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There are many integral representations for $\zeta(3)$

Some lesser known are for instance :

$$\int_0^1\frac{x(1-x)}{\sin\pi x}\text{d}x= 7\frac{\zeta(3)}{\pi^3} $$

$$\int_0^1 \frac{\operatorname{li}(x)^3 \space (x-1)}{x^3} \text{d}x = \frac{\zeta(3)}{4} $$

$$\int_0^\pi x(\pi - x) \csc(x) \space \text{d}x = 7 \space \zeta(3) $$

$$ \int_0^{\infty} \frac{\tanh^2(x)}{x^2} \text{d}x = \frac{ 14 \space \zeta(3)}{\pi^2} $$

$$\int_0^{\frac{\pi}{2}} x \log\tan x \;\text{d}x=\frac{7}{8}\zeta(3)$$

$\zeta(2) $ also has many integral representations as does $ \frac{1}{\zeta(2)} $ , although this is probably because $\frac{1}{\pi}$ and $\frac{1}{\pi^2} $ have many. Well I suspect that because I know no simple integral expression for $\frac{1}{\zeta(3)} $.

My question is: is there some interesting integral $^*$ whose result is simply $\frac{1}{\zeta(3)}$?

Note

$^*$ Interesting integral means that things like

$$\int\limits_0^{+\infty} e^{- \zeta(3) \space x}\ \text{d}x = \frac{1}{\zeta(3)} $$

are not a good answer to my question.

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    $\begingroup$ A (fluke?) example found using Mathematica is $$\frac{1}{\zeta (3)}=\frac{3}{2} \int_1^{\infty } \frac{\log ^2(x)}{(x+1) \left(-2 \text{Li}_3(-x)+2 \text{Li}_2(-x) \log (x)+\log (x+1) \log ^2(x)\right){}^2} \, dx$$ If you integrate (indefinitely), invert and then differentiate you will find the starting function from which this originates. I inverted and swapped bounds. This method does not work in general, but remarkably in this case does seem to work for $\zeta(3)$, $\zeta(5)$ and perhaps all the other odd zeta constants. $\endgroup$ Commented Nov 2, 2018 at 2:52
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    $\begingroup$ Btw a strange observation is that most of the integral representations for $\zeta(3)$ are of the form $ 7 \space \zeta(3) $ times “ something “. This $7$ is somewhat mysterious since those Integral representations are just randomly and Naturally occuring with simple short integrands. I call this the “ holy phenomenon “ :) $\endgroup$
    – mick
    Commented Nov 3, 2018 at 1:10
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    $\begingroup$ @mick Do you happen to know if the $7$ is because $7=2^3-1$? So similar representations of $\zeta(5)$ will represent $31\zeta(5)$? $\endgroup$ Commented Nov 3, 2018 at 1:25
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    $\begingroup$ @Carl that might make sense. Ofcourse there are Many ways to arrive at the 7 because there are multiple ways to do the integral. But yes cases of “ $ 31 \zeta(5) $ ” occur and i even think $ 7 \zeta(3) + 31 \zeta(5) $. Btw i wonder about Integral representations of $1/( \zeta(3) + \zeta(5) )$. I considered posting that instead of the OP actually. Anyways about that 7 , I think it relates to the integrals being special cases of representations of zeta-like functions instead of zeta functions. Think alternating zeta , polylog , hurwitz etc. $\endgroup$
    – mick
    Commented Nov 3, 2018 at 1:56
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    $\begingroup$ @mick, but James' comment exactly tells how he found the integral. Numerically it's correct, I checked. I don't really see your point about the polylog, because its argument inside the integral is a variable. I think it's pretty amazing that any kind of definite integral was found for this number $\endgroup$
    – Yuriy S
    Commented Nov 3, 2018 at 2:56

1 Answer 1

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You could find an integral representation similar to one for the regular zeta function by using the mobius function. $$\Gamma(s)\frac{\mu(n)}{n^s}=\int_0^\infty x^{s-1}\mu(n)e^{-nx}dx$$

Summing over both sides we get $$\frac{\Gamma(s)}{\zeta(s)}=\int_0^\infty x^{s-1}\rho(x)dx$$

where $\rho(x)=\sum_{n=1}^\infty \mu(n)e^{-nx}$. However, I don't think that $\rho(x)$ has an expression in closed form.

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    $\begingroup$ For $\text{Re}(s) > 1$, a similar idea gives $1/\zeta(s) = s \int_1^\infty M(x)x^{-s-1} dx$, where $M(x)$ is the Mertens function. $\endgroup$
    – Greg Hurst
    Commented Nov 2, 2018 at 1:20
  • $\begingroup$ It is very very unlikely that $\rho(x)=\sum_{n=1}^\infty \mu(n)e^{-nx}$ has a closed form. Both from reasons of analysis , differential galois theory and number theory. In fact if it did then solving the prime twins conjecture seems much easier. “ Parity problem “ comes to mind. Just a comment. $\endgroup$
    – mick
    Commented Nov 3, 2018 at 2:08

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