Tight upper bound on the growth of the integral for $n$-th moment of normal distributions with increasing mean $\theta \to \infty, n\to \infty?$ Consider the growth rate integral:
$$I(\theta,n):= \int_{0}^{\infty}s^{n}e^{-(s-\theta)^2}ds, \theta > 0.$$
Thanks to  the answer given by Kavi Ram Murthy, $C_n := \frac{I(\theta,n)}{\theta^n}$ clearly can't be independent of $n.$
So my question is: what can we say about the bounds of $C_n$ in terms of $n,$ where $I(\theta,n) \le C_n \theta^n?$ In general, how to determine the optimal growth rate of $I(\theta,n)$ in terms of $\theta$ and $n$ respectively.
In order to solve this problem, I wrote:
$I(\theta, n)= \sum_{r=0}^{n} {C(n,r)}\theta^{n-r} \left[  \int _{-\theta} ^{\infty} t^r e^{-t^2}\right]$
Since we're dealing with large $\theta$ here, I replace the $-\theta$ in the above integral by just $-\infty$ instead. This gives us:
$I(\theta, n)= \sum_{r=0}^{n} {C(n,r)}\theta^{n-r} \Gamma(\frac{r+1}{2}),$ where $\Gamma(.)$ denotes the Gamma function.
Intuitively, here's where my problem starts: how to optimally combine the growth rates of the Gamma function $\Gamma(\frac{r+1}{2})$ and that of the combination symbol $C(n,r)?$ We know that the Gamma function grows like factorials, and a bit of online search found that (e.g. this note) $C(n,r)$ can have at most growth of $2^n n ^{-1/2}, $ which happens when $r$ is closest to $n/2.$ But the optimal growth in terms of $\theta$ (i.e. at the order of $\theta^n$) happens only when $r=0,$ which only gives $C(n,r)=1.$ So how do we combine the two growth rates? Below is one effort to do so, but perhaps not the optimal one:
$I(\theta, n)= \sum_{r=0}^{n} {C(n,r)}\theta^{n-r} \Gamma(\frac{r+1}{2}) \le  \sum_{r=0}^{n} {C(n,r)}\theta^{n-r} \Gamma(\frac{n}{2}) = (1 + \theta)^{n}\Gamma(\frac{n}{2}) \le C (1 + \theta)^{n} (n/2 -1)^{n/2 - 1}. \sqrt{n/2 - 1} $ (using this approximation for Gamma function). But this isn't a tight upper bound for $I(\theta,n)$ as in the second inequality, we replaced all $\Gamma(\frac{r+1}{2}) \forall r \in \{0,1,...n\}$ by $\Gamma(\frac{n}{2}),$ making this bound not to be the desired one. This is where I need help - to find a better bound.
 A: $\frac {I(\theta,n)} {\theta^{n}} \geq \int_{2\theta}^{\infty} (\frac  {s} {\theta})^{n}e^{-(s-\theta)^{2}}ds \ge 2^n \int_{2\theta}^{\infty}e^{-(s-\theta)^{2}}ds $.
By Monotone Convergence Theorem this integral goes to $\infty$ as
$ n \to \infty$. Hence no such constant $C$ independent of $n$ can exist.
A: If we rearrange a bit:
$$(s-\theta)^2=s^2-2\theta s+\theta^2$$
and then rewrite our integral:
$$\frac{1}{\theta^n}\int_0^\infty s^ne^{-(s-\theta)^2}ds=\frac{e^{-\theta^2}}{\theta^n}\int_0^\infty s^ne^{-s^2}e^{2\theta s}ds>\frac{e^{-\theta^2}}{\theta^n}\Gamma(n-1)$$
If we look at it another way:
$$\int_0^\infty s^ne^{-(s-\theta)^2}ds=\int_0^\theta s^ne^{-(s-\theta)^2}ds+\int_\theta^\infty s^ne^{-(s-\theta)^2}ds=\int_0^\theta s^ne^{-(s-\theta)^2}ds+\int_0^\infty (s+\theta)^ne^{-s^2}ds$$
$$\ge\sum_{k=0}^n{n\choose k}\theta^k\int_0^\infty s^{n-k}e^{-s^2}ds=\sum_{k=0}^n{n\choose k}\theta^k(n-k)!$$
for $n$ an integer obviously. So if n is an integer we have the bounds:
$$\frac{e^{-\theta^2}n!}{\theta^n}\le\sum_{k=0}^n{n\choose k}(n-k)!\theta^{k-n}\le\frac{1}{\theta^n}\int_0^\infty s^ne^{-(s-\theta)^2}ds$$
