Asymptotics of $\int_0^\infty \frac{x^{2z}}{\Gamma(1+z)}\,dz$ I'm interested in the asymptotics of
$$\int_0^\infty \frac{x^{2z}}{\Gamma(1+z)}\,dz$$
as $x\to\infty$. I expect the results to behave similarly to $e^{x^2}=\sum_{k\ge 0}\frac{x^{2k}}{k!}$. However, I'm not quite sure how to develop the leading asymptotics of the integrals. I first thought that for large $x$, the integral should be dominated by its integral over a small ball around the maximum of the integrand $\frac{e^{2z\log x}}{\Gamma(1+z)}$.
To find this maximum, I computed
$$f'(z)=\left(2\log(x)-\psi^{(0)}(1+z)\right)f(z)$$
with $\psi^{(0)}(z)$ the logarithmic derivative of $\Gamma(z)$. Since $f$ never vanishes, the maximum must occur at $z\in(0,\infty)$ such that $\psi^{(0)}(1+z)=2\log x$. Assuming that we will take $x\to\infty$ as well as the asymptotics $\psi(z)=\log z +O(1/z)$ for large and positive $z$, we seek to solve $2\log x= \log(1+z) +O(1/z)$. Exponentiating, we find that $x^2=1+z+O(1)$. Thus, we have that $\operatorname{arg max} f(z)\sim x^2$ is asymptotically correct for large $x$. Let $z_0=x^2$. We can now rewrite the integral as dominated by
$$\frac{e^{2x^2\log x}}{\Gamma(1+x^2)}\int_{z_0-\epsilon}^{z_0+\epsilon} e^{2(z-z_0)\log x} \frac{\Gamma(1+z_0)}{\Gamma(1+z)}\,dz.$$
However, I'm not sure what size to take $\epsilon$ as a function of $x$. I do know that the expression outside of the integral is asymptotic to
$$(2\pi)^{-1/2} \frac{e^{x^2}}{x}.$$
I'm not sure how to deal with the actual integral though. Input is much appreciated.
 A: Denote your integral by $I(x)$. As you observed, for large $x$, the main contribution to the integral will come from large values of $z$. Therefore, we first approximate $\Gamma(1+z)$ by Stirling's formula and replace $z$ by $\mathrm{e}x^2 t$. Thus,
$$
I(x) = \int_0^{ + \infty } {\frac{{x^{2z} }}{{\Gamma (1 + z)}}{\rm d}z}  \sim \frac{1}{{\sqrt {2\pi } }}\int_0^{ + \infty } {\left( {\frac{{{\rm e}x^2 }}{z}} \right)^z \frac{{{\rm d}z}}{{\sqrt z }}}  = \sqrt {\frac{{\rm e}}{{2\pi }}} x\int_0^{ + \infty } {t^{ - {\rm e}x^2 t} \frac{{{\rm d}t}}{{\sqrt t }}}
$$
for $x\gg 1$. Now we set $t = {\rm e}^s$ and obtain
$$
I(x) \sim \sqrt {\frac{{\rm e}}{{2\pi }}} x\int_{ - \infty }^{ + \infty } {\exp ( - x^2 {\rm e}^{s + 1} s){\rm e}^{s/2} {\rm d}s}  = \sqrt {\frac{{\rm e}}{{2\pi }}} x{\rm e}^{x^2 } \int_{ - \infty }^{ + \infty } {\exp ( - x^2 ({\rm e}^{s + 1} s + 1)){\rm e}^{s/2} {\rm d}s} .
$$
Applying Laplace's method about $s=-1$,
$$
\int_{ - \infty }^{ + \infty } {\exp ( - x^2 ({\rm e}^{s + 1} s + 1)){\rm e}^{s/2} {\rm d}s}  \sim \frac{1}{x}\sqrt {2\pi } {\rm e}^{ - 1/2} , \quad x\to +\infty.
$$
Hence, $I(x) \sim {\rm e}^{x^2 }$ as $x\to +\infty$.
Addendum. According to this answer
$$
\int_0^{ + \infty } {\frac{{x^{2z} }}{{\Gamma (1 + z)}}{\rm d}z}  = {\rm e}^{x^2 }  - \int_{ - \infty }^{ + \infty } {\frac{{{\rm e}^{ - x^2 {\rm e}^t } }}{{t^2  + \pi ^2 }}{\rm d}t} .
$$
