I am struggling to solve this expression. I want to show that, $$\frac{1}{p}\nabla_{j}\int e^{ipR\cos(\theta)} dT=i\int \hat{p_{j}} e^{ipR\cos(\theta)} dT$$ here, $dT=d(\cos(\theta))d\phi$
I tried to solve this as below, $$\frac{1}{p}\nabla_{j}\int e^{ipR\cos(\theta)} dT=\frac{1}{p}\int \frac{\partial}{\partial R_{j}} e^{ipR\cos(\theta)} dT$$ $$=\frac{1}{p}\int e^{ipR\cos(\theta)}ip\cos(\theta)\frac{\partial R}{\partial R_{j}}dT$$
I know that, $\frac{\partial R}{\partial R_{j}}=\frac{R_{j}}{R}=\frac{1}{\hat{R_{j}}}$. Thus, by applying this to the above expression yields, $$=i\int e^{ipR\cos(\theta)}\frac{\cos(\theta)}{\hat{R_{j}}} dT$$ However, I am not able to get $\hat{p_{j}}$ term in to the expression. As the only $p$ term inside the integration cancels out to the $\frac{1}{p}$. Can anyone please help me on this.