$\int_{-\infty}^{\infty}{x\choose t}dt=2^x$ for $x\geq0$?? I noticed that on Desmos, for $m>0$ and $x\geq0$, $$\int_{-m}^{m}{x\choose t}dt$$ closer and closer approximated $2^x$. So, does
$$\int_{-\infty}^{\infty}{x\choose t}dt=2^x$$
Assuming that ${x\choose t}=\frac{\Gamma(x+1)}{\Gamma(t+1)\Gamma(x-t+1)}$.
I know that $\sum_{k\geq0}{x\choose k}=2^x$, but there are, of course, differences between $\int_{-\infty}^{\infty}$ and $\sum_{k\geq0}$. 
Can someone prove/disprove this?
 A: I once proved a more general result in my old posting. Here is another, shorter proof.

Claim. Let $\alpha$ be a complex number with $\operatorname{Re}(\alpha) > 0$. Then
$$ \forall z \in \mathbb{C} \ : \quad \int_{-\frac{1}{2}}^{\frac{1}{2}} \left(1 + e^{2\pi i \theta}\right)^{\alpha} e^{-2\pi i z \theta} \, d\theta = \binom{\alpha}{z}. \tag{*} $$

Before proving this claim, let us show that this leads to the desired conclusion. Indeed, it is easy to check that $x \mapsto \binom{\alpha}{x}$ is square-integrable on $\mathbb{R}$. So by the Fourier inversion formula,
$$ \int_{-\infty}^{\infty} \binom{\alpha}{x} e^{-2\pi i \xi x} \, dx
= \begin{cases}
\left(1 + e^{-2\pi i \xi}\right)^{\alpha}, & |\xi| < \frac{1}{2} \\
0, & \text{otherwise}
\end{cases} $$
Plugging $\xi = 0$ proves OP's claim.

Proof of Claim. We write $I(\alpha, z)$ for the integral in $\text{(*)}$. Since both sides of $\text{(*)}$ are holomorphic functions in $z$, it suffices to prove that $\text{(*)}$ holds for $z \in (-\infty, 0)$ by the principle of analytic continuation.
Let $z \in (-\infty, 0)$ and write $\gamma$ for the curve $e^{2\pi i \theta}$ with $-\frac{1}{2} \leq \theta \leq \frac{1}{2}$. Then with $\xi = e^{2\pi i \theta}$,
\begin{align*}
I(\alpha, z)
&= \frac{1}{2\pi i} \int_{\gamma} \frac{\left(1 + \xi \right)^{\alpha}}{\xi^{z+1}} \, d\xi,
\end{align*}
Deforming the contour $C$ to wrap the branch cut $(-\infty, 0]$ of the principal logarithm,
\begin{align*}
I(\alpha, z)
&= \frac{1}{2\pi i} \left( \int_{-1-0^+i}^{-0^+i} \frac{\left(1 + \xi \right)^{\alpha}}{\xi^{z+1}} \, d\xi + \int_{0^+i}^{-1+0^+i} \frac{\left(1 + \xi \right)^{\alpha}}{\xi^{z+1}} \, d\xi \right) \\
&= \frac{e^{i\pi(z+1)} - e^{-i\pi(z+1)}}{2\pi i} \int_{0}^{1} \frac{(1-x)^{\alpha}}{x^{z+1}} \, dx \\
&= \frac{\sin((z+1)\pi)}{\pi} \frac{\Gamma(\alpha+1)\Gamma(-z)}{\Gamma(\alpha+1-z)} \\
&= \binom{\alpha}{z}.
\end{align*}
In the last two lines, we utilized the beta function identity and Euler's reflection formula. Therefore $\text{(*)}$ holds for $z \in (-\infty, 0)$ and hence for all $z \in \mathbb{C}$. ////
