Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

$$\int_0^{2\pi}\left|\frac{(1-i\cos(\phi))}{(1+a-i\cos(\phi))}\right|^2d\phi$$

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?

share|improve this question

2 Answers 2

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

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

share|improve this answer

$$ \begin{align} \int_0^{2\pi}\left|\frac{(1-i\cos(\phi))}{(1+a-i\cos(\phi))}\right|^2\,\mathrm{d}\phi &=\int_0^{2\pi}\frac{1+\cos^2(\phi)}{(1+a)^2+\cos^2(\phi)}\,\mathrm{d}\phi\\ &=2\pi-\int_0^{2\pi}\frac{(1+a)^2-1}{(1+a)^2+\cos^2(\phi)}\,\mathrm{d}\phi\\ &=2\pi-2\int_0^\pi\frac{(1+a)^2-1}{2(1+a)^2+1+\cos(2\phi)}\,\mathrm{d}(2\phi)\\ &=2\pi-\oint\frac{4(1+a)^2-4}{4(1+a)^2+2+\left(z+\frac1z\right)}\frac{\mathrm{d}z}{iz}\\ &=2\pi+i\oint\frac{4(1+a)^2-4}{z^2+(4(1+a)^2+2)z+1}\,\mathrm{d}z\\ &=2\pi+i(2\pi i)\frac{(1+a)^2-1}{|1+a|\sqrt{(1+a)^2+1}}\\ &=2\pi\left(1-\frac{(1+a)^2-1}{|1+a|\sqrt{(1+a)^2+1}}\right) \end{align} $$ 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 $\frac{(1+a)^2-1}{|1+a|\sqrt{(1+a)^2+1}}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.