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Continuous binomial integral

Let n \in \mathbb{N}. Prove that

    \[\int_{-\infty}^{\infty} \binom{n}{x} \, \mathrm{d}x = 2^n\]

Solution

First of all,

    \begin{align*} \binom{n}{x} &= \frac{n!}{x! \left ( n-x \right )!} \\ &=\frac{n!}{\Gamma(x+1) \Gamma(n+1-x)} \\ &=\frac{n!}{\Gamma(1+x)(n-x)\left ( n-1-x \right )\cdots (1-x) \cdot \Gamma(1-x)} \\ &= \frac{n!}{x \Gamma(x)(n-x)\left ( n-1-x \right )\cdots (1-x) \Gamma(1-x) } \\ &= \frac{n!}{\pi} \cdot \frac{\sin \pi x}{(n-x)\left ( n-1-x \right )\cdots (1-x) \cdot x} \end{align*}

due to the well known formulae \Gamma(x+1) = x \Gamma(x) and \Gamma(x) \Gamma(1-x )  = \pi \csc \pi x.

However using the residue theorem we get that

    \[\frac{1}{(n-x)\left ( n-1-x \right )\cdots (1-x) \cdot x} = \sum_{k=0}^{n} \frac{1}{x-k} \cdot \frac{(-1)^k}{n!} \binom{n}{k}\]

Thus,

    \begin{align*} \int_{-\infty}^{\infty} \binom{n}{x} \, \mathrm{d}x &= \int_{-\infty}^{\infty} \frac{n!}{\pi} \cdot \frac{\sin \pi x}{(n-x)\left ( n-1-x \right )\cdots (1-x) \cdot x} \, \mathrm{d}x \\ &= \int_{-\infty}^{\infty} \frac{n!}{\pi} \cdot \sum_{k=0}^{n} \frac{\sin \pi x}{x-k} \cdot \frac{(-1)^k}{n!} \binom{n}{k} \, \mathrm{d}x \\ &=\frac{1}{\pi}\sum_{k=0}^{n} (-1)^k \binom{n}{k}\int_{-\infty}^{\infty} \frac{\sin \pi x}{x-k} \, \mathrm{d}x \\ &=\frac{1}{\pi} \sum_{k=0}^{n} (-1)^k \binom{n}{k} (-1)^k \pi \\ &= \sum_{k=0}^{n} \binom{n}{k} \\ &=2^n \end{align*}

due to the binomial theorem and the well known fact

    \[\int_{-\infty}^{\infty} \frac{\sin \pi x}{x-k} \, \mathrm{d}x = (-1)^k \pi\]

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