Evaluate $\int {x \choose n} \ dx$ (Problem 798 Crux Mathematicorum)

Question

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.

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}

Answer 2

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).

Answer 3

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 $$ I_n=\int {x \choose n} dx=\frac{1}{n!}\sum_{i=0}^n s(n,i) \frac{x^{k+1}}{k+1}+C. $$