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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?

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    $\begingroup$ 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. $\endgroup$ – Paul Sinclair Oct 1 '19 at 4:00
  • $\begingroup$ Yes, it is in fact true that $\Vert\vec{a}\Vert < 1$ (strictly smaller than). $\endgroup$ – fromGiants Oct 1 '19 at 17:31
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    $\begingroup$ Without loss of generality, you can assume $\phi = 0$. $\endgroup$ – eyeballfrog Oct 2 '19 at 14:49
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    $\begingroup$ @fromGiants Looks like you have lost interest in your integral?~! $\endgroup$ – Dr Zafar Ahmed DSc Oct 7 '19 at 8:49
  • $\begingroup$ @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. $\endgroup$ – fromGiants Oct 7 '19 at 11:41
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$$I=\int_{0}^{2\pi} ln[\sqrt{1-a^2\cos^2 x}+ai \cos x]~ dx= 2\int_{0}^{\pi} ln[\sqrt{1-a^2\cos^2 x}+ai \cos x]~ dx.$$ 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 $$I=0.$$

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