# Confusion about angular integration in spherical polar coordinates

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?

• Without loss of generality, you can pick your coordinate axes in such a way that $\mathbf{B}$ lays on the $z$ axis while $A$ lays in the $xz$ plane. This is $\mathbf{B}=B\, \mathbf{\hat{k}}$ , $\mathbf{A}=A\, \left[ \sin(\theta_A) \mathbf{\hat{i}} + \cos(\theta_A) \mathbf{\hat{k}} \right]$ and $\mathbf{p} = p \, \left[ \sin(\theta) \cos(\phi) \mathbf{\hat{i}} + \sin(\theta) \sin(\phi) \mathbf{\hat{j}} + \cos(\theta) \mathbf{\hat{k}} \right]$. Integration over $\phi$ and $\theta$ gives you something proportional to $Apf(p)J_1(Bp)\cos(\theta_A)$, with $J_n$ the Bessel function of the first kind. Apr 8, 2020 at 8:42
• @secavara Thanks! I see the key insight is that $\mathbf{A}$ can be taken to lie in the $xz$ plane. Apr 8, 2020 at 8:48
• No worries. Actually I just noticed that I didn't integrate properly. Taking into account the $p^2\sin(\theta)$ form the $d \mathbf{p}^3$, you get something proportional to $\frac{Af(p)\cos(\theta_A)\left(Bp\cos(Bp)-\sin(Bp)\right)}{B^2}$. Apr 8, 2020 at 8:58

The comment by @secavara is indeed correct, but there is a nice and simple trick as well to do this kind of integration in which you don't have to worry of the direction of $$\mathbf{A}$$, just follow, \begin{align} &\int\frac{d^3\mathbf{p}}{(2\pi)^3}f(|\mathbf{p}|)\mathbf{\hat{p}}\cdot\mathbf{A}\exp(i\mathbf{p}\cdot\mathbf{B})\\ &=-i\mathbf{A}\cdot\boldsymbol{\nabla}_{\mathbf{B}}\int\frac{d^3\mathbf{p}}{(2\pi)^3}f(|\mathbf{p}|)\frac{1}{|\mathbf{p}|}\exp(i\mathbf{p}\cdot\mathbf{B}) \end{align} Now one can consider $$\mathbf{B}$$ to be along $$z$$ direction and continue to do the integration as usual. The integral would be a function of $$|\mathbf{B}|$$ and in the final step you must calculate the gradient of $$g(|\mathbf{B}|)$$ with respect to $$\boldsymbol{\nabla}_\mathbf{B}$$ which, I presume, wouldn't be that difficult to calculate.