integral representation for $\sum_{k=0}^{x}k^{p}$

Question

How the following integral representation can be derived? $$\sum_{k=0}^{x}k^{p}=\int_{0}^{x+1}B_{p}\left(t\right)dt=\frac{B_{p+1}\left(x+1\right)-B_{p+1}}{p+1}$$

I know Faulhaber's formula which is as follows: $$\sum_{k=0}^{x}k^{p}=\frac{1}{p+1}\sum_{j=0}^{p}B_{j}{{p+1}\choose{j}}\left(N\right)^{\left(p+1-j\right)}$$

where $N=x+1$

or another formula: $$\sum_{p=1}^{k}p^{n}=\sum_{m=1}^{p}{{k+1}\choose{m+1}}{n\brace m}m!$$

but I don't know if they are useful or not.

Answer 1

We use the generating function \begin{align*} \frac{te^{tx}}{e^t-1}=\sum_{n\geq 0}B_n(x)\frac{t^n}{n!}\tag{1} \end{align*} of Bernoulli polynomials to derive the integral representation.

From (1) we can show for $n\geq 1$ \begin{align*} B_{n}(x+1)-B_n(x)&=nx^{n-1}\tag{2} \end{align*} as follows:

We obtain \begin{align*} \sum_{n\geq 0}\left(B_n(x+1)-B_n(x)\right)\frac{t^n}{n!} &=\frac{te^{t(x+1)}}{e^t-1}-\frac{te^{tx}}{e^t-1}\tag{3}\\ &=\frac{te^{tx}e^t}{e^t-1}-\frac{te^{tx}}{e^t-1}\\ &=te^{tx}\\ &=t\sum_{n\geq 0}x^n\frac{t^n}{n!}\\ &=\sum_{n\geq 1}nx^{n-1}\frac{t^n}{n!}\tag{4} \end{align*}

Comparison of the coefficient of $t^p$ in (3) and (4) gives for $p\geq 1$: \begin{align*} B_{p}(x+1)-B_p(x)=px^{p-1} \end{align*} and the claim (2) follows. Division by $p$ and shifting $p$ by one gives \begin{align*} x^p=\frac{B_{p+1}(x+1)-B_{p+1}(x)}{p+1}\tag{5} \end{align*}

We obtain from (5) for $N\geq 0, p\geq 0$: \begin{align*} \color{blue}{\sum_{k=0}^Nk^p}&=\sum_{k=0}^N\frac{B_{p+1}(k+1)-B_{p+1}(k)}{p+1}\\ &=\frac{B_{p+1}(N+1)-B_{p+1}(0)}{p+1}\tag{6}\\ &\,\,\color{blue}{=\frac{B_{p+1}(N+1)-B_{p+1}}{p+1}}\tag{7}\\ \end{align*}

In (6) and (7) we use the telescoping property and $B_p(0)=B_p, p\geq 0$.

We obtain from (1) \begin{align*} \sum_{n\geq 0}B_n^{\prime}(x)\frac{t^n}{n!}&=\frac{d}{dx}\left(\frac{te^{tx}}{e^t-1}\right)\tag{8}\\ &=\frac{t^2e^{tx}}{e^t-1}\\ &=\sum_{n\geq 0}^{\infty}B_n(x)\frac{t^{n+1}}{n!}\tag{9} \end{align*}

In the following we use the coefficient of operator $[t^p]$ to denote the coeffficient of $t^p$ in a series. Coefficient comparison of $t^p$ in (8) and (9) gives for $p\geq 1$: \begin{align*} p![t^p]\sum_{n\geq 0}B_n^{\prime}(x)\frac{t^n}{n!}&=B_p^{\prime}(x)\tag{10}\\ p![t^p]\sum_{n\geq 0}B_n(x)\frac{t^{n+1}}{n!}&=p!B_{p-1}(x)\frac{1}{(p-1)!}\\ &=pB_{p-1}(x)\tag{11} \end{align*}

We obtain from (10) and (11) \begin{align*} \int_{0}^{x+1}B_{p+1}^{\prime}(u)du&=B_{p+1}(x+1)-B_{p+1}(0)\\ &=(p+1)\int_{0}^{x+1}B_{p}(u)du\\ \end{align*} from which \begin{align*} \color{blue}{\int_{0}^{x+1}B_{p}(u)du=\frac{B_{p+1}(x+1)-B_{p+1}}{p+1}} \end{align*} follows.