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

