This question is heavily related to this old post of mine Integral over angular spherical coordinates. However, now I have a different integral, which is, I believe, convergent. I write down the integral and then I will explain why I believe that it is convergent. The integral is
$$\int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta d\theta
\frac{(\sin\theta\cos\phi-\sin\psi)}{(1-\sin\psi\sin\theta\cos\phi-\cos\psi\cos\theta)}$$
where $0<\psi<\pi$.
My line of reasoning is as follows:
-
Observe that the integral has a single pole at $(\theta,\phi)=(\psi,0)$
-
Break down the integral, according to
$$\begin{align*}
& \int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta d\theta
\frac{(\sin\theta\cos\phi-\sin\psi)}{(1-\sin\psi\sin\theta\cos\phi-\cos\psi\cos\theta)}
\\
&=\int_0^{2\pi}d\phi\int_0^{\psi}\sin\theta d\theta
\frac{(\sin\theta\cos\phi-\sin\psi)}{(1-\sin\psi\sin\theta\cos\phi-\cos\psi\cos\theta)}\\
&\quad +
\int_0^{2\pi}d\phi\int_{\psi}^{\pi}\sin\theta d\theta
\frac{(\sin\theta\cos\phi-\sin\psi)}{(1-\sin\psi\sin\theta\cos\phi-\cos\psi\cos\theta)}
\end{align*}$$ -
Consider the first integral:
$$I_1(\psi)=
\int_0^{2\pi}d\phi\int_0^{\psi}\sin\theta d\theta
\frac{(\sin\theta\cos\phi-\sin\psi)}{(1-\sin\psi\sin\theta\cos\phi-\cos\psi\cos\theta)}$$
and shift variables, according to $\theta=\psi-\epsilon$, and then
$$I_1(\psi)=
\int_0^{2\pi}d\phi\int_0^{\Delta_1}\sin(\psi-\epsilon) d\epsilon
\frac{\Big[\sin(\psi-\epsilon)\cos\phi-\sin\psi\Big]}{\Big[1-\sin\psi\sin(\psi-\epsilon)\cos\phi-\cos\psi\cos(\psi-\epsilon)\Big]}$$ -
Now the pole is shifted to $(\epsilon,\phi)=(0,0)$. Expanding the integrand around this point , assuming, hence, that the two angular variables $\epsilon$ and $\phi$ are small, yields
$$I_1(\psi)=\sin\psi
\int_0^{\Delta_2}d\phi\int_0^{\Delta_1} d\epsilon
\frac{2\epsilon\cos\psi}{(\epsilon^2+\phi^2\sin^2\psi)}$$
where I have restricted the integration interval, such that it makes sense to expand and neglect higher order terms (as they are smaller). -
Those integrals can actually be done
$$\begin{align*}
I_1(\psi)&=\sin(2\psi)\int_0^{\Delta_2}d\phi\frac{1}{2}\ln\bigg(1+\frac{\Delta_1^2}{\phi^2\sin^2\psi}\bigg)\\
&=\frac{1}{2}\sin(2\psi)\bigg[
\Delta_2\ln\bigg(\frac{\Delta_1^2/\sin^2\psi+\Delta_2^2}{\Delta_2^2}\bigg)+
2\frac{\Delta_1}{\sin\psi}\tan^{-1}\Big(\frac{\Delta_2\sin\psi}{\Delta_1}\Big)
\bigg]
\end{align*}$$
which is finite. Similar steps can be taken to show that $I_2(\psi)$ is finite…
Is this line of reasoning correct? Meaning, if there is no divergence near the pole, there should not be a divergence anywhere in the interval of integration, right?
Any help will be appreciated.
Best Answer
If you are only interested in convergence, once you remark that there is a single pole at $\left({\theta} , {\phi}\right) = \left({\psi} , 0\right)$, you can prove convergence by a simple Taylor expansion around the pole. The denominator has the expansion
\begin{equation}1-\sin {\psi} \sin {\theta} \cos {\phi}-\cos {\psi} \cos {\theta} = \frac{1}{2} {\left({\theta}-{\psi}\right)}^{2}+\frac{1}{2} {\sin }^{2} \left({\psi}\right) {{\phi}}^{2}+o \left({r}^{2}\right)\end{equation}
where $ r = \sqrt{{\left({\theta}-{\psi}\right)}^{2}+{{\phi}}^{2}}$ and for the numerator,
\begin{equation}\sin {\theta} \cos {\phi}-\sin {\psi} = \cos \left({\psi}\right) \left({\theta}-{\psi}\right)+o \left(r\right)\end{equation}
It follows that the function below the integral is
\begin{equation}F \left({\theta} , {\phi}\right) = O \left(\frac{1}{r}\right) \in {L}_{\text{loc}}^{1} \left(d {\theta} d {\phi}\right)\end{equation} which is locally integrable around the pole.