13
$\begingroup$

The volume of a unit $n$-dimensional ball (in Euclidean space) is

$$V_n = \frac{\pi^{n/2}}{\frac{n}{2}\Gamma(\frac{n}{2})}$$

The completed Riemann zeta function, or Riemann xi function, is

$$\xi(s) = (s-1) \frac{\frac{s}{2}\Gamma(\frac{s}{2})}{\pi^{s/2}} \zeta(s)$$

Save for the $(s-1)$, the extra factor is exactly the inverse of $V_s$.

Is there any explanation for this, or is it just a funny coincidence?

$\endgroup$
8
$\begingroup$

When we prove the functional equation, usually we start by proving the Mellin transform $$\Gamma\left(\frac{s}{2}\right)\pi^{-s/2}\zeta(s)=\int_{0}^{\infty}\psi(x)x^{s/2-1}dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ where $$\psi(x)=\sum_{n=1}^{\infty}e^{-\pi n^{2}x}.$$ This is where the factor of $\Gamma(s/2)\pi^{-s/2}$ comes from, and this can be proven by writing down the definition of $\Gamma(s/2)$, making a change of variable, and summing over $n$. Instead, when $k$ is an integer we can prove this identity in a different way that makes it clear that this factor of $\Gamma(s/2)\pi^{-s/2}$ is really $A_{k-1}/2$ where $A_{k-1}$ is the surface area of the $k$-dimensional ball.

We have that

$$\int_{-\infty}^{\infty}e^{-\pi n^{2}x^{2}}dx=\frac{1}{n},$$ and so $$\zeta(k)=\sum_{n=1}^{\infty}\frac{1}{n^{k}}=\sum_{n=1}^{\infty}\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}e^{-\pi n^{2}(x_{1}^{2}+\cdots+x_{k}^{2})}dx_{1}\cdots dx_{k}.$$ Switching to spherical coordinates and letting $r^{2}=x_{1}^{2}+\cdots+x_{k}^{2}$, a shell of radius $r$ has size $A_{k-1}r^{k-1}$ and so $$\zeta(k)=A_{k-1}\int_{0}^{\infty}\sum_{n=1}^{\infty}e^{-\pi n^{2}r^{2}}r^{k-1}dr,$$ and then by letting $t=r^{2},$ we then have that $$\frac{2\zeta(k)}{A_{k-1}}=\int_{0}^{\infty}\psi(t)t^{k/2-1}dt.$$

Modifying this proof, one can show directly that $$A_{k-1}=\frac{2\pi^{k/2}}{\Gamma(k/2)}.$$

$\endgroup$
  • $\begingroup$ I like this derivation. But I think it is only valid if $k$ is a positive integer, because otherwise you would need a non-integer or even negative number of gaussian integrals, correct? $\endgroup$ – asmaier Jun 7 '16 at 18:15
  • $\begingroup$ @asmaier: Definitely. The volume in $n$-dimensional euclidean space doesn't really make sense unless $n$ is an integer, and so the proof of the functional equation for $s$ not an integer is slightly different. $\endgroup$ – Eric Naslund Jun 8 '16 at 12:45
  • $\begingroup$ May I draw you attention then to another very similar question of mine: math.stackexchange.com/questions/1792755/… ? Maybe it is easy for you to answer it and you would even get a bounty ;-) $\endgroup$ – asmaier Jun 8 '16 at 20:10
  • $\begingroup$ @asmaier: I don't think there is a geometric reason beyond what I wrote here, and I posted that as an answer over there. (These questions should be linked) It may not be what you are looking for though. $\endgroup$ – Eric Naslund Jun 13 '16 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.