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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.

Kamal Saleh
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mick
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  • Some information on this constant is in the OEIS; however, no integral form is given. – Jair Taylor Nov 01 '18 at 23:03
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    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. – James Arathoon Nov 02 '18 at 02:52
  • @JamesArathoon, fluke? I think this is pretty amazing! Why don't you put this in an answer? – Yuriy S Nov 03 '18 at 01:02
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    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 “ :) – mick Nov 03 '18 at 01:10
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    @mick Do you happen to know if the $7$ is because $7=2^3-1$? So similar representations of $\zeta(5)$ will represent $31\zeta(5)$? – Carl Schildkraut Nov 03 '18 at 01:25
  • James his integral is nice looking. But is it even correct ? Do we have a proof ? How is it proven ? Also notice $\frac{1}{Li_3(1)} = \frac{1}{ \zeta(3) }$ so in a way this is partially not in the spirit of the OP. I assume contour integration or hypergeometric is used for computing that integral. – mick Nov 03 '18 at 01:43
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    @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. – mick Nov 03 '18 at 01:56
  • ... adding parameters and doing differentiation Under the integral sign might show that ... – mick Nov 03 '18 at 02:01
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    @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 – Yuriy S Nov 03 '18 at 02:56
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    There are not many non-trivial integrals that have $\pi$ in the denominator. See also here: https://math.stackexchange.com/q/2958600/515527 – Zacky Nov 03 '18 at 20:08
  • Speculating here ... $\zeta(3)$ is a period. Presumaby $1/\zeta(3)$ is not. Which would mean that $1/\zeta(3)$ cannot be written as a certain type of integral ... https://en.wikipedia.org/wiki/Ring_of_periods – GEdgar Nov 06 '18 at 19:45
  • Yes GEdgar i considered it too. And I assume that is correct. But no proof ? – mick Nov 07 '18 at 22:42
  • The maths underlying my comment above appear to be much simpler than I first thought - https://math.stackexchange.com/questions/2991661/understanding-when-the-multiplication-of-two-definite-integrals-gives-unity – James Arathoon Nov 09 '18 at 20:45
  • James , i upvoted your question. The integrand is probably never elementary with that method. And the fraction Will probably never cancel. It seems like there is bijection between representations of zeta(3) and its inverse because simplifications seem unlikely. Assuming there is no other representation method that is. – mick Nov 12 '18 at 01:00

1 Answers1

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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.

aleden
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    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. – Greg Hurst Nov 02 '18 at 01:20
  • 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. – mick Nov 03 '18 at 02:08