An Integral of Expectation in the Birthday Problem When working on a problem about an expectation in the birthday problem, I came across an integral that I am having trouble approximating.  Can anyone point me in the right direction for approximating the following integral for large $\ell$ ($m$ is a given positive integer):
$$\int_0^\infty e^{-t} t^m \left(1+\frac t\ell\right)^\ell\, dt$$
I've tried using Laplace's method, but it didn't seem to fit. Any help would be greatly appreciated.
 A: Note that 
$$\lim_{\ell\rightarrow \infty}\left(1+\frac t\ell\right)^\ell = e^t$$
So, for "large $\ell$"
$$\int_0^\infty e^{-t} t^m \left(1+\frac t\ell\right)^\ell\, dt \approx \int_0^\infty e^{-t} t^m e^{t}\, dt = \int_0^\infty t^m \, dt $$
But is this useful to your task?
ADDENDUM
The problem with the asymptotic result is that the resulting integral is divergent.
But note that the expression $\left(1+\frac t\ell\right)^\ell$ approaches its limit from below. Therefore, for any finite $\ell$, it will be smaller than $e^t$, say $e^{a_{\ell}t},\;\; a_{\ell}<1$ where we index $a$ by $\ell$ to indicate that it will depend on the value of $\ell$.
So we can write the whole expression as 
$$I=\int_0^\infty e^{-t} t^m \left(1+\frac t\ell\right)^\ell\, dt = \int_0^\infty e^{-(1-a_{\ell})t} t^m dt$$
which is a simple Mellin transform and
$$I = \frac {\Gamma(m+1)}{(1-a_{\ell})^{m+1}}$$
where $\Gamma()$ is the Gamma function.
Now I don't know what "large" means in the context of the OP's problem, but I note that for $\ell = 50$ $a_{\ell}$ reaches $0.99$, and for $\ell = 100$, $a_{\ell}$ reaches $0.995$. Still, the integral remains convergent for any finite $\ell$. Of course its actual value may be very large, (depending also on the value of $m$, since moreover the Gamma function is exponentially increasing in $m$), but it will remain finite.
A: Using the method that I give in this answer, you can find the leading term
in the asymptotics, which is:
$${\ell}^{\,m/2}\,2^{(m-1)/2}\,\Gamma\left({m+1\over 2}\right). $$
