# How prove or disprove is positive define matrix?

prove or disprove following matrix is positive definite matrix ?

$$\begin{bmatrix} \dfrac{\sin(a_1-a_1)}{a_1-a_1}&\cdots&\dfrac{\sin(a_1-a_n)}{a_1-a_n}\\ \vdots& &\vdots\\ \dfrac{\sin(a_n-a_1)}{a_n-a_1}&\cdots&\dfrac{\sin(a_n-a_n)}{a_n-a_n} \end{bmatrix}_{n\times n}$$ and where we defind $$\dfrac{\sin 0}{0}=1$$

Maybe this problem can use Integral to solve it. since $$\frac{1}{x} = \int_0^\infty e^{-tx} t \, dt$$? Thank you

Let $b_{kl}=\frac{\sin{(a_k-a_l)}}{a_k-a_l}$. The symmetric $n\times n$ matrix $M=(b_{kl})$ is positive definite, if and only if $(a_1,\ldots,a_n)$ are distinct.
Note that $$\frac{\sin a}{a}=\frac{1}{2}\int_{-1}^1e^{iat}dt$$ Thus, for ${\bf x}=(x_1,\ldots,x_n)\in\Bbb{C}^n$ we have \eqalign{ {\bf x}^*M{\bf x}&=\frac{1}{2}\int_{-1}^1\sum_{k,l=1}^nx_k\overline{x_l}e^{i(a_k-a_l)t}dt\cr &=\frac{1}{2}\int_{-1}^1\left\vert\sum_{k=1}^n{x_k}e^{ia_k t}\right\vert^2 dt\geq0\tag{1} } Thus $M$ is positive. Moreover, Suppose that ${\bf x}^*M{\bf x}=0$ we want to prove that $$x_1=x_2=\cdots=x_n=0.$$ From $(1)$ we conclude that the continuous function $t\mapsto \sum_{k=1}^n{x_k}e^{ia_k t}$ is zero on $[-1,1]$. Now, consider the function $f$ defined by $f(z)=\sum_{k=1}^n{x_k}e^{ia_k z}$, this is an analytic function in $\Bbb{C}$ that is equal to $0$ for $z\in[-1,1]$, so it must be identically zero.
Consider $j\in\{1,\ldots,n\}$, we have $$\forall\,t\in \Bbb{R},\quad \sum_{k=1}^nx_ke^{i(a_k-a_j)t}=0$$ hence, for $T>0$, $$\sum_{k=1}^nx_k\left(\frac{1}{2T}\int_{-T}^Te^{i(a_k-a_j)t}dt\right)=0$$ Letting $T$ tend to $\infty$ we conclude that $x_j=0$. Here we use that fact that $$\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^Te^{i wt}dt=\left\{\matrix{1&w=0\cr 0&w\ne 0}\right.$$ and the announced conclusion follows.
Are there any other assumptions you forgot to add? A counter-example which comes to mind is to take $n = 2$ and $a_1 = a_2 = 1$ so that the matrix is just $$\left( \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \right),$$ which is positive semidefinite, but not positive definite since one of its eigenvalues is $0$.