If $\overline{X}$ and $S^2$ be the usual sample mean and sample variance based in a random sample of $n$ observation $N(\mu,\sigma^{2})$ and $T=\frac{(\overline{X}-\mu) \sqrt{n}}{S}$ ` prove that $Cov(\overline{X}, T)=\sigma\frac{\sqrt{n-1}\Gamma\left(\frac{n-2}{2}\right)}{\sqrt{2n}\Gamma\left(\frac{n-1}{2}\right)}$.
I tried to solve the question using expectation but I don't have any idea about how to proceed further, can someone please help in proving it.
First, $\bar{X}_n$ and $S_n$ are independent (see, e.g., Basu's theorem). Then $$ \mathsf{E}\bar{X}_n(\bar{X}_n-\mu)/S_n=\operatorname{Var}(\bar{X}_n)\times \mathsf{E}S_n^{-1}=\frac{\sigma^2}{n}\times \frac{\sqrt{n-1}}{\sqrt{2}\sigma}\frac{\Gamma((n-2)/2)}{\Gamma((n-1)/2)} $$ because $\sigma S_n^{-1}/\sqrt{n-1}\sim\chi_{n-1}^{-1}$. Also note that $\mathsf{E}T_n=0$. Therefore, $$ \operatorname{Cov}(\bar{X}_n,T_n)=\frac{\sigma\sqrt{n-1}}{\sqrt{2n}}\frac{\Gamma((n-2)/2)}{\Gamma((n-1)/2)}. $$