# What is the name of the function $1/\Gamma(z) \int_0^\infty \frac{x^{z -1}}{e^w-1}dw$?

I recently saw

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?

• It is just the Riemann Zeta function Commented Feb 20, 2021 at 19:03
• 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. Commented Feb 20, 2021 at 19:10

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.

• You mean critical line, $\Re(s)=1/2$. Critical strip refers to $0<\Re(s)<1$. Commented Feb 20, 2021 at 19:11
• Yes ofc I'm sorry. I wrote "on" the strip, I thought of the right thing. Commented Feb 20, 2021 at 19:11

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.

• $\Gamma(s)$ not $\Gamma(s-1)$ Commented Feb 20, 2021 at 19:25
• thank you, let me fix that Commented Feb 20, 2021 at 19:27

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 $$z \mapsto F_-(z, 0)$$.