Integral of $n$-th Bernoulli polynomial of the second kind

Question

We have \begin{align} \int x(x−1)(x−2)...(x−n)\,dx=(n+1)!\cdot\psi_{n+2}(x), \end{align} where, $\psi_n(x)$ is the $n$-th Bernoulli polynomial of the second kind. We have \begin{align} I &= \int_0^{n} \underbrace{x}_{\color{red}{=\,u}}\cdot \underbrace{x(x-1)\cdots(x-n)}_{\color{red}{=\,v'}}\,d u \\ &= x\cdot (n+1)!\,\psi_{n+2}(x)\,\Big|_0^{n} - (n+1)! \int_0^{n} \psi_{n+2}(u)\,d u \\ &= (n+1)!\left[n\cdot\psi_{n+2}(n) - \int_0^n \psi_{n+2}(u)\,d u\right] \\ &= -(n+1)!\left[n\cdot|G_{n+2}| + \int_0^n \psi_{n+2}(u)\,d u\right] \end{align} What is $\displaystyle\int_0^n \psi_{n+2}(u)\,d u$ ?

Answer 1

It was so simple: \begin{align} (n+2)!\cdot\psi_{n+3}(x) &= \int x(x-1)(x-2)\cdots(x-n)(x-(n+1))\ dx \\ &=\underbrace{\int x \cdot x(x-1)(x-2)\cdots(x-n) \ dx}_{\color{red}{=\,I}} - (n+1) \underbrace{\int x(x-1)(x-2)\cdots(x-n) \ dx}_{\color{red}{=\,(n+1)!\,\cdot\,\psi_{n+2}(x)}} \\ &= I - (n+1)\cdot(n+1)!\,\cdot\,\psi_{n+2}(x) \end{align} Thus applies: \begin{align} I &= (n+1)\cdot(n+1)!\,\cdot\,\psi_{n+2}(x) + (n+2)! \cdot \psi_{n+3}(x) \color{white}{\frac{1}{2}} \\ &= (n+1)! \cdot \left[(n+1)\cdot \psi_{n+2}(x) + (n+2)\cdot \psi_{n+3}(x)\right] \color{white}{\frac{1}{2}} \end{align} Now the only question is what is $\psi_{n+3}(n)$....