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Does the fact that

$$\int_{-\infty}^{\infty}\exp\left(-\frac{1}{2}x^2\right)\mathrm{d}x=\sqrt{2\pi}$$

Have something to do with the fact that the regularized factorial of infinity is also $\sqrt{2\pi}$?

$$\infty!=\prod_{n=1}^\infty n=\exp\left(\sum_{n=1}^\infty\log n\right)=\exp(-\zeta'(0))=\exp\left(\frac{1}{2}\log2\pi\right)=\sqrt{2\pi}$$

If so, what is the connection between them?

Antonio Vargas
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user76284
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    You can also get the $\sqrt{2\pi}$ in Stirling's approximation (I wonder if that's related too). – Jam Jul 31 '14 at 03:22
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    @oliveeuler: Note that the $\sqrt{2\pi}$ in Stirling can be seen to arise from a steepest descent calculation, and that (with some embellishments) is essentially just a matter of approximating the integral with a Gaussian. So the $\sqrt{2\pi}$ in Stirling is morally the same as the first integral. ($\infty !$, on the other hand, I don't rightly know about.) – Semiclassical Jul 31 '14 at 03:53
  • I thought $\Gamma(\infty)$ was plain $\infty$? How do you go from a divergent series to $\zeta'$? Is that mathematically rigorous? – user_of_math Jul 31 '14 at 06:16
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    @user_of_math Indeed $\lim_{z\rightarrow\infty}\Gamma(z)$ diverges, but

    $$\frac{\mathrm{d}}{\mathrm{d}z}\zeta(z)=\frac{\mathrm{d}}{\mathrm{d}z}\sum_{n=1}^\infty n^{-z}=\sum_{n=1}^\infty\frac{\mathrm{d}}{\mathrm{d}z}n^{-z}=\sum_{n=1}^\infty-n^{-z}\log n$$

    Hence

    $$\zeta'(0)=-\frac{1}{2}\log{2\pi}=-\sum_{n=1}^\infty\log n$$

    Now

    $$\prod_{n=1}^\infty n=\exp\left(\log\prod_{n=1}^\infty n\right)=\exp\left(\sum_{n=1}^\infty\log n\right)=\exp(-\zeta'(0))=\exp\left(\frac{1}{2}\log{2\pi}\right)=\sqrt{2\pi}$$

     – user76284 Jul 31 '14 at 16:15
  • @user_of_math You can see this article for more information (it derives the regularized product of all prime numbers as $\infty#=4\pi^2$. – user76284 Jul 31 '14 at 16:18
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    It seems to me that the answer is yes if instead of $\zeta$-function regularization you instead consider the $\mathrm{\color{red} {regularization}}$ of the sum to be $$\sum_{n=1}^m \log n = m\log m - m + \frac{1}{2} \log m + \color{red} {\log\left(\int_{-\infty}^{\infty} e^{-x^2/2},dx\right)} + o(1),$$ where the asymptotic here comes from Stirling's formula. This regularization method often agrees with the $\zeta$-regularization method, and someone knowledgeable can probably explain why they agree here in particular. – Antonio Vargas Jul 31 '14 at 20:31
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    There is a physics connection between Gaussian integrals and $e^{-\zeta'(0)}$: they both describe functional determinants. However, (i) I am not sure how to view $\Bbb R$ as a function space for this interpretation to work, and (ii) I am not sure it's known why the two definitions of determinants agree with each other - see this question. – anon Aug 01 '14 at 00:37
  • This is one question where you might get better math help going over to the Physics Stack Exchange, amusingly enough. (For the simple reason that a substantial fraction of that population deals with such matters a lot.) – Semiclassical Aug 01 '14 at 00:53
  • For anyone who's interested in this kind of question, I've added a fresh bounty on the question which @blue linked. – Semiclassical Aug 01 '14 at 13:06

1 Answers1

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Since $\zeta(s)=\frac{\eta(s)}{1-2^{1-s}}$ and $\eta(0)=\frac{1}{2}$, we have $\zeta(0)=-\frac{1}{2}$. From: $$ \zeta'(s) = \zeta(s)\cdot\frac{d}{ds}\log\zeta(s) = \zeta(s)\left(\frac{\eta'(s)}{\eta(s)}-\frac{2^{1-s}\log(2)}{1-2^{1-s}}\right)\tag{1}$$we get $\zeta'(0)=-\eta'(0)-\log(2)$, and the problem boils down to showing: $$ \eta'(0) = \frac{1}{2}\log\frac{\pi}{2}\tag{2} $$ On the other hand, $\eta'(0)$ is directly related with the Wallis product through its series definition: $$ \eta'(0)=\frac{1}{2}\sum_{n\geq 1}(-1)^{n+1}\log\left(\frac{n+1}{n}\right)=\frac{1}{2}\log\prod_{n\geq 1}\left(1-\frac{1}{4n^2}\right)^{-1}\tag{3}$$ but: $$ \prod_{n=1}^{N}\left(1-\frac{1}{4n^2}\right)^{-1} = \frac{\Gamma\left(\frac{1}{2}\right)^2 \Gamma(N+1)^2}{2\,\Gamma\left(N+\frac{1}{2}\right)\Gamma\left(N+\frac{3}{2}\right)}\tag{4}$$ so by Gautschi's inequality the limit of the RHS of $(4)$ as $N\to +\infty$ is exactly $\frac{1}{2}\,\Gamma\left(\frac{1}{2}\right)^2$, and through the substitution $x=\sqrt{z}$ we have: $$ \int_{0}^{+\infty}e^{-x^2}\,dx = \frac{1}{2}\int_{0}^{+\infty}z^{-1/2}e^{-z}\,dx = \frac{1}{2}\,\Gamma\left(\frac{1}{2}\right).\tag{5}$$

So, a summary of the connection:

$$ \zeta'(0)\mapsto \eta'(0)\mapsto \text{Wallis product}\mapsto \Gamma\left(\frac{1}{2}\right)\mapsto \text{Gaussian integral}.$$

The arrows can be reversed as you like. We also have a detour, since the value of $(3)$ is gladly provided by the Weierstrass product for the sine function: $$ \frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right) \tag{6}$$ through an evaluation at $x=\frac{\pi}{2}$. No wonder, since $\Gamma(x)\,\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$.

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Jack D'Aurizio
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