Prove identity $\sum_{i=1}^nx_i\left(\prod_{j\neq i}\left(1-\frac{t}{x_j-x_i}\right)-1\right)={n\choose 2}t$?

Question

Consider the coefficient of $t$ in LHS: $$\displaystyle\sum_{i=1}^n x_i\sum_{j\neq i}-\frac{1}{x_j-x_i}=\sum_{i=1}^n\sum_{j\neq i}\frac{x_i}{x_i-x_j}=\sum_{i<j}\frac{x_i}{x_i-x_j}+\frac{x_j}{x_j-x_i}=\sum_{i<j}1={n\choose 2}$$

However, I don't know how to prove the coefficient of $t^s$ is $0$ for $s\geq 2$.

The identity comes from:

Show $$ \sum_{i=1}^n x_i \, \Delta(x_1, \cdots, x_i + t, \cdots, x_n) = \left( \sum_{i=1}^n x_i + \binom{n}{2} t \right) \Delta(x_1, \cdots, x_n) $$ where $$ \Delta(x_1, \cdots, x_n) := \prod_{1 \leq i < j \leq n} (x_i - x_j) $$

If we cancel $\Delta(x_1, \cdots, x_n)$ from both sides, then we get the identity in the title.

$\Delta(x_1, \cdots, x_n)$ can be represented by the determinant of corresponding Vandermonde matrix, I don't know whether it's helpful.

Answer 1

Let $f(z)=\frac{z}{t}\prod_{k=1}^{n}\left(1+\frac{t}{z-x_{k}}\right)$
By Residue theorem: $$\frac{1}{2\pi i}\oint_{\gamma} f(z)dz=\sum_{k=1}^{n}\textrm{Res}(f(z),x_{k})$$ $$=\sum_{k=1}^{n}\lim_{z\to x_k}(z-x_{k})f(z)=\sum_{k=1}^n x_k\prod_{j\neq k}\left(1-\frac{t}{x_j-x_k}\right) \tag{1}$$ Where $\gamma$ is a simple closed curve covering all $x_k$. Let $I(\gamma)$ the internal of curve $\gamma$. Since $I(\gamma)$ covers all singularities of $f(z)$, we can use residue at infinity: $$\frac{1}{2\pi i}\oint_{\gamma} f(z)dz=-\text{Res}(f(z),\infty)=\textrm{Res}\left(\frac{1}{z^2}f\left(\frac{1}{z}\right),0\right)$$ $$=\textrm{Res}\left(\frac{1}{tz^3}\prod_{k=1}^{n}\left(1+\frac{tz}{1-x_k z}\right),0\right)=\frac{1}{2\pi i}\oint_{\gamma'}\frac{1}{z^2}f\left(\frac{1}{z}\right)dz \tag{2}$$ Where $\gamma'$ is a simple closed curve around $0$ and $\gamma'\subset\{z,|x_k z|<1 \ \forall k\}$, using geometric expansion: $$\frac{1}{tz^3}\prod_{k=1}^{n}\left(1+\frac{tz}{1-x_k z}\right)=\frac{1}{tz^3}\prod_{k=1}^{n}(1+tz+tx_{k}z^2+tx_{k}^2z^3+\ldots) $$ $$=\frac{1}{tz^3}\left(1+ntz+\left({n \choose 2}t^2+t\sum_{k=1}^{n}x_{k}\right)z^2+O(z^3)\right) \tag{3}$$ $$=\frac{1}{tz^3}+\frac{n}{z^2}+\left({n \choose 2}t+\sum_{k=1}^{n}x_{k}\right)\frac{1}{z}+O(1)\tag{4}$$ As to calculate the residue, we only care about coefficient $z^{-1}$, from $(2)$ and $(4)$, we have: $$\sum_{k=1}^n x_k\prod_{j\neq k}\left(1-\frac{t}{x_j-x_k}\right)=\frac{1}{2\pi i}\oint_{\gamma} f(z)dz={n \choose 2}t+\sum_{k=1}^{n}x_{k}$$ Thus: $$\sum_{k=1}^n x_k\left(\prod_{j\neq k}\left(1-\frac{t}{x_j-x_k}\right)-1\right)={n \choose 2}t$$ as desired.

I'm not sure OP wanted to solve it this way but using residue theorem is a common technique to attack this kind of symmetric polynomial.