Integral of binomial coefficients

Solution 1:

There is indeed a relation with the Bernoulli numbers of second kind. If I'm right, this is the result :

Theorem : generating function of $f_n(x)$ $$\sum_{n=0}^{+\infty} f_n(x)z^n = \dfrac{(1+z)^{x-1}-1}{\log (1+z)}$$

Corrolary : $$f_n(x) = \sum_{k=0}^n {{x-1}\choose{k+1}} \dfrac{b_{n-k}}{(n-k)!}$$

I'll present here mainly the "formal" steps to get to this result, I'll skip the details of convergence, derivation under integral and all. Start with : $$f_n(x)=\dfrac{1}{n!} \int_1^x \frac{\Gamma(t)}{\Gamma(t-n)} \text{d}t=\dfrac{1}{n!} \int_1^x (t-1)(t-2) \cdots (t-n) \text{d}t$$

Now introduce $g(t,y)=y^{t-1}$. We have $\dfrac{\partial^n g}{\partial y^n}(t,y)=(t-1)(t-2) \cdots (t-n)y^{t-n-1}$. Thus : $$\begin{array}{rcl} f_n(x,y) & = & \displaystyle \frac{1}{n!} \int_1^x (t-1)(t-2) \cdots (t-n)y^{t-n-1} \text{d}t\\ & = & \displaystyle \frac{1}{n!} \dfrac{\partial^n \;}{\partial y^n} \left( \int_1^x y^{t-1} \text{d}t\right) \\ & = & \displaystyle \frac{1}{n!} \dfrac{\partial^n \;}{\partial y^n} \left( \frac{y^{x-1}-1}{\log(y)}\right) \\ & = & \end{array}$$

We have $f_n(x)=f_n(x,1)$. Thus, we get (modulo some good argument for the Taylor expansion) : $$\boxed{\sum_{n=0}^{+\infty} f_n(x)z^n = \sum_{n=0}^{+\infty} \displaystyle \left. \frac{1}{n!} \dfrac{\partial^n \;}{\partial y^n} \left( \frac{y^{x-1}-1}{\log(y)}\right) \right|_{y=1} z^n = \dfrac{(1+z)^{x-1}-1}{\log (1+z)}}$$

This demonstrates the theorem. The corollary is a direct consequences of the following series exapnsion : $$\displaystyle \dfrac{(1+z)^{x-1}-1}{z} = \sum_{n=0}^{+\infty} {{x-1}\choose{n+1}}z^n \; \; \; \; \; \; \; \text{and} \; \; \; \; \; \; \; \displaystyle \frac{z}{\log(1+z)} = \sum_{n=0}^{+\infty} \frac{b_n}{n!}z^n$$

$f_n(x)$ is then given by the Cauchy product : $$\boxed{f_n(x) = \sum_{k=0}^n {{x-1}\choose{k+1}} \dfrac{b_{n-k}}{(n-k)!}}$$

I'm not sure we can go any further than this.

By curiosity, how did you come across this problem ?

Solution 2:

Here's another variation of the theme based upon Stirling Numbers.

Starting from \begin{align*} f_n(x)&=\int_1^x\binom{t-1}{n}dt =\frac{1}{n!}\int_1^x{(t-1)}_ndt =\frac{1}{n!}\int_0^{x-1}{(u)}_ndu \end{align*} we can use the Stirling Numbers of the first kind $s(n,k)$ which can be defined for $n\geq 0$ and $0\leq k \leq n$ by \begin{align*} {(u)}_n=\sum_{k=0}^ns(n,k)u^k\tag{1} \end{align*}

Using this relationship we obtain for $n\geq 0$

\begin{align*} f_n(x)&=\frac{1}{n!}\int_0^{x-1}\sum_{k=0}^ns(n,k)u^kdu\\ &=\frac{1}{n!}\sum_{k=0}^ns(n,k)\int_0^{x-1}u^kdu\\ &=\frac{1}{n!}\sum_{k=0}^ns(n,k)\left.\frac{1}{k+1}u^{k+1}\right|_0^{x-1}\\ &=\frac{1}{n!}\sum_{k=0}^ns(n,k)\frac{1}{k+1}(x-1)^{k+1} \end{align*}

We conclude

\begin{align*} f_n(x)= \frac{1}{n!}\sum_{k=0}^n\frac{s(n,k)}{k+1}(x-1)^{k+1}&\qquad n\geq 0\tag{2}\\ \end{align*}

An exponential generating function $\sum_{n=0}^{\infty}f_n(x)\frac{z^n}{n!}$ can be derived from the generating function of the Stirling Numbers of the first kind.

Generating function: In accordance to the answer of @SylvainL. we claim

The following is valid

\begin{align*} \sum_{n=0}^{\infty}f_n(x)\frac{z^n}{n!}=\frac{(1+z)^{x-1}-1}{\log(1+z)} \end{align*}

We obtain from (1) \begin{align*} (1+z)^u&=\sum_{n=0}^{\infty}\binom{u}{n}z^n\\ &=\sum_{n=0}^{\infty}{(u)}_n\frac{z^n}{n!}\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^ns(n,k)u^k\right)\frac{z^n}{n!}\tag{3} \end{align*}

In the following we consider $f_n(x+1)$ instead of $f_n(x)$.

We observe

\begin{align*}\sum_{n=0}^{\infty}f_n(x+1)\frac{z^n}{n!} &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}\frac{s(n,k)}{k+1}x^{k+1}\right)\frac{z^n}{n!}\tag{4}\\ &=\int_0^x\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}s(n,k)u^{k}\right)\frac{z^n}{n!}du\\ &=\int_0^{x}(1+z)^udu\tag{5}\\ &=\int_0^xe^{u\log(1+z)}du\\ &=\left.\frac{1}{\log(1+z)}e^{u\log(1+z)}\right|_0^x\\ &=\frac{(1+z)^x-1}{\log(1+z)}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box \end{align*} and the claim follows.

Comment:

• In (4) we use the representation (2) of $f_n(x)$

• In (5) we use the generating function of $s(n,k)$ from (3)

Calculation for small values $n=0,1,2$:

Since $$s(0,0)=1,s(1,1)=1,s(2,1)=-1,s(2,2)=1$$ we obtain \begin{align*} f_0(x)&=s(0,0)(x-1)=x-1\\ f_1(x)&=s(1,1)\frac{1}{2}(x-1)^2=\frac{1}{2}(x-1)^2\\ f_2(x)&=\frac{1}{2}\left(s(2,1)\frac{1}{2}(x-1)^2+s(2,2)\frac{1}{3}(x-1)^3\right)\\ &=-\frac{1}{4}(x-1)^2+\frac{1}{6}(x-1)^3\\ &=\frac{1}{12}(2x^3-9x^2+12x-5) \end{align*}

Note: We don't need analytical considerations for the calculations above, since we can state that all calculations are done within the ring of formal power series.