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I recently saw

enter image description here

on the internet, and was curious what the name of the "pineapple" function was?

$$1/\Gamma(z) \int_0^\infty \frac{\color{red}{x}^{z-1}}{e^w-1}dw$$

I recognize the "mango" function as $\Gamma(z)$ the gamma function. Also, does anyone know what theorem this refers to?

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    $\begingroup$ It is just the Riemann Zeta function $\endgroup$ Commented Feb 20, 2021 at 19:03
  • $\begingroup$ The functional equation sends points with $\Re(s)>1$ to $\Re(s)<0$, so for points in the critical strip $0<\Re(s)<1$ no definition is given. Thus, this answer to the question is that there is no solution since there is no notion of a function have zeros in an interval where it is not defined. $\endgroup$
    – Milo Moses
    Commented Feb 20, 2021 at 19:10

3 Answers 3

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One representation of the Riemann zeta-function $$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s}, \quad \operatorname{Re}s>1.$$

This theorem refers to the Riemann Hypothesis, one of the greatest unsolved problem in mathematics. The RH states:

All non-trivial zeros of the Riemann Zeta-Function lie on the critical line.

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  • $\begingroup$ You mean critical line, $\Re(s)=1/2$. Critical strip refers to $0<\Re(s)<1$. $\endgroup$
    – Milo Moses
    Commented Feb 20, 2021 at 19:11
  • $\begingroup$ Yes ofc I'm sorry. I wrote "on" the strip, I thought of the right thing. $\endgroup$
    – vitamin d
    Commented Feb 20, 2021 at 19:11
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It possible rewrite Gamma function such as $$\Gamma(s)= \int_{0}^{\infty} e^{-x} x^{s-1} dx = \int_{0}^{\infty} e^{-nu} (nu)^{s-1}ndu = n^s \int_{0}^{\infty} e^{-nu} u^{s-1} du $$ Then \begin{align*} \Gamma(s)\zeta(s) = \sum_{n=0}^{\infty} \frac{\Gamma(s)}{n^s} = \sum_{n=0}^{\infty} \int_{0}^{\infty}e^{-nx}x^{s-1} dx &= \int_{0}^{\infty} x^{s-1} \left (\sum_{n=0}^{\infty} e^{-nx} \right) dx \\ & = \int_{0}^{\infty} \frac{x^{s-1}}{e^x-1} dx \end{align*} Hence, that is equivalent to Riemann Hypothesis.

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  • $\begingroup$ $\Gamma(s)$ not $\Gamma(s-1)$ $\endgroup$
    – reuns
    Commented Feb 20, 2021 at 19:25
  • $\begingroup$ thank you, let me fix that $\endgroup$ Commented Feb 20, 2021 at 19:27
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It's a special case of the Bose-Einstein integral $F_-$. In (statistical) physics, we define the Bose-Einstein integral ($F_-$) and Fermi-Dirac integral ($F_+$) as follows: $$F_{\mp}(s, \alpha)=\frac{1}{\Gamma(s)} \int_0^{+\infty}\frac{x^{s-1}}{e^{x+\alpha}\mp 1}\, \mathrm{d}x$$

So your function is just $z \mapsto F_-(z, 0)$.

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