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.

Edit: Math Stack Exchange cross-post