Unproved identity for computing $\zeta (s)$ at negative odd integers

Question

I stumbled upon this interesting identity, but I cannot find the proof of it: $$\zeta (-s)=\int_{-1}^0\sum_{k=1}^n k^s \mathrm dn ,\, s=2r+1,\, r\in\mathbb N_0 .$$ For example, setting $s=1$ yields $\zeta (-1)=\int_{-1}^0 \frac{n^2+n}{2}\, \mathrm dn =-\frac{1}{12}$. Is the formula for general negative odd integers true? If yes, can someone refer me to the proof?

Answer 1

\begin{align} \int_{-1}^0\sum_{k=1}^nk^s\mathrm{d}n &=\int_{-1}^0\frac1{s+1}\sum_{j=0}^s\binom{s+1}{j}B_j^+n^{s+1-j}\mathrm{d}n\\ &=\frac1{s+1}\sum_{j=0}^s\binom{s+1}{j}B_j^+\int_{-1}^0n^{s+1-j}\mathrm{d}n\\ &=\frac1{s+1}\sum_{j=0}^s\binom{s+1}{j}B_j^+\left[\frac{n^{s+2-j}}{s+2-j}\right]_{-1}^0\\ &=\frac1{s+1}\sum_{j=0}^s\binom{s+1}{j}B_j^+\frac{(-1)^{s+1-j}}{s+2-j}\\ &=\frac{(-1)^{s+1}}{s+1}\sum_{j=0}^s\binom{s+1}{j}B_j^+\frac{(-1)^{j}}{s+2-j}\\ &=\frac{(-1)^{s+1}}{s+1}\sum_{j=0}^s\binom{s+1}{j}\frac{B_j^-}{s+2-j}\\ &=\frac{(-1)^{s+1}}{s+1}(-B_{s+1}^-)\\ &=-\frac{B_{s+1}^+}{s+1}\\ &=\zeta(-s)\\ \end{align}