# Asymptotics of $\int_0^\infty \frac{x^{2z}}{\Gamma(1+z)}\,dz$ for large $x$

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

• Consider $g(x)=\int_0^\infty\frac{x^t}{\Gamma(t+1)}dt$. You differentiate and then split the integral at 1, you find that $g(x)/e^x$ has bounded derivative.
– user473423
Dec 31, 2022 at 8:06
• Ramanujan observed that $$\int_0^{ + \infty } {\frac{{x^{2z} }}{{\Gamma (1 + z)}}{\rm d}z} = {\rm e}^{x^2 } - \int_{ - \infty }^{ + \infty } {\frac{{\exp ( - x^2 {\rm e}^t )}}{{t^2 + \pi ^2 }}{\rm d}t} .$$
– Gary
Feb 8 at 6:13

## 2 Answers

$$\newcommand\Ga\Gamma$$Let $$g(x)$$ denote your integral. Then $$g(x)\sim e^{x^2} \tag{1}\label{1}$$ as $$x\to\infty$$.

Proof:
$$g(x)=\int_0^\infty dz\,\frac{x^{2z}}{\Ga(1+z)}. \tag{1.5}\label{1.5}$$ So, for $$x>0$$, $$g'(x)=\int_0^\infty dz\,\frac{2zx^{2z-1}}{\Ga(1+z)} =\int_0^\infty dz\,\frac{2x^{2z-1}}{\Ga(z)} \\ =2\int_{-1}^\infty dt\,\frac{x^{2t+1}}{\Ga(1+t)} =2x\int_{-1}^\infty dz\,\frac{x^{2z}}{\Ga(1+z)}.$$ So (and this is the key), $$g$$ is a solution of the ODE $$g'(x)=2xg(x)+h(x),$$ where $$h(x):=2x\int_{-1}^0 dz\,\frac{x^{2z}}{\Ga(1+z)}.$$ Also, $$g(0)=0$$. So, $$g(x)=e^{x^2}G(x), \tag{2}\label{2}$$ where $$G(x):=\int_0^x du\, e^{-u^2}h(u). \tag{3}\label{3}$$ As $$x\to\infty$$, $$G(x)\to\int_0^\infty du\, e^{-u^2}h(u) =\int_0^\infty du\, e^{-u^2}2u\int_{-1}^0 dz\,\frac{u^{2z}}{\Ga(1+z)} \\ =\int_0^\infty dv\, e^{-v}\int_{-1}^0 dz\,\frac{v^z}{\Ga(1+z)} =\int_{-1}^0 \frac{dz}{\Ga(1+z)}\int_0^\infty dv\, v^z e^{-v} \\ =\int_{-1}^0 \frac{dz}{\Ga(1+z)}\,\Ga(1+z)=1.$$ Now \eqref{1} follows from \eqref{2}. $$\quad\Box$$

Remark: Note that $$-\Ga'(1)$$ is Euler's $$\gamma>0$$. Therefore and because $$\Ga$$ is log convex on $$(0,\infty)$$, we see that $$\Ga$$ is decreasing on $$(0,1]$$. So, for $$x\ge1$$ we have $$h(x)<\frac{2x}{\Ga(1)}\int_{-1}^0 dz\,x^{2z}<2x.$$ So, by \eqref{2}--\eqref{3}, for $$x\ge1$$ the relative error of the asymptotic approximation \eqref{1} is $$0 which goes to $$0$$ very fast as $$x\to\infty$$.

This bound on the relative error seems hard (if at all possible) to get by the Laplace method/Watson lemma applied to the original integral expression \eqref{1.5} of $$g(x)$$, which gives an asymptotic expansion in integral powers of $$x$$.

However, one can apply the Watson lemma to the expression \eqref{2} of $$g(x)$$ to get the following asymptotic expansion of the absolute error of approximation \eqref{1}, in integral powers of $$\ln x$$: $$E(x):=e^{x^2}-g(x) =e^{x^2}\int_x^\infty du\, e^{-u^2}h(u) \sim\sum_{k\ge0}\frac{(-1)^k c_k}{(2\ln x)^{k+1}} \tag{4}\label{4}$$ as $$x\to\infty$$, where $$c_k:=\frac{d^k}{dz^k}\frac1{\Ga(1+z)}\big|_{z=0}$$, so that $$c_0=1$$. $$\quad\Box$$

For an illustration, below are the graphs $$\{(x,g(x)/e^{x^2})\colon0\le x\le 4\}$$ (solid black), $$\{(x,E(x)/E_0(x))\colon5\le x\le100\}$$ (solid red), $$\{(x,E(x)/E_1(x))\colon5\le x\le100\}$$ (solid green), and $$\{(x,E(x)/E_2(x))\colon5\le x\le100\}$$ (solid blue), where (cf. \eqref{4}) $$E_m(x):=\sum_{k=0}^m\frac{(-1)^k c_k}{(2\ln x)^{k+1}}$$:

The saddle point of the integrand is at $$z=x^2$$ for large $$x$$, this gives the saddle point approximation $$\int_0^\infty \frac{x^{2z}}{\Gamma(1+z)}\,dz\rightarrow \frac{x^{2x^2}}{\Gamma(1+x^2)}.$$ You may then obtain higher order terms by performing the gaussian integral around the saddle point,$$^\ast$$ which gives a pre-exponential factor $$\sqrt{2\pi}\,x$$. Substituting also the large-$$x$$ asymptotics of the $$\Gamma$$ function, I arrive at $$\int_0^\infty \frac{x^{2z}}{\Gamma(1+z)}\,dz\rightarrow e^{x^2}.$$

The $$e^{x^2}$$ approximation is already quite good for relatively small values of $$x$$; see the plot (blue curve = exact, orange curve = approximation):

$$^\ast$$ The integrand $$e^{F(z)}$$ with $$F((z)=2z\ln x-\ln\Gamma(1+z)$$ is expanded to second order about the saddle point $$z_0$$, where the first derivative vanishes, $$F'(z_0)=0$$. For $$x\gg 1$$ one has $$z_0=x^2$$ and $$F(z)\approx2z_0\ln x-\ln\Gamma(1+z_0)-\tfrac{1}{2}x^{-2}(z-z_0)^2.$$ The integral $$\int_{-\infty}^\infty e^{F(z)}\,dz$$ then gives the asymptotic approximation $$e^{x^2}$$.

• Re: The discriminatory power of graphical comparisons, have you tried plotting $e^{x^2}$ or $x^{-1} e^{x^2}$ just for fun as well? Dec 31, 2022 at 14:54
• Incidentally, I cross posted the question to math.stack.exchange (with the link to the post now in my mathoverflow post), and an answer there finds the asymptotics to be purely $e^{x^2}$. Dec 31, 2022 at 22:19
• What tool do you use for plotting? Dec 31, 2022 at 22:29
• the plots are Mathematica output... Dec 31, 2022 at 22:53
• Based on your Taylor expansion I would expect a prefactor of $\sqrt{2\pi} x$ rather than $\sqrt{2\pi x}$, which should explain the discrepancy with the other answers. Jan 1, 2023 at 2:29