Difficult Laplace Transform Type of Integral Good afternoon.  I have the following integral that I need help integrating;
$$
\mathrm{F}\left(x\right) =
\int_{0}^{\infty}\mathrm{e}^{s\left(j - 1/x\right)}\,
\left[\mathrm{T}_{N}\left(s\right)\right]^{k - j}\,\mathrm{d}s
$$
where $\mathrm{T}_{N}\left(s\right)$ is the truncated exponential function
$$
\mathrm{T}_{N}\left(s\right) = \sum_{n = 0}^{N}\frac{s^{n}}{n!}
$$
and $j,k$ are whole numbers.
I figured that tackling this using integration by parts is the best choice, but naturally this could take a while given that my choice of $u_{1}=\mathrm{T}_{N}^{k - j}\left(s\right)$ yields a $\mathrm{d}u_{1}$ of
$$
\mathrm{d}u_{1} =
\left(k - j\right)\mathrm{T}_{N}^{k-j-1}\left(s\right)
\mathrm{T}_{N - 1}\left(s\right)\,\mathrm{d}s
$$
since $\mathrm{d}\mathrm{T}_{N}\left(s\right)/\mathrm{d}s = \mathrm{T}_{N - 1}\left(s\right)$.  This does not seem to be simplifying the problem though as the next iteration would require me to then choose my $u_{2}$ to be
$$
u_{2} =
\frac{1}{k - j}\,\frac{\mathrm{d}u_{1}}{\mathrm{d}s} =
\mathrm{T}_{N}^{k - j - 1}\left(s\right)
\mathrm{T}_{N - 1}\left(s\right)
$$
Are there any other approaches to this without this drawn out IBP method or without considering expanding the truncated exponential ?.

I have only a partial answer: it pretty much just consists of repeated integration by parts, so maybe this isn't what you're looking for.
Write $A(n_1, \dots, n_r) = \int_0^\infty e^{-as} \prod_i T_{n_i}(s) \,ds$. Then integration by parts gives 
\begin{align*}
A(n_1, \dots, n_r) 
&= -\frac{1}{a}\bigg[e^{-as}\prod_i T_{n_i}(s)\bigg]\bigg|_0^\infty + \frac{1}{a}\sum_{j : n_j > 0} \int_0^\infty e^{-as} T_{n_j - 1}(s)\prod_{i \neq j} T_{n_i}(s) \,ds \\
&= \frac{1}{a} + \frac{1}{a} \sum_{j : n_j > 0} A(n_1, \dots, n_j - 1, \dots, n_r)
\end{align*}
when all $n_i \geq 0$. 
To make things a little easier, let's take $T_n(s) = 0$ when $n < 0$ (this is consistent with the identity $T_n'(s) = T_{n-1}(s)$), so we'll have $A(n_1, \dots, n_r) = 0$ when some $n_i < 0$, and so we can write
$$A(n_1, \dots, n_r) = \frac{1}{a}\bigg(\mathbb{1}_{(n_i)} + \sum_i A(n_1, \dots, n_i - 1, \dots, n_r)\bigg)$$
for any integers $n_1, \dots, n_r$, where $\mathbb{1}_{(n_i)} = 1$ if all $n_i \geq 0$ and is $0$ otherwise. 
Letting $f(x_1, \dots, x_r) = \sum A(n_1, \dots, n_r) x_1^{n_1} \cdots x_r^{n_r}$ be the generating function of $A$, this identity gives
$$af = \frac{1}{(1-x_1) \cdots (1-x_r)} + (x_1 + \cdots + x_r)f$$
hence
\begin{align*}
f 
&= \frac{1}{(1 - x_1) \cdots (1 - x_r)(a - x_1 - \cdots - x_r)} \\
&= \frac{1}{a} \sum_{l_1, \dots, l_r, m_1, \dots, m_r} a^{-m_1 - \cdots - m_r}\binom{m_1 + \cdots + m_r}{m_1, \dots, m_r} x_1^{l_1 + m_1} \cdots x_r^{l_r + m_r} \\
&= \frac{1}{a} \sum_{n_1, \dots, n_r} \sum_{0 \leq m_i \leq n_i} a^{-m_1 - \cdots - m_r}\binom{m_1 + \cdots + m_r}{m_1, \dots, m_r} x_1^{n_1} \cdots x_r^{n_r} \\
\end{align*}
and finally,
$$A(n_1, \dots, n_r) = \frac{1}{a} \sum_{0 \leq m_i \leq n_i} a^{-m_1 - \cdots - m_r}\binom{m_1 + \cdots + m_r}{m_1, \dots, m_r} $$
(this could alternatively be found by repeatedly expanding the recurrence). 
The case relevant to the question is the one where $n_1 = \cdots = n_r = N$, $r = k-j$, $a = 1/x - j$, but I'm not sure how to simplify this sum. I've been thinking about it in terms of lattice paths in $r$ dimensions: in particular the coefficient of $a^{-t}$ is the number of lattice paths of length $t$ where we have at most $n_i$ steps in the $i$-th direction for each $i$, but that doesn't seem to help simplify any further.
As noted by DinosaurEgg, we're only interested in 
$$A(N, \dots, N) = \frac{1}{a} \sum_{0 \leq m_i \leq N} a^{-m_1 - \cdots - m_r}\binom{m_1 + \cdots + m_r}{m_1, \dots, m_r}$$
but this sum still seems difficult to simplify to me, because of the $m_i \leq N$ constraints.
