Integral of Binomial $f(z,r) = \int_0^\infty {\binom{ r}{t} z^{t} dt} $ The binomial theorem
$$
(1 + z)^r = \sum_{k \geq 0}\binom{r}{k}z^{k} , \quad |z|<1,
$$
deals with the sum of the binomial coefficient times $z^{k}$ .
What function would instead result from taking the integral
$$
f (z,r) =
\int_{0}^{\infty}\binom{r}{t}z^{t}\,\mathrm{d}t\ ? 
$$
(intending the binomial expressed through the Gamma function:
$\displaystyle\binom{r}{t} = \frac{\Gamma(r+1)}{\Gamma(t+1)\Gamma(r-t+1)}$.)
I am just at the beginning of analyzing this curiosity of mine, and before attempting that
I would ask your advise, thoughts and references if any already exist.
-- update  --
Following the interesting @metamorphy's answer, I found another way to demonstrate that,
for the bilateral integral
$$
I_B (w,z) = \int\limits_{ - \infty }^\infty 
 {\left( \matrix{  z \cr   s \cr}  \right)w^{\,\,s} ds} 
$$
since
$$
I_B (1,0) = \int\limits_{ - \infty }^\infty  {\left( \matrix{  0 \cr   s \cr}  \right)ds}
  = \int\limits_{ - \infty }^\infty  {{{\sin \left( {\pi s} \right)} \over {\,\pi s}}ds}  = 2^{\,0}  = 1
$$
then, by the Vandermonde convolution we get
$$
\eqalign{
  & I_B (1,z) = \int\limits_{ - \infty }^\infty 
 {\left( \matrix{  z \cr   s \cr}  \right)w^{\,\,s} ds} \quad \left| {\;0 \le } \right.
{\mathop{\rm Re}\nolimits} \left( z \right)\quad  =   \cr 
  &  = \int\limits_{ - \infty }^\infty  {\sum\limits_k
 {\left( \matrix{  z \cr   k \cr}  \right)\;\left( \matrix{  0 \cr  s - k \cr}  \right)} \;ds}
  = \sum\limits_k {\left( \matrix{  z \cr   k \cr}  \right)\int\limits_{ - \infty }^\infty
  {\left( \matrix{  0 \cr   s - k \cr}  \right)\;ds} \;}  =   \cr 
  &  = \sum\limits_k {\left( \matrix{  z \cr   k \cr}  \right)\;}  = 2^{\,z}  \cr} 
$$
where for the convolution to be valid it is required that $0 \le \Re (z)$, while the approach
used by @metamorphy requires that $-1< \Re (z)$.
In any case it seems that the bilateral integral be more interesting to analyze, but only formally, because it converges only for $|z|=1$.
 A: Not an answer, but too long for a comment: I wouldn't expect a closed form.
If $r$ is a nonnegative integer, then (as I show in this answer) $$\binom{r}{t}=\frac{\sin\pi t}{\pi}\sum_{k=0}^r\binom{r}{k}\frac{(-1)^k}{t-k};$$ thus, even at $z=1$, $\int_0^\infty\binom{r}{t}\,dt$ is a combination of values of the sine integral.
Let's replace $z$ with $e^{-z}$ (now we assume $\Re z>0$). So, if $r$ is a nonnegative integer, $\int_0^\infty\binom{r}{t}e^{-zt}\,dt$ expresses in terms of $\int_0^x\frac{\sin\pi t}{\pi t}e^{zt}\,dt$ at integer values of $x\leqslant r$ (and some elementary extra things). Clearly, the situation when $r$ is not an integer is even more complicated. Anyway, alternative integral representations come from the ones for $\binom{r}{t}$, that is, for the "reciprocal beta".
Say, if $\Re(a+b)>-1$ and $0<c<1$, we have $$\int_{c-i\infty}^{c+i\infty}w^{-a-1}(1-w)^{-b-1}\,dw=\frac{2\pi i\,\Gamma(a+b+1)}{\Gamma(a+1)\Gamma(b+1)},\tag{I}\label{intrep}$$ which gives a representation for $\binom{r}{t}$, $\Re r>-1$ if we put $a=t$ and $b=r-t$; this results in $$\int_0^\infty\binom{r}{t}e^{-zt}\,dt=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}w^{-1}(1-w)^{-r-1}\big(z+\log w-\log(1-w)\big)^{-1}dw$$ for $1/2<c<1$ (at least). If we deform the path of integration to encircle any of the branch cuts of the integrand, we obtain real alternative integral representations. The same can be done for $\Re r\leqslant-1$.

Update. The integral representation \eqref{intrep} can be used to show that
$$\int_{-\infty}^\infty\binom{s}{x}\,dx=2^s\qquad(\Re s>-1)$$ (stated in this answer). We take $c=1/2$ and substitute $w=\displaystyle\frac12\left(1-i\tan\frac\phi2\right)$: $$\binom{s}{x}=\frac{2^s}{2\pi}\int_{-\pi}^\pi\left(1+i\tan\frac\phi2\right)^{-s}e^{ix\phi}\,d\phi,$$ which, integrated over $x\in(-y,y)$, after substituting $\phi=t/y$, gives $$\int_{-y}^y\binom{s}{x}\,dx=\frac{2^s}{\pi}\int_{-y\pi}^{y\pi}\left(1+i\tan\frac{t}{2y}\right)^{-s}\frac{\sin t}{t}\,dt.$$ Now take $y\to\infty$ (DCT is clearly applicable here).
