10
$\begingroup$

Evaluate the integral $\int_{-\infty}^{\infty}\binom{n}{x}dx$.

This question came in Cambridge Integration Bee and I have no clue what to do in this.

I rewrote $\binom{n}{x}$ as $\frac{n!}{{x!}{(n-x)!}}$ but I don't know how to integrate those factorials. Also what to do if $n$ isn't an integer as no information regarding it is given?

$\endgroup$
7
  • 3
    $\begingroup$ I think what is meant is $$ \int_{-\infty}^{\infty} \frac{\Gamma(n+1)}{\Gamma(x+1)\Gamma(n-x+1)}dx $$ since $k!=\Gamma(k+1)$ and $\binom{n}{x}=\frac{n!}{x!(n-x)!}$. $\endgroup$ Commented Apr 24 at 7:41
  • 1
    $\begingroup$ P.S. And the answer is probably $2^n$.... $\endgroup$ Commented Apr 24 at 7:44
  • $\begingroup$ @vanderWolf what should I do after gamma function? $\endgroup$ Commented Apr 24 at 8:33
  • $\begingroup$ @MathStackexchangeIsNotSoBad If $n$ is an integer, you can use the identities $\Gamma (x+1)=x\Gamma (x)$ and $\Gamma (x)\Gamma (1-x)=\frac{\pi}{\sin \pi x}$ $\endgroup$
    – Cyankite
    Commented Apr 24 at 8:43
  • $\begingroup$ @Cyankite Hey how did you arrive at the last identity...ive never heard of it $\endgroup$ Commented Apr 24 at 8:45

3 Answers 3

26
$\begingroup$

Since $$\dbinom {n}{x}=\dbinom{n-1}{x}+ \dbinom{n-1}{x-1}$$ $$I_n:= \int_{-\infty} ^ \infty \dbinom{n}{x}dx= \color{red}{\int_{-\infty} ^ \infty \dbinom{n-1}{x}dx}+ \color{blue}{\int_{-\infty} ^ \infty \dbinom{n-1}{x-1}dx}$$

And by using a substitution $x=t-1$ then replacing $t$ with $x$ one gets $$I_{n-1}= \color{red}{\int_{-\infty} ^ \infty \dbinom{n-1}{x}dx}= \color{blue}{\int_{-\infty} ^ \infty \dbinom{n-1}{x-1}dx}$$

then $I_n = 2I_{n-1}$ so $I_n= 2^n I_0 $

$$I_0= \int_{-\infty} ^ \infty \dbinom{0}{x}dx= \int_{-\infty} ^ \infty \frac{1}{(x)!(-x)!}dx$$

$$\Gamma(1+x)=x!$$ $$\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$$

$$I_0 =\int_{-\infty} ^ \infty \frac{\sin(\pi x)}{\pi x}dx= \frac{1}{\pi} \int_{-\infty} ^ \infty \frac{\sin( x)}{x}dx $$

and since $\displaystyle \int_{-\infty} ^ \infty \frac{\sin( x)}{x}dx = \pi$ so $I_0 =1 $ and $I_n = 2^n $

$\endgroup$
6
  • $\begingroup$ Pls prove that formula of pi/sin(pi x)... How did that come up?? $\endgroup$ Commented Apr 24 at 8:57
  • $\begingroup$ Nice answer. What if $n$ is not an integer? $\endgroup$
    – Cyankite
    Commented Apr 24 at 9:01
  • $\begingroup$ @MathStackexchangeIsNotSoBad This is a known formula there are may different to prove this see this for a proof $\endgroup$
    – pie
    Commented Apr 24 at 9:03
  • $\begingroup$ @Cyankite I don't think a closed form exists if $n$ is not an integer. $\endgroup$
    – pie
    Commented Apr 24 at 9:04
  • 2
    $\begingroup$ This is more than elegant ! (+1) for sure. The formula seems to work for non integer values of $n$. $\endgroup$ Commented Apr 24 at 10:01
12
$\begingroup$

You get the result from one of the Ramanujan's Papers, formula 1.2, by setting $n=0$, $\alpha=0$, $\beta=n+2$, and using the fact that $$ \binom{n}{x}=\frac{\Gamma(n+1)}{\Gamma(n-x+1)\Gamma(x+1)} $$ so that \begin{align*} \int \binom{n}{x} dx &=\int \frac{\Gamma(n+1)}{\Gamma(n+2-y)\Gamma(y)} dy\\ &=\frac1{n+1}\int \frac{\Gamma(n+2)}{\Gamma(n+2-y)\Gamma(y)} dy\\ &=\frac1{n+1}\int \frac{dy}{\mathrm{B}(n+2-y,y)}. \end{align*}

$\endgroup$
4
  • 1
    $\begingroup$ How to integrate that beta-function integral? And your link is not available to me(HTTP ERROR 502) $\endgroup$
    – Cyankite
    Commented Apr 24 at 9:03
  • 1
    $\begingroup$ @Cyankite, could you check now for it seems to work (at least for me)? $\endgroup$ Commented Apr 25 at 11:59
  • $\begingroup$ @Cyankite, it's a "non-secure" connection, you should let your browser open it. $\endgroup$ Commented Apr 26 at 7:22
  • $\begingroup$ @vanderWolf It is OK now:) $\endgroup$
    – Cyankite
    Commented Apr 26 at 7:42
9
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\int_{-\infty}^{\infty}{n \choose x}\dd x} = \int_{-\infty}^{\infty}\bracks{\int_{-\pi}^{\pi}{\pars{1 + \expo{\ic\phi}}^{n} \over \expo{\ic x\phi}}{\dd\phi \over 2\pi}}\dd x \\[5mm] = & \ \int_{-\pi}^{\pi}{\pars{1 + \expo{\ic\phi}}^{n}\,\ \overbrace{\int_{-\infty}^{\infty}\expo{-\ic\phi x}\, {\dd x \over 2\pi}}^{\ds{\delta\pars{\phi}}}}\,\ \,\dd\phi = \pars{1 + \expo{\ic 0}}^{n} = \bbx{\color{#44f}{2^{n}}} \\ & \end{align}

$\endgroup$
3
  • 2
    $\begingroup$ Can you explain the formula for $\binom nx$? $\endgroup$
    – user
    Commented Apr 29 at 12:23
  • 1
    $\begingroup$ @user See en.wikipedia.org/wiki/Egorychev_method $\endgroup$
    – user170231
    Commented Apr 29 at 17:01
  • 3
    $\begingroup$ @user170231 I know this method being applied to integer $x$. Nothing else can be found in the reference you gave. So some additional work / explanation is needed. $\endgroup$
    – user
    Commented Apr 29 at 19:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .