Link between Integral and Product representation of Zeta Function?

Question

I watched a BlackPen RedPen video on the Riemann Zeta function in which he derived the Integral Representation of it from the Gamma Function, the link is here https://www.youtube.com/watch?v=ctG4YgMs74w, It is Called the Bose Integral and it appears in Statistical Mechanics.I also viewed the Wikipedia page https://en.wikipedia.org/wiki/Riemann_zeta_function, It shows the Euler Product formula proven by the legendary mathematician Leonhard Euler. But my question is this, How the Integral Representation of the Riemann Zeta Function is linked with the Product Formula of Euler??

Answer 1

If you question strictly speaks about any integral representation of Riemann zeta function, then there are many which includes the one you mentioned (Bose integral). Another integral representation is the one mentioned by Riemann in his famous paper "Ueber die anzahl der primzahlen unter einer gegenbenen grosse".

Actually, if you see the proof of Bose integral, it uses the following representation of zeta function. $$\zeta(s)=\sum_{k\ge 0}\frac{1}{k^s}$$ For all $\Re(s)\gt 0$. Then he does the substitution on integral definition of Gamma function and finally relates it with zeta function by above mentioned definition. Which is very useful for analytic continuation. But the Euler product formula is to be found from above mentioned definition of zeta function. So it's not possible that integral has some other possible relation with product formula. The only relation that exists will be established by that summation definition of zeta function only.

But, there are some other definitions of zeta function like, $$\ln\zeta(s)=\int_{0}^{\infty} x^{-s-1}\Pi(x) dx$$ Where, $\Pi(x)$ denotes the Riemann prime counting function. You can go through the paper i mentioned which makes it obvious that this is directly linked with Euler product formula.

Answer 2

I have seen the youtube video and it have some issues:

• first is just assumed that for real $x$ then $\sum\limits_{n=1}^{\infty} \frac{1}{n^x}$ will become the Riemann Zeta function just by changing the variable $x$ to a complex variable $s$, which is not accurate since the Riemann Zeta function is extended to the complex plane through a construction named analytical continuation, so for some values of $s$ the Riemann Zeta function is completely different from thinking about $\sum\limits_{n=1}^{\infty} \frac{1}{n^x}$
• and in this change of variable, an important issue happens when doing the exponential separation, since $e^{un}={(e^{u})}^n$ is only true if both $u$ and $n$ are reals, but in general, for a complex $s$ the $e^{sn}\neq {(e^{s})}^n$ because complex exponentiation is a multivalued operation, as it is explained here Complex Exponentiation.

So in general, the connection is because for positive real numbers the Riemann Zeta function could be represented through the summation $\sum\limits_{n=1}^{\infty} \frac{1}{n^x}$, and for real and positive valued variables $x$ the realation of the video is true, but I think caution must be taken for applying it to complex variables since could arise issues because of the step of splitting the complex exponentiation.

added later In the same link you give from the Riemann Zeta Function from Wikipedia it is said that the equivalence between the Riemann Zeta Function and the integral of the video is only valid for $\Re{(s)} = \sigma > 1$... I am not completly sure if that resolve the problem of the multivalued condition of complex exponentials, but at least if you a choose a real $s \cong x>1$ it going to work as intended on the video.

From your last question, in the same link there is another Wiki where it is explained exactly what you are asking for, is a bit long to replicate it here, so I left you the link.