白给恒等式
Elegia
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个人记录
\begin{aligned}
&\quad \sum_i\sum_j\binom ni \binom nj (-1)^{i+j} \int_0^{-i} x^{n-1}(x+i-j)^{n+1}\,\mathrm{d} x \\
&{\color{red} =0}
\end{aligned}
简要解释：
\begin{aligned}
&\quad \sum_i\sum_j\binom ni \binom nj (-1)^{i+j} \int_0^{-i} x^{n-1}(x+i-j)^{n+1}\,\mathrm{d} x \\
&= \sum_i\sum_j\binom ni \binom nj (-1)^{i+j} \int_0^{-i} x^{n-1}\sum_k \binom{n+1}k x^k(i-j)^{n+1-k}\,\mathrm{d} x\\
&= \sum_i\sum_j\binom ni \binom nj (-1)^{i+j} \sum_k \binom{n+1}k \frac{(-i)^{n+k}}{n+k}(i-j)^{n+1-k}\\
&= \sum_i\binom ni (-1)^{i} \sum_{k\le 1} \binom{n+1}k \frac{(-i)^{n+k}}{n+k} f_{n,k}(i)\\
f_{n,1}(i)&= \Delta^n (-j)^n\\
&= (-1)^{n} \Delta^n j^{n}\\
&= n!\\
f_{n,0}(i) &= \Delta^n \left[(-j)^{n+1} + (n+1)i(-j)^n\right]\\
&= -\binom{n+1}2 n!+(n+1)!i\\
&= (n+1)!(i-n/2)\\
V&=\sum_i\binom ni (-1)^{i} \left( \frac{(-i)^n}{n}(n+1)!(i-n/2) + (-i)^{n+1} n!\right)\\
&= - (n+1)!n!/2 -\Delta^n(-i)^{n+1}n!\frac 1n\\
&= - (n+1)!n!/2 + n!^2\binom{n+1}2\frac 1n\\
&= - (n+1)!n!/2 + n!^2 (n+1)/2\\
&=0
\end{aligned}