# Why is this function a really good asymptotic for exp(x)sqrt(x) f(x)=sum_(n=0)^(oo) a_n x^n \ a_n = (1)/(Gamma(n+0.5)) exp(x)sqrt(x), where for large positive numbers, f(x)exp(−x)~~sqrt x?

Why is this function a really good asymptotic for $\mathrm{exp}\left(x\right)\sqrt{x}$
$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}{x}^{n}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{a}_{n}=\frac{1}{\mathrm{\Gamma }\left(n+0.5\right)}$
Why is this entire function a really good asymptotic for $\mathrm{exp}\left(x\right)\sqrt{x}$, where for large positive numbers, $f\left(x\right)\mathrm{exp}\left(-x\right)\approx \sqrt{x}$?
As |x| gets larger, the error term is asymptotically $f\left(x\right)-\mathrm{exp}\left(x\right)\sqrt{x}\approx \frac{1}{x\cdot \mathrm{\Gamma }\left(-0.5\right)}$, and the error term for $f\left(x\right)\mathrm{exp}\left(-x\right)-\sqrt{x}\approx \frac{\mathrm{exp}\left(-x\right)}{x\cdot \mathrm{\Gamma }\left(-0.5\right)}$. If we treat $f\left(x\right)$ 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 $f\left(x\right)$, we have for any value of real r:
${a}_{n}=\oint {x}^{-n}f\left(x\right)={\int }_{-\pi }^{\pi }\frac{1}{2\pi }\left(r{e}^{-ix}{\right)}^{-n}f\left(r{e}^{ix}\right)\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}$
The conjecture is that this is an equivalent definition for ${a}_{n}$, where $f\left(x\right)↦\mathrm{exp}\left(x\right)\sqrt{x}$ and $x↦r{e}^{ix}$
${a}_{n}=\underset{r\to \mathrm{\infty }}{lim}{\int }_{-\pi }^{\pi }\frac{1}{2\pi }\left(r{e}^{-ix}{\right)}^{-n}\mathrm{exp}\left(r{e}^{ix}\right)\sqrt{r{e}^{ix}}\right)\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}x=\frac{1}{\mathrm{\Gamma }\left(n+0.5\right)}$
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oldgaffer1b
Repeated integrations by parts show that, for every positive $a$ and $x$
${\int }_{0}^{x}{\mathrm{e}}^{-t}{t}^{a-1}\mathrm{d}t=\mathrm{\Gamma }\left(a\right){\mathrm{e}}^{-x}\sum _{n⩾0}\frac{{x}^{n+a}}{\mathrm{\Gamma }\left(n+a+1\right)}.$
When $x\to \mathrm{\infty }$, the LHS converges to $\mathrm{\Gamma }\left(a\right)$, hence the series in the RHS is equivalent to ${\mathrm{e}}^{x}$. Now,
$\sum _{n⩾0}\frac{{x}^{n}}{\mathrm{\Gamma }\left(n+a\right)}=\frac{1}{\mathrm{\Gamma }\left(a\right)}+{x}^{1-a}\sum _{n⩾0}\frac{{x}^{n+a}}{\mathrm{\Gamma }\left(n+a+1\right)}$
hence
$\sum _{n⩾0}\frac{{x}^{n}}{\mathrm{\Gamma }\left(n+a\right)}\sim {x}^{1-a}{\mathrm{e}}^{x}.$
For $a=\frac{1}{2}$, this is the result mentioned in the question.
An exact formula using the incomplete gamma function (that is, the LHS of the first identity in this answer) is
$\sum _{n⩾0}\frac{{x}^{n}}{\mathrm{\Gamma }\left(n+a\right)}=\frac{\gamma \left(a,x\right)}{\mathrm{\Gamma }\left(a\right)}{x}^{1-a}{\mathrm{e}}^{x}+\frac{1}{\mathrm{\Gamma }\left(a\right)}.$
Edit: ...And this approach yields the more precise expansion, also mentioned in the question,
$\sum _{n⩾0}\frac{{x}^{n}}{\mathrm{\Gamma }\left(n+a\right)}={x}^{1-a}{\mathrm{e}}^{x}+\frac{1-a}{\mathrm{\Gamma }\left(a\right)}\frac{1}{x}+O\left(\frac{1}{{x}^{2}}\right).$
More generally, for every nonnegative integer $N$ and every noninteger $a$.
$\sum _{n⩾0}\frac{{x}^{n}}{\mathrm{\Gamma }\left(n+a\right)}={x}^{1-a}{\mathrm{e}}^{x}+\frac{\mathrm{sin}\left(\pi a\right)}{\pi }\sum _{k=1}^{N}\frac{\mathrm{\Gamma }\left(k+1-a\right)}{{x}^{k}}+O\left(\frac{1}{{x}^{N+1}}\right).$
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KesseTher12
In fact the story starts with finding an asymptotic expansion for the function
$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{x}^{n}}{\mathrm{\Gamma }\left(n+1/2\right)}=\frac{\sqrt{\pi }\phantom{\rule{thinmathspace}{0ex}}\sqrt{x}{\mathrm{e}}^{x}\mathrm{e}\mathrm{r}\mathrm{f}\left(\sqrt{x}\right)+1}{\sqrt{\pi }}$
which is given by
$f\left(x\right)\sim \sqrt{x}\phantom{\rule{thinmathspace}{0ex}}{e}^{x}$