# Solving $\int^{2\pi}_0 d\theta \ln\left(\sqrt{1 - a^2\cos^2\theta} + ia\cos\theta\right)$

I'm trying to take the following integral. $$I = \int_0^{2\pi}d\theta \ln\left[\sqrt{1 - (\hat{\rho}\cdot\vec{a})^2} + i\hat{\rho}\cdot\vec{a}\right],$$ where $$\hat{\rho} = (\cos\theta,\sin\theta)$$ is the unit vector in polar coordinates, and $$\vec{a} = (a_x,a_y) = \Vert a \Vert(\cos\phi,\sin\phi)$$ is some arbitrary vector, and it is known that $$a \equiv \Vert\vec{a}\Vert < 1$$. We can choose $$\phi=0$$, such that $$I = \int^{2\pi}_0 d\theta \ln\left(\sqrt{1 - a^2\cos^2\theta} + ia\cos\theta\right).$$

I have tried change of variables, factorization etc, but get nowhere. Is there any hope of being able to solve it?

• Is $\|\vec a\| \le 1$? When it is greater than $1$, the whole character of this problem changes. In any case, rewrite $\vec a = \|a\|(\cos\phi, \sin\phi)$. It will accomplish a useful reduction of complexity. – Paul Sinclair Oct 1 '19 at 4:00
• Yes, it is in fact true that $\Vert\vec{a}\Vert < 1$ (strictly smaller than). – fromGiants Oct 1 '19 at 17:31
• Without loss of generality, you can assume $\phi = 0$. – eyeballfrog Oct 2 '19 at 14:49
• @fromGiants Looks like you have lost interest in your integral?~! – Dr Zafar Ahmed DSc Oct 7 '19 at 8:49
• @DrZafarAhmedDSc no I was just not sure about the rewriting of the integral in your solution, namely when you split the integrand into two logarithms. – fromGiants Oct 7 '19 at 11:41

$$I=\int_{0}^{2\pi} ln[\sqrt{1-a^2\cos^2 x}+ai \cos x]~ dx= 2\int_{0}^{\pi} ln[\sqrt{1-a^2\cos^2 x}+ai \cos x]~ dx.$$ Next use $$\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx$$. Then $$2I=2\int_{0}^{\pi}\left( \ln[\sqrt{1-a^2\cos^2 x}+ai \cos x]+\ln[\sqrt{1-a^2\cos^2 x}-ai \cos x]\right)~ dx.$$ $$2I=2\int_{0}^{\pi} \ln[ 1-a^1\cos^2 x+ a^2 \cos^2 x]~dx=2\int_{0}^{\pi} \ln 1 ~dx =0$$ Hence $$I=0.$$