Does an integral of a polar function from $0$ to infinity have to diverge? This is more of a theoretical question, but I was curious if a polar equation automatically diverges as it goes to infinity? After all, the area will just be the area in the polar graph added to itself infinitely, right?
 A: I understand what you're asking now. If you want to integrate a function $f\colon [0,\infty) \to [0,\infty)$, then you can write $\int_{0}^{\pi/2} \frac12 r^2(\theta) d\theta$. You're asking if $\int_{0}^{\infty} \frac12 r^2(\theta) d\theta=\infty$. Well unless $r=0$ then yes of course. You're adding up a positive number infinitely many times. You are correct.
A: No. $$\int 0 = 0$$ no matter of the domain.
A: If you want your integral to have the meaning of an area, then you should not integrate the polar function itself but its square:
$$\frac12\int_{\phi=0}^{2\pi}r(\phi)^2d\phi$$
Of course, if you decide to evaluate that integral over a larger interval then you are running around the origin multiple times. For a closed curve and an integration interval that spans an integer multiple of $2\pi$ the integral would be the same integer multiple of the original area.
A: Hint:
I'm not sure to interpret correctly your question, but, for $r=e^{-\theta^2}$ we have:
$$
\int_0^\infty r d\theta=\frac{\sqrt{\pi}}{2}
$$
and also
$$
\int_0^\infty r^2 d\theta=\frac{1}{2}\sqrt{\frac{\pi}{2}}
$$
