1

It is well-known applying the monotone convergence theorem that for all $x>1$ we have the following functional equation : $\zeta(x)\Gamma(x)= \displaystyle \int \limits_{[0,+\infty]} \dfrac{t^{x-1}}{\exp(t)-1}\mathrm{d}\lambda(t)$ or $\zeta(x)=\dfrac{1}{\Gamma(x)}\displaystyle \int \limits_{[0,+\infty]} \dfrac{t^{x-1}}{\exp(t)-1}\mathrm{d}\lambda(t) \ $ (where $\lambda$ is the Lebesgue's measure).

Recall that : $\Gamma(x) = \displaystyle \int \limits_{[0,+\infty]}\exp(-t)t^{x-1}\mathrm{d}\lambda(t)$ and $\zeta(x)= \displaystyle \sum\limits_{n=1}^{+\infty} \dfrac{1}{n^x}$ and are well-defined for all $x>1$.

Is it possible to find a link between this functional equation and Bernoulli's numbers ?

I was thinking about the following result (which is not trivial) : $\dfrac{t}{\exp(t)-1} = \displaystyle \sum\limits_{n=0}^{+\infty}B_n \dfrac{t^n}{n!} \ $ where the sequence $(B_n)_{n\ge 0}$ represents the Bernoulli's numbers.

Then we will obtain for all $x > 1$ : $\zeta(x)\Gamma(x)= \displaystyle \int \limits_{[0,1]}t^{x-2}\left( \displaystyle \sum\limits_{n=0}^{+\infty}B_n \dfrac{t^n}{n!} \right) \mathrm{d}\lambda(t) + \displaystyle \int \limits_{[1,+\infty]} \dfrac{t^{x-1}}{\exp(t)-1} \mathrm{d}\lambda(t) \\ = \displaystyle \int \limits_{[0,1]}\left( \displaystyle \sum\limits_{n=0}^{+\infty}B_n \dfrac{t^{n+x-2}}{n!} \right) \mathrm{d}\lambda(t) + \displaystyle \int \limits_{[1,+\infty]} \dfrac{t^{x-1}}{\exp(t)-1} \mathrm{d}\lambda(t) \\ = \displaystyle\sum \limits_{n=0}^{+\infty} \dfrac{B_n}{n!(n+x-1)} + \displaystyle \int \limits_{[1,+\infty]} \dfrac{t^{x-1}}{\exp(t)-1} \mathrm{d}\lambda(t)$

The last equality is obtained using the fact that the series converges uniformly on every compact set included in $[-2\pi, 2\pi]$.

I doubt that we could go further ?

Thnaks in advance !

Maman
  • 3,300
  • 1
    The values $\zeta(-n)$ for $n=0,1,2,3,\dots$ are definitely related to Bernoulli numbers. – Greg Martin Oct 09 '21 at 18:58
  • I suspect it will be difficult as the Mellin transform of $t^{n-1}$ is a distribution: $\mathcal{M}_t\leftt^{n-1}\right=\int\limits_0^\infty t^{n-1},t^{s-1},dt=2 \pi \delta (i,(s+n-1))$. – Steven Clark Oct 09 '21 at 21:43
  • @StevenClark thank you but could you be more explicit ? – Maman Oct 09 '21 at 22:03
  • 1
    With respect to my first comment, I believe the answer I posted on MathOverflow at https://mathoverflow.net/q/294849 provides some insight, but the question here seems to overlap the question and answer at https://math.stackexchange.com/q/855740 which perhaps provides insight more relevant to your question. – Steven Clark Oct 11 '21 at 16:42
  • @StevenClark Thank you for clarifications and for the second link ! – Maman Oct 11 '21 at 16:44
  • $t/(e^t-1)=\sum_{n=0}^{\infty}B_n t^n/n!$ holds only when $|t|<2\pi$, so it's nonsense to use this for $t\in(0,+\infty)$; and even if this series converged everywhere, what would you expect to get by integrating it over $(0,+\infty)$?.. – metamorphy Oct 17 '21 at 09:57
  • @metamorphy Indeed that's why a reasonable thing is to cut the integral into two parts with $[0,1]$ and $[1,+\infty)$ and then we just have the Bernoulli numbers in the first part ! I just wanted a functional equation which would make appear these three well-knows objects nothing more ! – Maman Oct 17 '21 at 11:03

1 Answers1

1

My own opinion, but still worth posting as an answer, not a comment. The identities $$\zeta(-n)=(-1)^n\frac{B_{n+1}}{n+1},\qquad\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}\qquad(n\in\mathbb{Z}_{\geqslant0})$$ look like a complete story (along the lines of the question).

With Riemann's functional equation at hand, each one is a consequence of the other; so it's the connection to the (definition of) Bernoulli numbers that deserves a discussion (although I'm sure it's done in a number of other questions here on MSE).

Taking the generating function $z/(e^z-1)=\sum_{n=0}^\infty B_n z^n/n!$ as the definition, the second of the identities is obtained from $$\sum_{n=1}^\infty\zeta(2n)z^{2n}=\sum_{k,n=1}^\infty\left(\frac zk\right)^{2n}=\sum_{k=1}^\infty\frac{z^2}{k^2-z^2}=\frac12(1-\pi z\cot\pi z)$$ using a known consequence of the infinite product for the sine function, and then the complex-exponential form of $\cot\pi z$, making a direct link to the definition of Bernoulli numbers as stated.

An alternative way to get the first of the identities above (i.e., not via the functional equation) is to play Ramanujan's master theorem back, starting with (your well-known representation) $$\Gamma(s)\zeta(s)=\int_0^\infty\frac{x^{s-1}\,dx}{e^x-1}.$$

metamorphy
  • 39,111