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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?

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    $\begingroup$ You can also get the $\sqrt{2\pi}$ in Stirling's approximation (I wonder if that's related too). $\endgroup$
    – Jam
    Commented Jul 31, 2014 at 3:22
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    $\begingroup$ @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.) $\endgroup$ Commented Jul 31, 2014 at 3:53
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    $\begingroup$ @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 $$\prod_{n=1}^\infty n=\exp\left(\log\prod_{n=1}^\infty n\right)=\exp\left(\sum_{n=1}^\infty\log n\right)=\exp(-\zeta'(0))=\exp\left(\frac{1}{2}\log{2\pi}\right)=\sqrt{2\pi}$$ $\endgroup$
    – user76284
    Commented Jul 31, 2014 at 16:15
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    $\begingroup$ 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. $\endgroup$ Commented Jul 31, 2014 at 20:31
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    $\begingroup$ 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. $\endgroup$
    – anon
    Commented Aug 1, 2014 at 0:37

1 Answer 1

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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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