Prove

\(I= \displaystyle\int_{-\infty }^{\infty}\sum_{n=0}^{\infty } \frac{\left(-x^2\right)^n }{n!^{2 s}}\; \mathrm{d}x= \pi^{1-s}\).

We can start as follows, by transforming it into a generalized hypergeometric function:

\(I=\displaystyle\int_{-\infty }^{\infty }\, _0F_{2 s-1} (\overbrace{1,1,1,…,1}^{2 s-1 \text{times}}; -x^2)\mathrm{d}x\), since, from the series expansion of the generalized hypergeometric function, \(\, _pF_q\left(a_1,a_p;b_1,b_q;z\right)=\sum_{k=0}^{\infty } \frac{\prod_{j=1}^p \left(a_j\right)_k z^k}{\prod_{j=1}^q k! \left(b_j\right)_k}\), where \((.)_k\) is the Pochhammer symbol \((a)_k=\frac{\Gamma (a+k)}{\Gamma (a)}\).

Now the integrand function does not appear to be convergent numerically, except for \(s= \frac{1}{2}\) where it becomes the Gaussian integral, and the case of \(s=1\) where it becomes a Bessel function. For \(s=\frac{3}{2}\) and \( x=10^{19}\), the integrand takes values of \(10^{1015852872356}\) (serious). Beyond that the computer starts to produce smoke. Yet it eventually converges as there is a closed form solution. It is like saying that it works in theory but not in practice!

For, it turns out, under the restriction that \(2 s\in \mathbb{Z}_{>\, 0}\), we can use the following result:

\(\int_0^{\infty } t^{\alpha -1} _pF_q \left(a_1,\ldots ,a_p;b_1,\ldots ,b_q;-t\right) \, dt=\frac{\Gamma (\alpha ) \prod {k=1}^p \Gamma \left(a_k-\alpha \right)}{\left(\prod {k=1}^p \Gamma \left(a_k\right)\right) \prod {k=1}^q \Gamma \left(b_k-\alpha \right)}\)

Allora, we can substitute \(x=\sqrt(u)\), and with \(\alpha =\frac{1}{2},p=0,b_k=1,q=2 s-1\), given that \(\Gamma(\frac{1}{2})=\sqrt(\pi)\),

\(I=\frac{\sqrt{\pi }}{\prod _{k=1}^{2 s-1} \sqrt{\pi }}=\pi ^{1-s}\).

So either the integrand eventually converges, or I am doing something wrong, or both. Perhaps neither.

Maestro, I think that the identity follows from the Ramanujan’s master theorem, see https://en.wikipedia.org/wiki/Ramanujan%27s_master_theorem

Let $latex F(x)= \sum\limits_{k=0}^\infty\frac{(-x)^n}{(n!)^{2s})}$,

Mellin transform is $latex M[F(x)](z) = \frac{\Gamma(z)}{\Gamma(1-z)^{2s-1}}$,

$latex I = 2M[F(x^2)](1) = M[F(x)](z/2) = \frac{\Gamma(1/2)}{\Gamma(1/2)^{2s-1}}$.

Thanks a million!

No, thank you, Maestro. I only learned about the theorem while trying to prove one of the previous identities, that you posted on Twitter some time ago.

BTW, the paper at the end of that wiki article is worth reading, or at least skimming through.