I'm having to perform an integral of the following form, $$\int\frac{d^3\mathbf{p}}{(2\pi)^3}f(|\mathbf{p}|)\mathbf{\hat{p}}\cdot\mathbf{A}\exp\left(i\mathbf{p}\cdot\mathbf{B}\right)$$ Here, $\mathbf{A}$ is a $\mathbf{p}$ independent vector quantity. Except for the $\mathbf{\hat{p}}\cdot\mathbf{A}$ part this is a common type of integral in Field Theory calculations.
Generally one chooses the angle between $\mathbf{p}$ and $\mathbf{B}$ to be $\theta$. With that one writes $\exp\left(i\mathbf{p}\cdot\mathbf{B}\right)=\exp\left(i|\mathbf{p}||\mathbf{B}|\cos\theta\right)$. And we have a $d\theta$ integration coming from $d^3\mathbf{p}$ because $\hat{p},\hat{\theta}$ and $\hat{\phi}$ are being varied. Now my question is what happens to $\mathbf{\hat{p}}\cdot\mathbf{A}$?
My first guess was to write $\mathbf{\hat{p}}\cdot\mathbf{A}=\sin\theta\cos\phi A_x+\sin\theta\sin\phi A_y+\cos\theta A_z$. But what's bugging me is that the angle between $\mathbf{p}$ and $\mathbf{B}$ is $\theta$ why should that be same for the case of $\mathbf{\hat{p}}\cdot\mathbf{A}$. Moreover, while doing the angular integration the angle $\theta$ is measured with respect to the $z$-axis. Which direction should be fixed as $z$? Along $\mathbf{B}$ or along $\mathbf{A}$? Can someone help me reduce the the the angular part of the integration?