## A Peculiar Integral

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.