Another Sophomore's Dream: $\int_{-\infty}^{\infty}\binom{n}{x}dx=\sum_{i=0}^{n} \binom{n}{i}$

Question

I found an identity $$\int_{-\infty}^{\infty}\binom{n}{x}dx=\sum_{i=0}^{n} \binom{n}{i}$$

where LHS can be calculated by the Reflection relation and Dirichlet integral.

The result is $2^n$, which is apparently equal to RHS.

This identity is in the form "integration=summation", which is similar to the "Sophomore's dream", $\int_0^1 x^{-x}dx = \sum_{n=1}^\infty n^{-n}$.

Is this another coincidence? If not, what is the reason behind it to make it true.

Answer 1

A very heuristic argument shows that this is not a mere coincidence. Indeed, we begin with a more general setting and later discuss how this is related to OP's identity.

Let $f(\theta)$ be a locally integrable $1$-periodic function on $\mathbb{R}$. Define $\hat{f}(x)$ by

$$ \hat{f}(x) = \int_{-1/2}^{1/2} f(\theta) e^{-2\pi i x \theta} \, \mathrm{d}\theta. $$

Note that $\hat{f}(x)$ is precisely the Fourier series coefficient of $f$ when $x \in \mathbb{Z}$, but here we allow $x$ to take values in $\mathbb{R}$. (Alternatively speaking, we are considering a class of functions $\hat{f}(x)$ whose Fourier transform is supported on $[-\frac{1}{2}, \frac{1}{2}]$.)

We also note that, informally, we have

$$ \int_{-\infty}^{\infty} e^{-2\pi i x \theta} \, \mathrm{d}x = \delta_0(\theta). $$

This is just a formal way of stating the Fourier inversion theorem. Combining these together, without caring mathematical rigor, we expect to have

\begin{align*} \int_{-\infty}^{\infty} \hat{f}(x) \, \mathrm{d}x &= \int_{-1/2}^{1/2} f(\theta) \left( \int_{-\infty}^{\infty} e^{-2\pi i x \theta} \, \mathrm{d}x \right) \, \mathrm{d}\theta \\ &= \int_{-1/2}^{1/2} f(\theta) \delta_0(\theta)\, \mathrm{d}\theta = f(0) = \sum_{k=-\infty}^{\infty} \hat{f}(k). \end{align*}

Of course, we are ignoring several technical issues, such as when we can apply Fourier inversion pointwise, when the Fourier series converges to the original function pointiwse, etc.

However, we can show that all these issues can be properly handled in the case $f(\theta) = (1 + e^{2\pi i\theta})^n$ together with the following theorem:

Theorem. For any $x \in \mathbb{C}$ and $n \in \mathbb{C}$ with $\operatorname{Re}(n) > -1$, we have $$ \binom{n}{x} = \int_{-1/2}^{1/2} (1 + e^{2\pi i\theta})^n e^{-2\pi i x \theta} \, \mathrm{d}\theta. $$

This is easily proved by expanding $(1+z)^n$ using binomial theorem/series when $x \in \mathbb{Z}$. What is remarkable is that this equality continues to hold for any $x \in \mathbb{C}$. Let me omit the proof of this theorem, but if I remember correctgly, this has been discussed several times in this community.