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How to show that: $$ \int_{0}^{\infty}\frac{1}{e^x-1}\left[\frac{12\,e^x}{(e^x-1)^2}-\frac{12}{x^2}+1\right]\,dx=\frac{5}{2}+\frac{\zeta'(2)}{\zeta(2)}-\frac{\zeta'(0)}{\zeta(0)}\tag{1} $$ Is it possible to split the integral into two integrals representing ${\zeta'}/{\zeta}(0)$ and ${\zeta'}/{\zeta}(2)$.


The logarithmic derivative of Riemann Zeta function is mainly defined by the infinte series: $$ \log'\left[\zeta(s)\right]=\frac{\zeta'}{\zeta}(s)=-\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s} $$ Where $\Lambda(n)$ is Von Mangoldt function. ${\zeta'}/{\zeta}$ has some intresting idendities, like: $$ \frac{\zeta'}{\zeta}(s)+\frac{\zeta'}{\zeta}(1-s)=\log\pi-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{s}{2}\right)-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{1-s}{2}\right) $$ Does ${\zeta'}/{\zeta}$ has an integral representation? Thanks.

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  • $\begingroup$ Where did your integral come from? :) $\endgroup$
    – Wolfgang
    Commented Apr 13, 2022 at 7:19
  • $\begingroup$ With respect to your final question, the integral representation is $\frac{\zeta'(s)}{\zeta(s)}=-s\,\mathcal{M}_x[\psi (x)](-s)=-s\int\limits_0^\infty \psi(x)\,x^{-s-1}\,dx$ which is valid for $\Re(s)>1$ where $\psi(x)=\sum\limits_{n=1}^x \Lambda(n)$ is the second Chebyshev function (see en.wikipedia.org/wiki/Chebyshev_function). $\endgroup$ Commented Apr 20, 2022 at 20:17

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Let us consider the integral $$I_{1}\left(s\right)=\int_{0}^{\infty}\frac{x^{s-1}e^{x}}{\left(e^{x}-1\right)^{3}}dx,\,\mathrm{Re}\left(s\right)>3.$$ Then, by the dominated convergence theorem, we get $$I_{1}\left(s\right)=-\frac{1}{2}\sum_{k\geq2}k\left(k-1\right)\int_{0}^{\infty}x^{s-1}e^{-kx}dx$$ $$=-\frac{1}{2}\Gamma\left(s\right)\left(\zeta\left(s-2\right)-\zeta\left(s-1\right)\right).$$ In a similar fashion we have $$I_{2}\left(s\right)=\int_{0}^{\infty}\frac{x^{s-1}}{e^{x}-1}dx=\Gamma\left(s\right)\zeta\left(s\right),\mathrm{Re}\left(s\right)>1$$ so we can conclude that $$I_{3}\left(s\right)=\int_{0}^{\infty}\frac{x^{s-1}}{e^{x}-1}\left(\frac{12e^{x}}{\left(e^{x}-1\right)^{2}}-\frac{12}{x^{2}}+1\right)dx$$ $$=-\frac{12}{2}\Gamma\left(s\right)\left(\zeta\left(s-1\right)-\zeta\left(s-2\right)\right)-12\Gamma\left(s-2\right)\zeta\left(s-2\right)+\Gamma\left(s\right)\zeta\left(s\right),\mathrm{Re}\left(s\right)>3$$ but, using the asymptotic expansion of $\Gamma\left(s\right)$ at $s=0$ and $\zeta\left(s\right)$ at $s=1$ we can see that $I\left(s\right)$ is defined for $\mathrm{Re}\left(s\right)\geq1$ and $$\lim_{s\rightarrow1}-\frac{12}{2}\Gamma\left(s\right)\left(\zeta\left(s-1\right)-\zeta\left(s-2\right)\right)-12\Gamma\left(s-2\right)\zeta\left(s-2\right)+\Gamma\left(s\right)\zeta\left(s\right)$$ $$=\color{red}{\frac{5}{2}-\gamma+12\log\left(A\right)}$$ where $A$ is the Glaisher–Kinkelin constant and $\gamma$ is the Euler-Mascheroni constant, which is equivalet to your result.

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    $\begingroup$ Well Done! Thanks Marco. $\endgroup$ Commented Sep 9, 2018 at 13:18
  • $\begingroup$ For the record: the pattern does not persist if we go higher: The next one is $$\int_{0}^{\infty}\frac{1}{e^x-1}\left[\frac{24\,e^x}{(e^x-1)^3}-\frac{24}{x^3}+\frac{12}{x^2}-1\right]\,dx=\frac12\frac{ζ(3)}{ζ(2)}-\frac{ζ'(2)}{ζ(2)}+\log(2\pi)-3 $$ Wolfram, and after that comes $240 \log(A)-120ζ'(-3)-360 ζ'(-2)-775/12-19 \gamma$, so either $\log(A)$ or $ \gamma $ will survive... $\endgroup$
    – Wolfgang
    Commented Apr 13, 2022 at 7:18
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This is to address your question about splitting the integral into two integrals representing ${\zeta'}/{\zeta}(0)$ and ${\zeta'}/{\zeta}(2)$. I have been intrigued by this kind of question already before encountering your question (see my edits to this answer), wondering whether ${\zeta'}/{\zeta}(2k)$ alone can be an "exponential period". But recently, after figuring out that ${\zeta'}/\zeta(0)=\log({2\pi})$ and googling on approach0.xyz for integrals with that kind of value, I found that there are plenty, some even of not too different shapes, e.g. this one: $$\int_0^1\left(\frac{1}{\log y} + \frac{1}{1-y}\right)^2\;dy = \frac{\zeta'(0)}{\zeta(0)} - \frac{3}{2}.$$ Substituting $y=e^{-x}$, this becomes $$\int_0^\infty\left[-\frac{1}{x} + \frac{1}{1- e^{-x}}\right]^2e^{-x}~dx =\int_0^\infty e^{x}\left[\frac{1}{x~e^x} - \frac{1}{e^x-1}\right]^2~dx,$$ which is quite close in style to the original one, so you may quite reasonably say that there is a way of splitting that one accordingly.

And that also provides a positive answer to your final question, at least for even arguments, $\dfrac{\zeta'(2)}{\zeta(2)}, \dfrac{\zeta'(4)}{\zeta(4)}$ etc.

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