Gaussian integral

Generalization of Gaussian integral

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— 級数bot (@infseriesbot) June 9, 2021

We are going to prove this result posted by @infeseriesbot that is a generalization of the Gaussian integral $(s=\frac{1}{2})$. \[\int_{-\infty}^{\infty} \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!^{2s}} dx = \pi^{1-s} \quad \operatorname{Re}(s)>0 \] It was previously discussed in Twitter a few months ago and turned out that the proof is a consquence of the Ramanujan's master theorem. However, other proofs were provided using the bracketts method, the log-gamma function and contour integration. In the proof we also find the values of $s$ that guarantee the convergence of the integral $(\operatorname{Re}(s)>0)$.

Proof. Recall the Ramanujan's master theorem: If a function $f(x)$ has an expansion of the form \[ f(x) = \sum_{k=0}^{\infty} \frac{\phi(k)}{k!} (-x)^k\] for some function (say analytic or integrable) $\varphi(k)$, then the Mellin transform of $f(x)$ is given by \[\{\mathscr{M}f(x)\}(t) = \int_{0}^{\infty} x^{t-1} f(x) dx = \Gamma(t)\varphi(-t) \] In this particular case \begin{align*} \int_{-\infty}^{\infty} \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!^{2s}} dx =& 2\int_{0}^{\infty} \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!^{2s}} dx \\ =& \int_{0}^{\infty}\sum_{n=0}^{\infty} w^{\frac{1}{2}-1} \frac{(-w)^n}{n!^{2s}} dw \quad (w \mapsto x^2)\\ =& \int_{0}^{\infty} w^{\frac{1}{2}-1} \sum_{n=0}^{\infty } \frac{(-w)^n}{n!^{2s}} dw \\ =& \int_{0}^{\infty} w^{\frac{1}{2}-1} \sum_{n=0}^{\infty } (n!)^{1-2s} \frac{(-w)^n}{n!} dw \end{align*} this is a Mellin transform $\displaystyle \{\mathscr{M}f(w)\}\left(\frac{1}{2}\right) = \int_{0}^{\infty} w^{\frac{1}{2}-1} f(w)dw$ \[ \textrm{ where } f(w) = \sum_{n=0}^{\infty } (n!)^{1-2s}\frac{(-w)^n}{n!} \quad \textrm{and } \; \phi(-t) = (-t)!^{1-2s} = \left[\Gamma(1-t)\right]^{1-2s}\] By the Ramanujan's master theorem \[\int_{0}^{\infty} w^{\frac{1}{2}-1} \sum_{n=0}^{\infty } (n!)^{1-2s}\frac{(-w)^n}{n!} dw = \Gamma\left(\frac{1}{2}\right) \left[\Gamma\left(\frac{1}{2}\right)\right]^{1-2s} = \pi^{1-s} \] The result is valid for $\operatorname{Re}(s)>0$. To show this, lets review the convergence of $\displaystyle \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!^{2s}} dx$. We can use the ratio test for the sequence $ \displaystyle \left(\frac{(-x^2)^n}{n!^{2s}}\right)_{n}$ \begin{align*} \lim_{n\to \infty}\left| \frac{\frac{(-x^2)^{n+1}}{(n+1)!^{2s}}}{\frac{(-x^2)^{n}}{n!^{2s}}}\right| =& \lim_{n\to \infty} \left| \frac{n!^{2s}}{(n+1)!^{2s}}\right||x^2| \\ =& \lim_{n\to \infty} \left| \frac{n!^{2s}}{(n+1)!^{2s}}\right||x^2| \\ =& \lim_{n \to \infty } \exp\left(2\operatorname{Re}(s\ln n!)-2\operatorname{Re}(s\ln(n+1)!)\right) |x^2|\\ =& \lim_{n \to \infty } \exp\left(2\operatorname{Re}(s) \left(\ln n!-\ln(n+1)!\right)\right) |x^2|\\ =& \exp\left( \lim_{n \to \infty } \operatorname{Re}(s) 2\left(\ln n!-\ln(n+1)!\right)\right) |x^2|\\ =& \exp \left(\operatorname{Re}(s) \lim_{n \to \infty}2\ln\left(\frac{n!}{(n+1)!}\right) \right) |x^2|\\ =& \exp \left(\operatorname{Re}(s) (-\infty) \right) |x^2| \end{align*} \[ \lim_{n\to \infty}\left| \frac{\frac{(-x^2)^{n+1}}{(n+1)!^{2s}}}{\frac{(-x^2)^{n}}{n!^{2s}}}\right| = \begin{cases} 0 \quad \textrm{ if } \operatorname{Re}(s) >0 \\ |x^2| \quad \textrm{ if } \operatorname{Re}(s) =0\\ \infty \quad \textrm{ if } \operatorname{Re}(s) <0 \end{cases} \] Therefore \[ \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!^{2s}} dx <\infty \quad \textrm{if} \begin{cases} \operatorname{Re}(s)>0\\ \operatorname{Re}(s)=0 \wedge |x|<1 \end{cases} \] Then, we have obtained the desired result: \[ \boxed{ \int_{-\infty}^{\infty} \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!^{2s}} dx = \pi^{1-s} \quad \operatorname{Re}(s)>0} \]