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I'm trying to improve my integration skills and was wondering if any of you have some neat ways to prove this. Thanks.

$$\int_{0}^{\infty} \frac{\zeta(\pi \cdot s) - \zeta(e \cdot s)}{ \zeta(\pi \cdot s) \zeta(e \cdot s) \cdot s }\,\mathrm{d}s = 3 + \ln\left(\frac{1}{\pi^3}\right)$$

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    $\begingroup$ @reuns I find this kind of question super annoying as well. However, Mathematica does seem to agree that this is correct numerically... $\endgroup$
    – Igor Rivin
    Mar 28, 2021 at 5:37
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    $\begingroup$ The constants are my own, but it's from a calc textbook. I can give a solution if no one gives one. I was just hoping to see the people who are smarter than me (everyone here) give an elegant proof. $\endgroup$ Mar 28, 2021 at 5:38
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    $\begingroup$ This is just a Frullani integral for $f=1/\zeta$. Use $\zeta(0)=-1/2$ and $\zeta(+\infty)=1$. That's it. $\endgroup$
    – metamorphy
    Mar 28, 2021 at 5:49
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    $\begingroup$ In such a case the best way is to say upfront (in question body) that you have a solution (and maybe include your own solution) and you want to see alternative approaches. $\endgroup$
    – Paramanand Singh
    Mar 28, 2021 at 5:56
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    $\begingroup$ Agree with Paramanand Singh. It would be great if you provided the solution as well so that this thread can be a better source for future readers. $\endgroup$
    – VIVID
    Mar 28, 2021 at 5:58

2 Answers 2

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As also mentioned by metamorphy, we can use Frullani's formula here: $$\begin{align} \int_{0}^{\infty} \frac{\zeta(\pi x) - \zeta(e x)}{ \zeta(\pi x) \zeta(e x) \cdot x }\,\mathrm{d}x &= \int_0^\infty \frac{\frac{1}{\zeta(ex)} - \frac{1}{\zeta(\pi x)}}{x}\mathrm dx \\ &= \left(\frac{1}{\zeta(\infty)} - \frac{1}{\zeta(0)}\right)\ln\frac{e}{\pi} \\ &= \left(\frac{1}{1} - \frac{1}{-1/2}\right)\left(1 + \ln\frac{1}{\pi}\right) \\ &= 3 + \ln\frac{1}{\pi^3}\end{align}$$

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Not really an answer but too long for a comment:

The integrand equals $$\frac1s \left( \frac1{\zeta(e s)} - \frac1{\zeta(\pi s)}\right).$$

As we know, $$\frac1{\zeta(s)} = \sum_{n=1}^\infty \mu(n) n^{-s},$$ so the integral equals $$\sum \frac{\mu(n)}s \left(n^{- e s} - n^{-\pi s}\right).$$ Presumably this gives us a series of something reasonable.

Now, $$\int_0^\infty(a^s-b^s)/s d s = \frac{1}{2} \left(\log \left(\frac{1}{\log (a)}\right)-\log (\log (a))-\log \left(\frac{1}{\log (b)}\right)+\log (\log (b))\right)\text{ if }\Re(\log (a))\leq 0\land \Re(\log (b))\leq 0\land \frac{\log (b)}{\log (a)}<1$$ which gives us a faint hope of a reasonable sum (though the $\mu(n)$ is not helpful).

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