Quantify how small are the integrals $\int_0^N e^{-Nx}\binom{x}{N}dx $, as $N\to\infty$ While I was reading the Wikipedia entry for Gregory coefficients I've thought that should be very nice and fun calculate definite integrals involving binomial coefficients.
This is a simple exercise that I've thought after I did some experiments using Wolfram Alpha online calculator with codes like these:
integrate   e^(-35 x) Binomial[x,35] dx, from x=0 to x=35
integrate   e^(-200 x) Binomial[x,200] dx, from x=0 to x=200
I believe that the absolute value of integrals is small.

Question. (Being $N\geq 1$ integer) I would like to know how to quantify how small are these integrals. Does exist 
  $$\lim_{N\to\infty}\int_0^N e^{-Nx}\binom{x}{N}dx?$$ Alternatively, quantify $$ \left| \int_0^N e^{-Nx}\binom{x}{N}dx \right|$$ as $N$ tends to infinite.  Thanks in advance.

 A: I did some calculations and it seems like your integral, call it $f(N)$, is approximately $-(-1)^N/N^3$ as $N$ gets large. You may be able to get more accurate approximation.
A: Here is a confirmation of @Somos' computation. Notice that
$$ \binom{x}{N} = \frac{x(x-1)\cdots(x-N+1)}{N!} = \sum_{k=0}^{N} (-1)^{N-k} \frac{1}{N!}\left[ {N \atop k} \right] x^k, $$
where $\left[ {N \atop k} \right]$ is the unsigned Stirling numbers of the first kind. Plugging this back and computing,
\begin{align*}
I_N := \int_{0}^{N} e^{-Nx} \binom{x}{N} \, \mathrm{d}x
&= \sum_{k=0}^{N} (-1)^{N-k} \frac{1}{N!}\left[ {N \atop k} \right] \int_{0}^{N} x^k e^{-Nx} \, \mathrm{d}x \\
&= \sum_{k=0}^{N} (-1)^{N-k} \frac{1}{N!}\left[ {N \atop k} \right] \frac{k!}{N^{k+1}}(1 - \epsilon_{N,k}),
\end{align*}
where $\epsilon_{N,k} = \int_{N^2}^{\infty} \frac{x^k}{k!} e^{-x} \, dx $. 
Estimation of error term. We first note that there exists a constant $C_1 > 0$ satisfying
$$ \epsilon_{N,k} \leq \epsilon_{N,N} \leq C_1 e^{-N}$$
for all $1 \leq k \leq N$. The first inequality is easily proved under certain probabilistic interpretation. Let $T_1, T_2, \cdots$ be independent random variables having exponential distributions. Then we can write $ \epsilon_{N,k} = \Bbb{P}(T_1 + \cdots + T_{k+1} > N^2 ) $. This proves that $\epsilon_{N,k}$ is monotone increasing in $k$. Next, apply the substitution $ x \mapsto x + N$ to write
$$ \epsilon_{N,N} = e^{-N} \int_{N^2 - N}^{\infty} \frac{(x+N)^N}{N!} e^{-x} \, dx. $$
Now notice that $x + N \leq \frac{N}{N-1} x$ for $x \geq N^2 - N$. Using this,
$$ \epsilon_{N,N} \leq
e^{-N} \left(\frac{N}{N-1}\right)^N \int_{0}^{\infty} \frac{x^N}{N!} e^{-x} \, dx
\leq C_1 e^{-N} $$
for some $C_1 > 0$. This bound is somewhat crude, but it is enough for our purpose. Next we recall the following identity
$$ \sum_{k=0}^{N} \left[ {N \atop k} \right] = N! $$
From this, we have
$$ \left| \sum_{k=0}^{N} (-1)^{N-k} \frac{1}{N!}\left[ {N \atop k} \right] \frac{k!}{N^{k+1}} \epsilon_{N,k} \right|
\leq C_1e^{-N}. $$
Extracting the leading term. We remark the following identities: if $N \geq 1$, then
$$ \left[ {N \atop 0} \right] = 0, \qquad \left[ {N \atop 1} \right] = (N-1)!, \qquad \left[ {N \atop 2} \right] = (N-1)!N_{N-1}. $$
Since $k!/N^k$ is decreasing in $k$, for $k \geq 3$ we have $ k!/N^k \leq 6/N^3$. So
\begin{align*}
&\sum_{k=0}^{N} (-1)^{N-k} \frac{1}{N!}\left[ {N \atop k} \right] \frac{k!}{N^{k+1}} \\
&\hspace{2em} = (-1)^{N-1} \frac{1}{N^3} + (-1)^{N-2} \frac{2H_{N-1}}{N^4} + \mathcal{O}\left( \sum_{k=3}^{N} \frac{1}{N!}\left[ {N \atop k} \right] \frac{6}{N^{4}} \right) \\
&\hspace{4em} = (-1)^{N-1} \frac{1}{N^3} + (-1)^{N-2} \frac{2H_{N-1}}{N^4} + \mathcal{O}\left( \frac{1}{N^{4}} \right).
\end{align*}
Conclusion. Combining both estimates, we obtain
$$I_N = (-1)^{N-1} \frac{1}{N^3} + (-1)^{N-2} \frac{2\log N}{N^4} + \mathcal{O}\left( \frac{1}{N^{4}} \right). $$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Note that $\ds{\pars{~\mbox{with}\ \verts{z} < 1~}}$
\begin{align}
\int_{0}^{N}\expo{-Nx}{x \choose N}\,\dd x & =
\bracks{z^{N}}\int_{0}^{N}\expo{-Nx}\pars{1 + z}^{x}\,\,\dd x =
\bracks{z^{N}}\int_{0}^{N}\bracks{\expo{-N}\pars{1 + z}}^{x}\,\,\dd x
\\[5mm] & =
\bracks{z^{N}}{1 \over \ln\pars{\expo{-N}\pars{1 + z}}}
\int_{0}^{N}\partiald{\bracks{\expo{-N}\pars{1 + z}}^{x}}{x}\,\,\dd x
\\[5mm] &=
\bracks{z^{N}}
{\bracks{\expo{-N}\pars{1 + z}}^{N} - 1 \over -N + \ln\pars{1 + z}}
\\[5mm] & =
-\expo{-N^{2}}
\braces{\bracks{z^{N}}{\pars{1 + z}^{N} \over N - \ln\pars{1 + z}}} +
\braces{\bracks{z^{N}}{1 \over N - \ln\pars{1 + z}}}
\end{align}

Can you take it from here ?.
  I'm still making some 'checkings' !!!.

