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.

  • $\begingroup$ Are you sure you mean $\binom{x}{N}$ instead of $\binom{N}{x}$? $\endgroup$ May 7, 2017 at 20:19
  • $\begingroup$ Many thanks for your attention and good afternoon @AntonioVargas . I've created this exercise as comparison with an integral representation related to Gregory coefficients (see the cited Wikipedia's entry), addtionally you can see the syntaxis of binomial numbers in Wolfram Alpha Language $\endgroup$
    – user243301
    May 7, 2017 at 20:34
  • $\begingroup$ I presume that my calculations with Wolfram Alpha are rights, but I know that usually binomial coefficients have integers arguments. Many thanks one more time @AntonioVargas $\endgroup$
    – user243301
    May 7, 2017 at 20:40
  • 1
    $\begingroup$ Possibly useful: $e^{-Nx}\binom{x}{N}dx \sim \frac{(-1)^{N-1}}{N}xe^{-Nx}$ as $x\approx0$. $\endgroup$
    – πr8
    May 7, 2017 at 23:10
  • 1
    $\begingroup$ I second @Somos' computation that my heuristic computation gives the estimate $$ \int_{0}^{N} e^{-Nx}\binom{x}{N} \, dx = (-1)^{N-1} \bigg( \frac{1}{N^3} - \frac{2H_{N-1}}{N^4} + \frac{3(H_{N-1}^2 - H_{N-1}^{(2)})}{N^5} + \cdots \bigg). $$ I may be able to improve my heuristics to a rigorous asymptotics if some nice estimates on the Stirling's numbers of the 1st kind are available to me. $\endgroup$ May 8, 2017 at 1:42

3 Answers 3


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.

  • $\begingroup$ Many thanks for your calculation and attention in my question, is much appreciated. $\endgroup$
    – user243301
    May 8, 2017 at 4:28

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

$$I_N = (-1)^{N-1} \frac{1}{N^3} + (-1)^{N-2} \frac{2\log N}{N^4} + \mathcal{O}\left( \frac{1}{N^{4}} \right). $$

  • $\begingroup$ Many thanks to you and the other users, especially to Somos. $\endgroup$
    – user243301
    May 8, 2017 at 11:15

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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' !!!.

  • $\begingroup$ Once again you, a magician of mathematics and integrals! I am going to study your answer, many thanks. $\endgroup$
    – user243301
    May 8, 2017 at 4:32
  • $\begingroup$ @user243301 Thanks. It still looks hard. $\endgroup$ May 8, 2017 at 5:13

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