# Elementary proof that $\sum_{j=1}^{n} \prod_{k \neq j} \frac{1}{1+(a_j - a_k + i)^2} \in \mathbb{R}$ for $a_1, \dots, a_n \in \mathbb{R}$ distinct

By a straightforward contour integral, one can show that for $$a_1, \dots, a_n \in \mathbb{R}$$ distinct, we have

$$\frac{1}{\pi} \int_{-\infty}^{\infty} \prod_{j=1}^{n} \frac{1}{1+(x-a_j)^2} \, dx = \sum_{j=1}^{n} \prod_{k \neq j} \frac{1}{1+(a_j - a_k + i)^2}$$

and therefore the latter sum is real-valued. Is there an elementary (or at least purely algebraic) way to show this, without using complex analysis?

• This should be related to Lagrange interpolation. Commented Dec 27, 2021 at 18:58
• after taking conjugates and some easy algebra, it suffices to show that: $$\sum\limits_{j=1}^n\prod\limits_{k\neq j}\dfrac{a_j-a_k}{1+(a_j-a_k)^2} = 0,$$ which is probably established with one nice trick with Lagrange Interpolation. Commented Dec 27, 2021 at 20:15
• If $p(x) = (x-a_1)\dots(x-a_n),$ then the above same as showing: $$\sum_{j=1}^n\dfrac{p'(a_j)}{p(a_j-i)p(a_j+i)} = 0.$$ maybe someone can finish this off. Commented Dec 27, 2021 at 20:30

Consider a more general problem. Suppose we started with $$\frac1{2\pi}\int_{-\infty}^{\infty}\frac{dx}{\prod_{k=1}^n(x-z_k)(x-\bar z_k)},$$ where all $$z_k$$'s are pairwise distinct and all $$\Im z_k>0$$. The analog of your problem would then be to show that $$\sum_{j=1}^n\frac{i}{(z_j-\bar z_j)\prod_{k\ne j}^n(z_j-z_k)(z_j-\bar z_k)}\; \in\mathbb R,$$ which is the same as to show that $$\sum_{j=1}^n\frac{1}{(z_j-\bar z_j)\prod\limits_{k\ne j}^n(z_j-z_k)(z_j-\bar z_k)}+\sum_{j=1}^n\frac{ 1}{(\bar z_j- z_j)\prod\limits_{k\ne j}^n(\bar z_j-z_k)(\bar z_j-\bar z_k)}=0\tag{1}$$ This is a special case of the identity $$\qquad\qquad\sum_{j=1}^N\frac{1}{\prod_{k\ne j}^N(u_j-u_k)}=0,\tag{2}$$ which corresponds to setting in (2) $$N=2n$$ and $$u_j=z_j$$, $$u_{j+n}=\bar z_j$$ for $$j=1,\ldots,n$$. It thus remains to prove (2).
Knowing the beginning of the story, one proof is straightforward: it suffices to compute $$\displaystyle\int_\Gamma\frac{du}{\prod_{k=1}^N(u-u_j)}$$ by residues in two different ways (here $$\Gamma$$ is a closed contour around $$u_1,\ldots,u_N$$). Shrinking the contour $$\Gamma$$ produces the left side of (2), while expanding it to infinity gives zero.
If instead we want an algebraic proof of (2), we may use Lagrange interpolation. Given $$f_1,\ldots,f_N\in\mathbb C$$, there exists a unique polynomial $$Q(x)$$ of degree $$N-1$$ such that $$Q(u_k)=f_k$$, and moreover $$Q(x)=\sum_{j=1}^Nf_j\prod_{k\ne j}^N\frac{x-u_k}{u_j-u_k}.$$ Choose $$f_1=\ldots=f_N=1$$ so that $$Q(x)=1$$. The previous formula then transforms into $$\sum_{j=1}^N\prod_{k\ne j}^N\frac{x-u_k}{u_j-u_k}=1.$$ Equating the coefficients of $$x^{N-1}$$ at both sides immediately yields (2).