What are the asymptotics of $\int_0^\infty \frac{x^z}{\Gamma(1+z^2)}\,dz$ for large $x$? First, let's fix some notation. I'll write $f(x)=o(g(x))$ if $\lim_{x\to\infty} \left|\frac{f(x)}{g(x)}\right|=0$. I'll also write $g(x)=\omega(f(x))$ if $f(x)=o(g(x))$, i.e. $\limsup_{x\to\infty} \left|\frac{g(x)}{f(x)}\right|=\infty$. I'll say $f(x)\sim g(x)$ as $x\to\infty$ if $f(x)=g(x)+o(g(x))$ for all large enough $x$, i.e. $f(x)/g(x)=1+o(1)$ for all large enough $x$.

We would like to study the asymptotics of the integral convergent for all real $x$,
$$\mathcal{S}(x)=\int_0^\infty \frac{x^z}{\Gamma(1+z^2)}\,dz$$
for $x\to\infty$. The integrand is rapidly decaying so we expect $\mathcal{S}$ to grow slowly at infinity in comparison with integrals with slower decay of integrands such as
$$\mathcal{A_\alpha}(x)=\int_0^\infty \frac{x^z}{\Gamma(1+z/\alpha)}\,dz$$
for arbitrary $\alpha>0$. The above integral can be shown to satisfy $\mathcal{A}_\alpha(x)\sim \exp\big(x^\alpha+\log\alpha\big)$ for $x\to\infty$ by similar methods as I will explore below to take care of $\mathcal{S}(x)$. We thus expect our integral $\mathcal{S}$ to satisfy $\mathcal{S}(x)=o\big(\mathcal{A}_\alpha(x)\big)$ for every $\alpha>0$. What is the asymptotic behavior of $\mathcal{S}(x)$ as $x\to\infty$?
 A: Indeed, we will find that the asymptotics of $\mathcal{S}(x)$ have a rich and interesting structure, made of a combination of several different growth hierarchies, unlike the asymptotics for $\mathcal{A}_\alpha(x)$, which are comparatively simpler in nature.
The first step is use a Laplace expansion in the exponent. Write
$$\mathcal{S}(x)=\int_0^\infty \frac{x^z}{\Gamma(1+z^2)}\,dz.$$
For large $x$, the integrand will be dominated by the behavior of the integrad near its maximum. We first write the integrand as
$$\exp\left(z\log(x)-\log \Gamma\big(1+z^2\big)\right):=\exp(f(z))$$
and work to find the maximum of $f(z)$. We compute
$$f'(z)=\log(x)- 2z\,\psi^{(0)}(1+z^2)$$
with $\psi^{(0)}$ the log-derivative of $\Gamma$. Of course, we'd like to compute the $z$ such that $f'(z)=0$. We will need to use the asymptotics of the digamma function to obtain an asymptotically correct expression for $z$. We have that
$$\psi^{(0)}(1+z^2)=2\log z +O\left(\frac{1}{z^2}\right),$$
hence
$$\frac{1}{4}\log(x)=z\log(z)+O\left(\frac{1}{z}\right)$$
Taking the Lambert function of both sides, then exponentiating, keeping track of error terms, and finally asymptotically expanding the resulting expression yields
$$z_0=\frac{\frac{1}{4}\log x}{\log\log x}\big(1+o(1)\big).$$
Now, expanding $f(z)$ in the exponent up to its quadratic Taylor polynomial, extending the range to all the real line, and integrating out the resulting gaussian, we find that
$$
\begin{aligned}
\mathcal{S}(x) &= \sqrt{2\pi}\,\frac{\exp \big(z_0\log x\big)}{\Gamma(1+z_0^2)}|f''(z_0)|^{-1/2}\big(1+o(1)\big).
\end{aligned}
$$
We now use the asymptotics of the Gamma function to obtain
$$
\begin{aligned}
\mathcal{S}(x) &= \frac{|f''(z_0)|^{-1/2}}{z_0}\frac{\exp\big(z_0\log x\big)}{ \exp\big(2 z_0^2\log z_0 -z_0^2\big)}\big(1+o(1)\big).
\end{aligned}
$$
We now compute that $f''(z_0)=2\log (z_0) \big(1+o(1)\big)$. Putting everything together, we finally arrive at
$$\mathcal{S}(x)=2\sqrt{2} \,\frac{\log^{1/2}\log x}{\log x}\exp\left(\frac{1}{8}\frac{\log^2 x}{\log\log x}\big(1+o(1)\big)\right)\big(1+o(1)\big).$$
With a bit more effort, one can also obtain more detailed asymptotics
