Can you prove/disprove $\int_{-\infty}^{\infty}{\frac{x!}{(x-t)!(t)!}dt} = 2^x$ I was studying something about binomial coefficients then I thought that whether I can expand the formulae $\sum_{i=0}^n {\binom{n}{i}}= 2^n$ to integrals. So I tried plugging in some values in calculator, and here are my results:-
★Sometimes the ratio is above 1
$$\frac {\int_{-170}^{170}{\frac{(3.9)!}{(3.9-t)!(t)!}dt}} {2^{3.9}} = 1.0000008068$$
★Sometimes the ratio is below 1
$$\frac {\int_{-170}^{170}{\frac{(3.1)!}{(3.1-t)!(t)!}dt}} {2^{3.1}} = 0.999999987055$$
★It is also worth noting that the ratio isn't close to 1 when we plug in negative values, may be because the function in integration diverges at both ends but I think it should converge when we go towards infinity
$$\frac {\int_{-100}^{100}{\frac{(-1.8)!}{(-1.8-t)!(t)!}dt}} {2^{-1.8}} = 221.136809313$$
 A: I think this is a rather well-known result, although my quick googling does not reveal any references. So here is my own proof of the following general result:

Theorem. Let $\alpha$ be a complex number away from negative integers, and denote by
$$
\binom{\alpha}{z} := \frac{\alpha!}{z!(\alpha-z)!} = \frac{\Gamma(\alpha+1)}{\Gamma(z+1)\Gamma(\alpha-z+1)}
$$
the extended binomial coefficient. Then for $\Re(\alpha) > 0$ and $x \in \mathbb{R}$, we have
$$
\binom{\alpha}{x}
= \sum_{n=0}^{\infty} \binom{\alpha}{n}\frac{\sin \pi(x-n)}{\pi(x-n)}
\tag{1}
$$
  and
  $$ \int_{\mathbb{R}}\binom{\alpha}{x} e^{-2\pi i \xi x} \, \mathrm{d}x
= \begin{cases}
\big(1 + e^{-2\pi i \xi}\big)^{\alpha} & |\xi| < \frac{1}{2} \\
0 & \text{otherwise}.
\end{cases}
\tag{2} $$

Remark. Your question corresponds to $\xi = 0$ case.
Proof. Define $I(\alpha, x)$ by
$$
I(\alpha, x)
= \int_{-\frac{1}{2}}^{\frac{1}{2}} \big(1 + e^{2\pi i \xi}\big)^{\alpha} e^{-2\pi i x \xi} \, \mathrm{d}\xi.
$$
We begin by proving that $I(\alpha, x)$ equals the right-hand side of $\text{(1)}$. Indeed, by the extended binomial theorem,
\begin{align*}
I(\alpha, x)
&= \int_{-\frac{1}{2}}^{\frac{1}{2}} \biggl( \sum_{n=0}^{\infty} \binom{\alpha}{n} e^{2\pi n i \xi} \biggr)e^{-2\pi i x \xi} \, \mathrm{d}\xi
= \sum_{n=0}^{\infty} \binom{\alpha}{n} \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{2\pi (n-x) i \xi} \, \mathrm{d}\xi \\
&= \sum_{n=0}^{\infty} \binom{\alpha}{n} \left[ \frac{e^{2\pi (n-x) i \xi}}{2\pi (n-x)} \right]_{-\frac{1}{2}}^{\frac{1}{2}}
= \sum_{n=0}^{\infty} \binom{\alpha}{n}\frac{\sin \pi(x-n)}{\pi(x-n)}
\end{align*}
This calculation is fully justified once we establish the validity of switching the order of the integration and summation. This is accomplished either by the Weierstrass $M$-test or by the Fubini's Theorem, together with the estimate:
$$
\left|\binom{\alpha}{n}\right| = O\left(\frac{1}{n^{1+\Re(\alpha)}}\right)
\qquad\text{as}\quad n\to\infty,
$$
which itself follows from the Stirling's approximation:
\begin{align*}
\left|\binom{\alpha}{n}\right|
&= \frac{1}{n!}\left| \prod_{k=1}^{n} (\alpha - k + 1) \right| = \frac{1}{n!}\left| \prod_{k=1}^{n} (k - \alpha - 1) \right| = \frac{1}{n!}\left| \frac{\Gamma(n-\alpha)}{\Gamma(\alpha)} \right| \\
&\leq C \left| \frac{n^{-1/2} \left(\frac{n-\alpha}{e}\right)^{n-\alpha}}{n^{1/2} \left(\frac{n}{e}\right)^n} \right|
\leq \frac{C}{n^{1+\Re(\alpha)}}.
\end{align*}
Next, we show the that
$$ I(\alpha, x) = \binom{\alpha}{x} \tag{3} $$
holds. Once this is established, $\text{(1)}$ is immediate and $\text{(2)}$ is a consequence of the Fourier Inversion Theorem.
To prove $\text{(3)}$, note that both sides of $\text{(3)}$ define entire functions in $x$. So, by the principle of analytic continuation, it suffices to verify the equality on a set having an accumulation point.
Now choose the principal branch cut for the complex logarithm and make the substitution $z = e^{2\pi i \xi}$. If $\mathcal{C}$ parametrizes the unit circle minus the principal branch cut in the counterclockwise orientation, then
\begin{align*}
I(\alpha, x)
&= \frac{1}{2\pi i} \int_{\mathcal{C}} (1+z)^{\alpha}z^{-x-1} \, \mathrm{d}z
\end{align*}
For the purpose of the proof, we restrict ourselves to the case $x < 0$. Then the integrand is analytic on $\mathbb{C}\setminus(-\infty, 0]$. By invoking the keyhole contour,
\begin{align*}
I(\alpha, x)
&= \frac{1}{2\pi i} \lim_{\epsilon \to 0^+} \biggl( \int_{-1-\epsilon i}^{-\epsilon i} (1+z)^{\alpha}z^{-x-1} \, \mathrm{d}z + \int_{\epsilon}^{-1+\epsilon i} (1+z)^{\alpha}z^{-x-1} \, \mathrm{d}z \biggr) \\
&= \frac{1}{2\pi i} \biggl( \int_{0}^{1} (1-t)^{\alpha}e^{-i\pi(-x-1)} t^{-x-1} \, \mathrm{d}t - \int_{0}^{1} (1-t)^{\alpha}e^{i\pi(-x-1)}t^{-x-1} \, \mathrm{d}t \biggr) \\
&= \frac{\sin(-\pi x)}{\pi} \int_{0}^{1} (1-t)^{\alpha} t^{-x-1} \, \mathrm{d}t.
\end{align*}
Finally, using the beta function identity and Euler's reflection formula, the above simplifies to
$$ I(\alpha, x) = \frac{1}{\Gamma(-x)\Gamma(x+1)} \cdot \frac{\Gamma(\alpha+1)\Gamma(-x)}{\Gamma(\alpha-x+1)} = \binom{\alpha}{x}. $$
Therefore $\text{(3)}$ holds for $ x < 0$, and thus for all of $x \in \mathbb{C}$, completing the proof.
A: We have:
$$\int_{-\infty}^\infty\binom0t~\mathrm dt=\int_{-\infty}^\infty\frac{\sin(\pi t)}{\pi t}~\mathrm dt=1$$
and
$$\int_{-\infty}^\infty\binom xt~\mathrm dt=\int_{-\infty}^\infty\binom{x-1}{t-1}+\binom{x-1}t~\mathrm dt=2\int_{-\infty}^\infty\binom{x-1}t~\mathrm dt$$
which proves this is indeed equivalent to $2^x$ for integer $x$ where it converges. Further we know that this is at the very least a piecewise exponential function, depending on what happens for $x\in(0,1)$.
