I was trying to evaluate $\int_{0}^{\infty}{\frac{e^x}{x!} dx}$ or approximate it (WFA only gives an approximate result so maybe there is no closed form). I tried the Fourier transform :
$$\mathcal{F}\left[\frac{e^x}{x!}\right]=\int_{-\infty}^{\infty}{\frac{e^x}{x!}e^{i\omega x}dx}$$
But setting up a semicircular contour for example, we have for the semicircle going anticlockwise ($\psi$) :
$$\begin{align}\int_{\psi}{\frac{e^z}{z!}e^{i\omega z }dz}=\int_{0}^{\pi}{\frac{e^{Re^{i\theta}}}{\Gamma(Re^{i\theta})}e^{i\omega Re^{i\theta}}iRe^{i\theta}d\theta}\end{align}$$
Then attempting to bound this (with this on the last step but not sure if the condtions are met):
$$\begin{align}\left|\int_{0}^{\pi}{\frac{e^{Re^{i\theta}}}{\Gamma(Re^{i\theta})}e^{i\omega Re^{i\theta}}iRe^{i\theta}d\theta}\right|&\leq\int_{0}^{\pi}{\left|\frac{e^{Re^{i\theta}}}{\Gamma(Re^{i\theta})}e^{i\omega Re^{i\theta}}iRe^{i\theta}d\theta\right|}\\ &=R\int_{0}^{\pi}{\frac{e^{R\cos\theta}}{|\Gamma(Re^{i\theta})|}e^{-i\omega R\sin\theta}d\theta}\\ &\stackrel{?}{\leq}R\int_{0}^{\pi}{\frac{e^{R\cos\theta}}{\Gamma(R\cos\theta)}\cosh(\pi R\sin\theta)d\theta}\end{align}$$
Can this be continued? If not, what method can be applied to at least approximate the original integral?
EDIT : Stirling's approximation from the beginning :
$$\int_{0}^{\infty}{\frac{e^x}{x!}dx}\approx\frac{1}{\sqrt{2\pi}}\int_{1}^{\infty}{\frac{e^{2x}}{x^x\sqrt{x}}dx}$$
Starting at 1 to avoid division by zero but problematic if the area between 0 and 1 was a large contribution
Best Answer
The Fransen-Robinson Wikipedia page gave a limit representation which is inferred to be derived by a Riemann sum on $[0,\infty)$. Here is an analogous one:
$$\int_0^\infty\frac{e^x}{x!}dx=\lim_{a\to\infty}\sum_{n=0}^\infty\frac{e^\frac na}{\Gamma\left(\frac na+1\right)}$$
the sum is the Mittag Leffler function $\operatorname E_a(x)$, so:
$$\int_0^\infty\frac{e^x}{x!}dx=\lim_{a\to0}a\operatorname E_a(e^a)$$