Evaluate $\int {x \choose n} \ dx$ (Problem 798 Crux Mathematicorum) Evaluate $$I_{n}= \int {x \choose n} \ dx$$ where $n$ is a non-negative integer.Any idea of what closed form $I_{n}$ will have.
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\begin{align}
{1 \over \pars{n + 1}!}\,\lim_{z \to 0}
\partiald[n + 1]{}{z}\bracks{{z\pars{1 + z}^{x} \over \ln\pars{1 + z}}}
+ \mbox{a constant.}
\end{align}
A: Well, let's see, the combinatorial way is to go: if $|\alpha| < 1$
$$(1+\alpha)^x=\sum_{n=0}^\infty {x\choose n}\alpha^n$$
Then integrating both sides with respect to $x$ gives
$${1\over\log(1+\alpha)}(1+\alpha)^{x}=\sum_{n=0}^\infty\alpha^n \int{x\choose n}\,dx$$
We know that
$$\log(1+\alpha)=\alpha\sum_{k=1}^\infty {(-1)^{k+1}\alpha^{k-1}\over k}$$ so inverting the power series gives:
$${1\over\log(1+\alpha)}={1\over \alpha}\sum_{k=0}^\infty b_k\alpha^k$$
where the $b_k$ satisfy the recurrence relation:
$$b_k=-\sum_{i=1}^kb_{k-i}{(-1)^{i}\over i+1}$$
From there you'd write
$${1\over\log(1+\alpha)}(1+\alpha)^x=\left(\sum_{k=0}^\infty b_k\alpha^k\right)\left(\sum_{n=0}^\infty {x\choose n}\alpha^n\right)=\sum_{m=0}^\infty\alpha^m\int {x\choose m}\, dx$$
and match coefficients on each side, an altogether unpleasant task to do, even formally. In practice it seems it would be much easier to have a computer do the computation and hand-integrate things.
I work through it specifically to highlight the points where the difficulties lies, computationally speaking. The kind of recurrence can also be nicely automated by a computer, or you can use some other formal identities for them. @Felix Marin's answer shows a nice, compact one which is really quite neat, but which I think obscures the difficulty in actually using it in practice (at least by-hand, there's always machines).
A: We have
$$
{x \choose n}=\frac{1}{n!} (x)_n=\frac{1}{n!}\sum_{i=0}^n s(n,i) x^k.
$$
Here $(x)_n$ is the Pochhammer symbol and $s(n,k)$  are the Stirling numbers of the first kind.
Thus 
$$
I_n=\int {x \choose n} dx=\frac{1}{n!}\sum_{i=0}^n s(n,i) \frac{x^{k+1}}{k+1}+C.
$$
