Complex integral over surface of sphere

Question

How do we go about computing the integral $$\int_{|x|=t} \frac{e^{ikx}}{|x|} d\sigma$$ where $d\sigma$ is the measure of the sphere of radius $t$ in $\mathbb{R}^3$? My approach thus far has been to rewrite $$\int_{|x|=t} \frac{e^{ikx}}{|x|} d\sigma = \int_{|x|=t} \frac{cos(kx)+isin(kx)}{t} d\sigma \\ = \int\int\int \frac{cos(kx)+isin(kx)}{t} (t^2)drd\theta d\phi$$ but am getting stuck reducing the calculation from here.

Answer 1

As explained in the comment, we choose spherical coordinates in "$k$-direction", so that $k\cdot x=|k||x|\cos(\theta)$. Then \begin{align} \int_{|x|=t} \frac{e^{ikx}}{|x|} d\sigma &= \frac{1}{t}\int\int t^2\sin(\theta)e^{i \cos(\theta)|k|t} d\phi d\theta \end{align} Note that he "$t^2\sin(\theta)$" came from the variable change. Furthermore:

• The $\phi$-integral is now trivial (nothing depends on $\phi$).
• we can substitute $\cos(\theta)=u$ with $\sin(\theta)d\theta=du$

Therefore the remaining integral will be $\int e^{ui|k|t}$. Given that $i|k|t$ is just a constant, this should be doable I hope ;)