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This problem becomes quite a bit more tedious to handle when instead we have $\cos(\theta-\theta_0)$

Following from Claude's advice

$$\tag{1}I(b) = \int_0^{2\pi} \ln\big(a-b\cos(\theta-\theta_0)\big)d\theta_0$$

because we are not integrating wrt b (and I'm pretty sure the bounds of integration not containing b is also a requirement, but I'll have to fact check that later), we are allowed to commute

$$\frac{\partial}{\partial b}\int_0^{2\pi} f(a,b,\theta,\theta_0)d\theta_0 = \int_0^{2\pi}\frac{\partial}{\partial b}f(a,b,\theta,\theta_0)d\theta_0$$

$$\tag{2} I'(b) = -\int_0^{2\pi} \frac{\cos(\theta-\theta_0)}{a-b\cos(\theta-\theta_0)}d\theta_0$$

We'll solve (2) by first using polynomial long division: $\int_{-\pi}^{\pi} -\frac{a}{b\big(a-b\cos(\theta-\theta_0)\big)}+\frac{1}{b}d\theta_0$ and from the tangent half-angle substitution, $\cos(\theta-\theta_0) = \frac{1-t^2}{1+t^2},d\theta_0 = \frac{2}{1+t^2}dt, t = \tan(\frac{\theta-\theta_0}{2}),t_0 = \tan(\frac{\theta+\pi}{2}), t_f = \tan(\frac{\theta-\pi}{2})$ and our integral becomes:

$$-\frac{2}{b}\int_{t_0}^{t_f}\frac{1}{(1+t^2)\big(a-b\frac{1-t^2}{1+t^2}\big)}dt_0+\frac{2\pi}{b} \\ = -\frac{2}{b}\int_{t_0}^{t_f}\frac{1}{\big((a-b)+(a+b)t^2\big)}dt+\frac{2\pi}{b} \\ = -\frac{2}{b(a-b)}\int_{t_0}^{t_f}\frac{1}{\big(1+\frac{a+b}{a-b}t^2\big)}dt+\frac{2\pi}{b} \\ = -\frac{2}{b\sqrt{a^2-b^2}}\int_{\sqrt{\frac{a+b}{a-b}}t_0}^{\sqrt{\frac{a+b}{a-b}}t_f}\frac{1}{\big(1+u^2\big)}du+\frac{2\pi}{b} \\ = -\frac{2}{b\sqrt{a^2-b^2}}\int_{\arctan\Big(\sqrt{\frac{a+b}{a-b}}t_0\Big)}^{\arctan\Big(\sqrt{\frac{a+b}{a-b}}t_f\Big)}\frac{\sec^2(x)}{\big(1+\tan^2(x)\big)}dx+\frac{2\pi}{b} \\ = -\frac{2}{b\sqrt{a^2-b^2}}\Bigg(\arctan\bigg(\sqrt{\frac{a+b}{a-b}}\tan\Big(\frac{\theta-\pi}{2}\Big)\bigg)-\arctan\bigg(\sqrt{\frac{a+b}{a-b}}\tan\Big(\frac{\theta+\pi}{2}\Big)\bigg)\Bigg) + \frac{2\pi}{b}$$

And we can see from this result that the inclusion of $\theta$ has resulted in a significantly more verbose integrand. I do not have a reference answer, so if anyone spots a mistake, please let me know.

The next step is to integrate $I'(b)$ to get $I(b)$, which works because of Fubini's theorem since $f'(a,b,\theta,\theta_0)$ is continuous from $0\le \theta_0 \le 2\pi$ and the bounds of integration are independent of b: $I(b) = \int_0^{2\pi} \int f'(a,b,\theta,\theta_0) db d\theta_0 = \int \int_0^{2\pi}f'(a,b,\theta,\theta_0)d\theta_0 db$. Or

$$\tag{3}I(b) = 2\int \frac{1}{b}\Bigg(\pi - \frac{1}{\sqrt{a^2-b^2}}\bigg(\arctan\Big(\sqrt{\frac{a+b}{a-b}}\tan\big(\frac{\theta-\pi}{2}\big)\Big)-\arctan\Big(\sqrt{\frac{a+b}{a-b}}\tan\big(\frac{\theta+\pi}{2}\big)\Big)\bigg)\Bigg) db$$

The resulting indefinite integral can have its constant, $C$, evaluated by using $I(b=0) = 2\pi \ln(a)$.

Does anyone have any issues with how I'm solving this problem? Maybe there's an even simpler way?

EDIT: My indefinite integral looks like it might violate it after all. Technically the indefinite integral would be the same as $\int db = \int_0^b db - F(0)$ so our bounds of integration for one of the 2 integrals would technically include the variable of integration. However, if I made up a brand new variable and definitely integrated it like Lai did here, THEN I think it'd be mathematically legal. Should still yield the same result unless I'm mistaken.

Researcher R
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    A faster route is to convince yourself that $\int_0^{2\pi}f(x-y)~dx=\int_0^{2\pi}f(x)~dx$ for any $2\pi$-periodic function $f$. – Semiclassical Aug 26 '24 at 04:55
  • Ah, okay so $f(x) = \ln\big(a-b\cos(x)\big)$, and $f(x-y) = \ln\big(a-b\cos(x-y)\big)$, and therefore $$\int_0^{2\pi}\ln\big(a-b\cos(\theta-\theta_0)\big)d\theta_0 = \int_0^{2\pi}\ln\big(a-b\cos(-\theta_0)\big)d\theta_0$$. I'd like to in some way keep the $\theta$ though i.e. $$I(a,b,\theta) = \int_0^{2\pi} f(a,b,\theta,\theta_0)d\theta_0$$, so would there be a way to reintroduce $\theta$ at a later step? – Researcher R Aug 26 '24 at 23:38
  • You can change the integral, of course, but you're not going to get around the basic fact: shifting a periodic function along its domain of periodicity doesn't change its integral over that domain. – Semiclassical Aug 26 '24 at 23:52
  • A few. This problem involves Dirichlet characters. See Terrence Tao's work. – suckling pig Aug 27 '24 at 00:10
  • @youthinasia Could you link to a specific resource? – Researcher R Aug 28 '24 at 20:26
  • @Semiclassical And that's if and only if the integral is over the full period, correct? – Researcher R Aug 28 '24 at 20:34
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    Yes. (Or over a multiple of the period, e.g., integrating over $[0,4\pi]$ would work too.) – Semiclassical Aug 28 '24 at 20:49
  • @ResearcherR sorry, I was just being facetious... – suckling pig Aug 29 '24 at 03:52
  • @Semiclassical Working through results for several different monovariable periodic functions, and yeah, no matter what the result, for definite integration over the full period, it always yields the same result whether $\theta$ is included in the function or not. hmmm. Well just to add context, what we're seeing here is the integral of Green's function obtained via method of images for the 2D Poisson's Equation in polar coordinates. With Green's function, we generally want a end result where the dummy variables like $\theta_0$ have been integrated out, and variables like $\theta$ are left – Researcher R Aug 30 '24 at 05:12
  • @Semiclassical So it's interesting that the end result might as well be independent of $\theta$. When I move to 3D Poisson's Equation in spherical coordinates and I perform the volume integral of Green's function, I'll have to see if this property comes into play for either the azimuthal or polar angles. – Researcher R Aug 30 '24 at 05:14

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