Simplification of an expression with products $\prod_{1\leq i<j\leq n}$ and $\prod_{1\leq i,j\leq n}$.

Question

The determinant of a matrix $C=(c_{ij})_{n\times n}$ whose entries have the form $c_{ij}=\frac{1}{a_i+b_j}$ is given by $$\det C=\frac{\prod_{1\leq i<j\leq n}(a_i-a_j)(b_i-b_j)}{\prod_{1\leq i,j\leq n}(a_1+b_i)}.$$ In these notes (p. 145), this formula is applied to certain matrices $G$ and $G_m$. The result is

$$\det G=\frac{\prod_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)^2}{\prod_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)},\quad \det G_m=\frac{\prod^{\prime}_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)^2}{\prod^{\prime}_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}\tag{1}$$ "where $^\prime$ means that the index $m$ has been skipped in the product".

The aim is to compute $\frac{\det G_m}{\det G}$. A direct substitution gives $$\frac{\det G_m}{\det G}=\frac{\prod^{\prime}_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)^2}{\prod^{\prime}_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}\cdot \frac{\prod_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}{\prod_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)^2}$$ which, according to the notes, should simplify to

$$\frac{\det G_m}{\det G}=2 m^2 \pi^2\underset{{1\leq k\leq n}}{{\prod}^{\prime}}\frac{(m^2+k^2)^2}{(m^2-k^2)^2}.\tag{2}$$

Question: How to manipulate $(1)$ properly in order to get $(2)$?

Answer 1

We have

$$\frac{\det G_m}{\det G}=\frac{\prod^{\prime}_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)^2}{\prod_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)^2} \cdot \frac{\prod_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}{\prod^{\prime}_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}.$$

And \begin{align*} \frac{\prod^{\prime}_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)}{\prod_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)} & = \frac{1}{\prod_{m< k \le n}(m^2\pi^2-k^2\pi^2) \prod_{1 \le k<m}(k^2\pi^2-m^2\pi^2)}, \\ & = \frac{(-1)^{m-1}}{\prod_{1 \le k \le n}^{\prime}(m^2\pi^2-k^2\pi^2)}. \end{align*} Similarly, $$\frac{\prod_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}{\prod^{\prime}_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}=\prod_{1 \le k \le n}^{\prime}(m^2\pi^2+k^2\pi^2)^2 \cdot 2m^2\pi^2.$$ The product $2m^2\pi^2$ above appears for $i=j=m$.

Hence, \begin{align*} \frac{\det G_m}{\det G} & =\left[ \frac{(-1)^{m-1}}{\prod_{1 \le k \le n}^{\prime}(m^2\pi^2-k^2\pi^2)} \right]^2 \cdot \prod_{1 \le k \le n}^{\prime}(m^2\pi^2+k^2\pi^2)^2 \cdot 2m^2\pi^2,\\ & =2m^2\pi^2\prod_{1 \le k \le n}^{\prime}\frac{(m^2+k^2)^2}{(m^2-k^2)^2}. \end{align*}