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?


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 $$A_{k-1}=\frac{2\pi^{k/2}}{\Gamma(k/2)}.$$

  • $\begingroup$ 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

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