Let $A$ be an $n\times n$ matrix whose entries are \begin{align*} a_{ij} = [\Gamma(\lambda_{i}+\mu_{j})] \end{align*} where $0 < \lambda_{1} < \ldots < \lambda_{n}$ and $0 < \mu_{1} < \ldots < \mu_{n}$ are real positive numbers and $\Gamma$ denotes the Gamma function given by $\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt$ for $\operatorname{Re}(z)>0$.
We need to show that matrix $A$ is non-singular.
I have no idea how to start. Any hint or solution will be appreciated.
For "good enough" functions $f_i,g_i:(a,b)\to\mathbb{R}$ (such that all the integrals below exist) $$\newcommand{\ndet}[1]{\det_{1\leqslant i,j\leqslant n}\left\{#1\right\}}\ndet{\int_a^b f_i(x)g_j(x)\,dx}=\idotsint\limits_{a<x_1<\dots<x_n<b}\ndet{f_i(x_j)}\ndet{g_i(x_j)}\,dx_1\cdots\,dx_n.$$ Indeed, the RHS is $\idotsint_{(a,b)^n}$ divided by $n!$, and if we expand $\det_{1\leqslant i,j\leqslant n}\{f_i(x_j)\}$ as a sum over the permutations of $(x_1,\dots,x_n)$, we get a sum of $n!$ integrals, equal to the LHS each.
Plugging in $f_i(x)=x^{\lambda_i-1}$ and $g_i(x)=x^{\mu_i}e^{-x}$ (say), we get $$\ndet{\Gamma(\lambda_i+\mu_j)}=\idotsint\limits_{0<x_1<\dots<x_n}\ndet{x_i^{\lambda_j-1}}\ndet{x_i^{\mu_j}}e^{-x_1-\dots-x_n}\,dx_1\cdots\,dx_n.$$ Finally, for $0<a_1<\dots<a_n$ and $b_1<\dots<b_n$, we have $\displaystyle\ndet{a_i^{b_j}}>0$.
(It appears already cross-posted to MO. Why no link in the OP?..)