Why is this function a really good asymptotic for $\mathrm{exp}(x)\sqrt{x}$

$f(x)=\sum _{n=0}^{\mathrm{\infty}}{a}_{n}{x}^{n}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{a}_{n}=\frac{1}{\mathrm{\Gamma}(n+0.5)}$

Why is this entire function a really good asymptotic for $\mathrm{exp}(x)\sqrt{x}$, where for large positive numbers, $f(x)\mathrm{exp}(-x)\approx \sqrt{x}$?

As |x| gets larger, the error term is asymptotically $f(x)-\mathrm{exp}(x)\sqrt{x}\approx \frac{1}{x\cdot \mathrm{\Gamma}(-0.5)}$, and the error term for $f(x)\mathrm{exp}(-x)-\sqrt{x}\approx \frac{\mathrm{exp}(-x)}{x\cdot \mathrm{\Gamma}(-0.5)}$. If we treat $f(x)$ as an infinite Laurent series, than it does not converge.

I stumbled upon the result, using numerical approximations, so I can't really explain the equation for the ${a}_{n}$ coefficients, other than it appears to be the numerical limit of a pseudo Cauchy integral for the ${a}_{n}$ coefficients as the circle for the Cauchy integral path gets larger. I suspect the formula has been seen before, and can be generated by some other technique. By definition, for any entire function $f(x)$, we have for any value of real r:

${a}_{n}=\oint {x}^{-n}f(x)={\int}_{-\pi}^{\pi}\frac{1}{2\pi}(r{e}^{-ix}{)}^{-n}f(r{e}^{ix}))\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}$

The conjecture is that this is an equivalent definition for ${a}_{n}$, where $f(x)\mapsto \mathrm{exp}(x)\sqrt{x}$ and $x\mapsto r{e}^{ix}$

${a}_{n}=\underset{r\to \mathrm{\infty}}{lim}{\int}_{-\pi}^{\pi}\frac{1}{2\pi}(r{e}^{-ix}{)}^{-n}\mathrm{exp}(r{e}^{ix})\sqrt{r{e}^{ix}})\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}x=\frac{1}{\mathrm{\Gamma}(n+0.5)}$

$f(x)=\sum _{n=0}^{\mathrm{\infty}}{a}_{n}{x}^{n}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{a}_{n}=\frac{1}{\mathrm{\Gamma}(n+0.5)}$

Why is this entire function a really good asymptotic for $\mathrm{exp}(x)\sqrt{x}$, where for large positive numbers, $f(x)\mathrm{exp}(-x)\approx \sqrt{x}$?

As |x| gets larger, the error term is asymptotically $f(x)-\mathrm{exp}(x)\sqrt{x}\approx \frac{1}{x\cdot \mathrm{\Gamma}(-0.5)}$, and the error term for $f(x)\mathrm{exp}(-x)-\sqrt{x}\approx \frac{\mathrm{exp}(-x)}{x\cdot \mathrm{\Gamma}(-0.5)}$. If we treat $f(x)$ as an infinite Laurent series, than it does not converge.

I stumbled upon the result, using numerical approximations, so I can't really explain the equation for the ${a}_{n}$ coefficients, other than it appears to be the numerical limit of a pseudo Cauchy integral for the ${a}_{n}$ coefficients as the circle for the Cauchy integral path gets larger. I suspect the formula has been seen before, and can be generated by some other technique. By definition, for any entire function $f(x)$, we have for any value of real r:

${a}_{n}=\oint {x}^{-n}f(x)={\int}_{-\pi}^{\pi}\frac{1}{2\pi}(r{e}^{-ix}{)}^{-n}f(r{e}^{ix}))\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}$

The conjecture is that this is an equivalent definition for ${a}_{n}$, where $f(x)\mapsto \mathrm{exp}(x)\sqrt{x}$ and $x\mapsto r{e}^{ix}$

${a}_{n}=\underset{r\to \mathrm{\infty}}{lim}{\int}_{-\pi}^{\pi}\frac{1}{2\pi}(r{e}^{-ix}{)}^{-n}\mathrm{exp}(r{e}^{ix})\sqrt{r{e}^{ix}})\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}x=\frac{1}{\mathrm{\Gamma}(n+0.5)}$