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I'm trying to find the analytical form of the following integral:


I've tried in Wolfram Alpha, which ran out of free calculation time, and in Mathematica I get the original expression for the indefinite integral and the program stays in the "Running..." state for several minutes, when evaluating the definite integral, and I stopped it.

However, I am reading a paper, which says there is an analytical expression for this integral, and also to me it seems rather straightforward. Any hints?

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up vote 2 down vote accepted

We have \begin{align*} \int_0^{2\pi} \left| \frac{1- i \cos(\phi)}{1+a-i\cos(\phi)} \right|^2 d\phi &= \int_0^{2\pi} \frac{1+\cos^2\phi}{(1+a)^2 + \cos^2\phi} d\phi\\ & = \int_0^{2\pi} \left[1 - \frac{a^2+2a}{(1+a)^2 + \cos^2\phi}\right]d\phi\\ & = 2\pi - 2(a^2+2a)\int_{-\pi/2}^{\pi/2} \frac{1}{(1+a)^2+\cos^2\phi}d\phi\\ & = 2\pi - 2(a^2+2a)\int_{-\pi/2}^{\pi/2} \frac{1}{(1+a)^2 \sin^2 \phi + [1+(1+a)^2]\cos^2\phi}d\phi\\ & = 2\pi - 2(a^2+2a)\int_{-\pi/2}^{\pi/2} \frac{\sec^2\phi}{(1+a)^2 \tan^2 \phi + [1+(1+a)^2]}d\phi. \end{align*}

Now let $u=\frac{|1+a|}{\sqrt{a^2+2a+2}}\tan\phi$, getting

$$ 2\pi - 2\frac{(a^2+2a)}{|1+a|\sqrt{a^2+2a+2}}\int_{-\infty}^{\infty} \frac{1}{u^2 + 1}du= 2\pi - \frac{2\pi(a^2+2a)}{|1+a|\sqrt{a^2+2a+2}}. $$

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$$ \begin{align} \int_0^{2\pi}\left|\frac{(1-i\cos(\phi))}{(1+a-i\cos(\phi))}\right|^2\,\mathrm{d}\phi &=\int_0^{2\pi}\frac{1+\cos^2(\phi)}{(1+a)^2+\cos^2(\phi)}\,\mathrm{d}\phi\\ &=2\pi-\int_0^{2\pi}\frac{(1+a)^2-1}{(1+a)^2+\cos^2(\phi)}\,\mathrm{d}\phi\\ &=2\pi-2\int_0^\pi\frac{(1+a)^2-1}{2(1+a)^2+1+\cos(2\phi)}\,\mathrm{d}(2\phi)\\ &=2\pi-\oint\frac{4(1+a)^2-4}{4(1+a)^2+2+\left(z+\frac1z\right)}\frac{\mathrm{d}z}{iz}\\ &=2\pi+i\oint\frac{4(1+a)^2-4}{z^2+(4(1+a)^2+2)z+1}\,\mathrm{d}z\\ &=2\pi+i(2\pi i)\frac{(1+a)^2-1}{|1+a|\sqrt{(1+a)^2+1}}\\ &=2\pi\left(1-\frac{(1+a)^2-1}{|1+a|\sqrt{(1+a)^2+1}}\right) \end{align} $$ Where $z=e^{i2\phi}$, the contour of integration is the unit circle, and the integrand has one singularity inside the unit circle at $z=-2(1+a)^2-1+2|1+a|\sqrt{(1+a)^2+1}$ with residue $\frac{(1+a)^2-1}{|1+a|\sqrt{(1+a)^2+1}}$

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