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

Question

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?

Answer 1

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 $

Answer 2

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*}

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}