Asymptotics of an integral in the coupon collector's problem I am interested in asymptotics of the integral $$I(h,n)=\int_0^\infty t(u) e^{-un}du$$ as $h\to \infty$ ($n$ is a fixed positive integer) where $t(u)$ is the functional inverse of the function
$$u(t)=-\ln\left(1-e^{-t}\left(1+t+\frac{t^2}{2}+\dots+\frac{t^{h-1}}{(h-1)!}\right)\right). $$ This integral comes up as an expectation in the coupon collector's problem. Based on numerical evidence, it seems that $I(h,n)$ is nearly linear in $h$. If someone could provide a quick derivation of the asymptotics of $I(h,n)$ as $h\to\infty$, that would be great.
Note that $1-e^{-t}\left(1+t+\frac{t^2}{2}+\dots+\frac{t^{h-1}}{(h-1)!}\right)$ increases from $0$ to $1$ for $t\in(0,\infty)$, so this function is invertible.
 A: Denote $\exp t$ by $E$ and the $m$th partial sum of its Taylor series by $E_m$ so that \begin{align}I(h,n)&=\int_0^\infty t(1-E^{-1}E_{h-1})^n\cdot\frac d{dt}(-\log(1-E^{-1}E_{h-1}))\,dt\\&=-\int_0^\infty tE^{-n}(E-E_{h-1})^n\cdot\frac{-E^{-1}E_{h-1}+E^{-1}E_{h-2}}{1-E^{-1}E_{h-1}}\,dt\\&=\frac1{(h-1)!}\int_0^\infty t^hE^{-n}(E-E_{h-1})^{n-1}\,dt\\&=\frac1{(h-1)!}\int_0^\infty t^{hn}e^{-nt}\left(\sum_{i=0}^\infty\frac{t^i}{(h+i)!}\right)^{n-1}\, dt.\end{align} Writing $$\left(\sum_{i=0}^\infty\frac{t^i}{(h+i)!}\right)^{n-1}=\sum_{j=0}^\infty c_jt^j$$ such that $c_0=1/(h!)^{n-1}$ and $c_j=h!/j\sum_{k=1}^j(kn-j)c_{j-k}/(h+k)!$ for all $j>0$, we obtain \begin{align}I(h,n)&=\frac1{(h-1)!}\int_0^\infty\sum_{j=0}^\infty c_jt^{j+hn}e^{-nt}\,dt\\&=\frac1{(h-1)!}\sum_{j=0}^\infty\frac{c_j(j+hn)!}{n^{j+hn+1}}\\&=\frac1{(h-1)!}\left(\frac{(hn)!}{(h!)^{n-1}n^{j+hn+1}}+\sum_{j=1}^\infty\frac{c_j(j+hn)!}{n^{j+hn+1}}\right)\\&=\frac hn\left(\frac{(hn)!}{(h!)^nn^{hn}}+\sum_{j=1}^\infty\frac{c_j(j+hn)!}{n^{j+hn}h!}\right).\end{align}
