \,dt$

In the above equation -2k represents the even negative integer, after some development using the complex form of Sin and Arctan I ended up at the following stalemate.

$\frac{1}{2i}\int_{0}^{\infty} \frac{(1+it)^{2k} - (1-it)^{2k}}{(e^{t\pi}+1)} \,dt = \frac{1}{4k+2}$

Willing to verify my result for $k=1 \implies \zeta(-2)$ I managed to find the following:

$\int_{0}^{\infty}\frac{2t}{(e^{t\pi}+1)} \,dt=\frac{1}{6}$

However my competences in calculus give me no clue on how to directly justify this last equality.

ReGramakor
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3 Answers3

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More generally, for $a > 1$,

$$ \eqalign{\int_0^\infty \frac{2t}{e^{at}+1}\; dt &= \sum_{k=1}^\infty (-1)^{k-1} \int_0^\infty \frac{2t}{e^{kat}}\; dt \cr &= \sum_{k=1}^\infty \frac{2 (-1)^{k-1}}{a^2 k^2}\cr &= \frac{\pi^2}{6a^2}}$$

Robert Israel
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Assuming you can interchange integrals and infinite sums, we have \begin{align*} 2\int_0^\infty\frac t{e^{\pi t}+1}dt&=2\int_0^\infty t\sum_{n\ge1}(-1)^{n-1}e^{-\pi n t} dt\\ &=2\sum_{n\ge1}(-1)^{n-1}\int_0^\infty te^{\pi n t} dt.\\ \end{align*} Here, by integration by parts, we have \begin{align*} \int_0^\infty te^{\pi n t} dt&=\frac1{\pi n}\int_0^\infty t(e^{\pi n t})' dt\\ &=\frac1{\pi n}\left(\left[te^{\pi nt}\right]_0^\infty-\int_0^\infty e^{\pi n t} dt\right)\\ &=\frac1{\pi^2 n^2}. \end{align*} Thus, we have that \begin{align*} 2\int_0^\infty\frac t{e^{\pi t}+1}dt&=\frac2{\pi^2}\sum_{n\ge1}(-1)^{n-1}\frac1{n^2}.\\ &=\frac2{\pi^2}\frac{\pi^2}{12}=\frac16. \end{align*}

Kenta S
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  • What about proving (instead of assuming) you can interchange integral and infinite sums? Then, it would be mathematics. – NoNames Feb 28 '21 at 18:07
  • That would follow from Tonelli's theorem. – Kenta S Feb 28 '21 at 18:08
  • Tonelli's theorem is about integrals, not sums, and (primarily) for non-negative integrands. So you'd need two more sentences for justification. :p – NoNames Feb 28 '21 at 18:13
  • Pairing every two summands, one can make the summands $e^{-\pi(2n-1)t}-e^{-\pi(2n)t}$, which are positive. You can apply Tonelli's theorem by using a discrete measure. – Kenta S Feb 28 '21 at 18:18
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Let $x=e^{-\pi t}$. Then

$$\int_{0}^{\infty}\frac{2t}{e^{t\pi}+1} \,dt =-\frac2{\pi^2}\int_0^1 \frac{\ln x}{1+x}dx =-\frac2{\pi^2}\cdot (-\frac{\pi^2}{12} )= \frac16 $$ where $\int_0^1 \frac{\ln x}{1+x}dx = -\int_0^1 \frac{\ln (1+x)}{x}dx = -\frac{\pi^2}{12}$

Quanto
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  • This is largely a link only answer, because the hardest you skip, and paste a link to another's work. – amWhy Feb 28 '21 at 19:04
  • You answer far to many PSQs (lhfs), and here you do so again, borrowing from others' solutions. – amWhy Feb 28 '21 at 19:36