Why is this function a really good asymptotic for $\exp(x)\sqrt{x}$ $$f(x)=\sum_{n=0}^{\infty} a_n x^n\;\;\;\;\; a_n = \frac{1}{\Gamma(n+0.5)}$$
Why is this entire function a really good asymptotic for $\exp(x)\sqrt{x}$, where for large positive numbers, $f(x)\exp(-x) \approx \sqrt{x}$?
As |x| gets larger, the error term is asymptotically $f(x)-\exp(x)\sqrt{x} \approx \frac{1}{x\cdot\Gamma(-0.5)}$, and the error term for $f(x)\exp(-x) - \sqrt{x} \approx \frac{\exp(-x)}{x\cdot \Gamma(-0.5)}$.  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:
$$a_n = \oint x^{-n} f(x) = \int_{-\pi}^{\pi} \frac{1}{2\pi} (re^{-ix})^{-n} f(re^{ix}) )\; \mathrm{d}x\;\;$$ 
The conjecture is that this is an equivalent definition for $a_n$, where $f(x) \mapsto \exp(x)\sqrt{x}$ and $x \mapsto re^{ix}$.
$$a_n =\lim_{r\to\infty} \int_{-\pi}^{\pi} \frac{1}{2\pi} (re^{-ix})^{-n}\exp(re^{ix})\sqrt{re^{ix}})\; \mathrm{d}x = \frac{1}{\Gamma(n+0.5)}   $$ 
 A: Repeated integrations by parts show that, for every positive $a$ and $x$, $$\int_0^x\mathrm e^{-t}t^{a-1}\mathrm dt=\Gamma(a)\mathrm e^{-x}\sum_{n\geqslant0}\frac{x^{n+a}}{\Gamma(n+a+1)}.$$ When $x\to\infty$, the LHS converges to $\Gamma(a)$, hence the series in the RHS is equivalent to $\mathrm e^x$. Now, $$\sum_{n\geqslant0}\frac{x^{n}}{\Gamma(n+a)}=\frac1{\Gamma(a)}+x^{1-a}\sum_{n\geqslant0}\frac{x^{n+a}}{\Gamma(n+a+1)}$$ hence $$\sum_{n\geqslant0}\frac{x^{n}}{\Gamma(n+a)}\sim x^{1-a}\mathrm e^x.$$ For $a=\frac12$, this is the result mentioned in the question.
An exact formula using the incomplete gamma function $\gamma(a,\ )$ (that is, the LHS of the first identity in this answer) is $$\sum_{n\geqslant0}\frac{x^{n}}{\Gamma(n+a)}=\frac{\gamma(a,x)}{\Gamma(a)} x^{1-a}\mathrm e^x+\frac1{\Gamma(a)}.$$
Edit: ...And this approach yields the more precise expansion, also mentioned in the question, $$\sum_{n\geqslant0}\frac{x^{n}}{\Gamma(n+a)}=x^{1-a}\mathrm e^x+\frac{1-a}{\Gamma(a)}\frac1x+O\left(\frac1{x^2}\right).$$ More generally, for every nonnegative integer $N$ and every noninteger $a$,  $$\sum_{n\geqslant0}\frac{x^{n}}{\Gamma(n+a)}=x^{1-a}\mathrm e^x+\frac{\sin(\pi a)}{\pi}\sum_{k=1}^N\frac{\Gamma(k+1-a)}{x^k}+O\left(\frac1{x^{N+1}}\right).$$
A: In fact the story starts with finding an asymptotic expansion for the function

$$ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{\Gamma(n+1/2)}= {\frac {\sqrt {\pi }\,\sqrt {x}{{\rm e}^{x}}
{{\rm erf}\left(\sqrt {x}\right)}+1}{\sqrt {\pi }}}$$

which is given by

$$ f(x) \sim \sqrt{x}\, e^{x} $$

