Tell me more ×
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.
up vote 2 down vote favorite

Given $a > b > 0$ what is the fastest possible way to evaluate the following integral using Residue theorem. I'm confused weather to take the imaginary part of $z^2$ or whole integral. $$\int_0^{2 \pi} {\cos^2 \theta \over a + b \cos \theta}\; d\theta$$

share|improve this question

3 Answers

up vote 2 down vote accepted

If $z=e^{i\theta}$, then $\cos\theta=\dfrac 1 2\left(z+\dfrac1z\right)$, and $dz=ie^{i\theta}\,d\theta/2=iz\,d\theta/2$, so $-2i\dfrac{dz}{z} = d\theta$. Then $$ \int_0^{2\pi} \frac{\cos^2\theta}{a+b\cos\theta} d\theta = \int\limits_\text{circle} \frac{\frac14\left(z+\frac1z\right)^2}{a+\frac b2\left(z+\frac1z\right)}(-i)\frac{dz}{z} = -i\int\limits_\text{circle} \frac{z^4+2z^2+1}{2z^2(2az+b(z^2+1))} dz. $$

This function has a double pole at $z=0$ and simple poles at $\dfrac{-a\pm\sqrt{a^2-b^2}}{b}$. So the question is: for which values of $a,b$ are the simple poles inside the circle? If there's just one simple pole inside the circle at $c$, then the integral becomes $$ \int\limits_\text{circle} \frac{g(z)}{z-c} dz = 2\pi i g(c), $$ where $g(z)$ is whatever's left after you've factored out $1/(z-c)$. If there's more than one simple pole, you need a sum: take values of $g$ at those points and sum them.

share|improve this answer
Actually I found I could substitute $ \cos^2 \theta = Im(z^2)$ and simplify things a lot :D the whole integral will be $ Im ( \int z^2/...)$ – hasExams Sep 1 '12 at 17:41
We have $e^{2i\theta}=\cos[2\theta]+i\sin[2\theta]$. You take the real part ($\cos^{2}[\theta]-\sin^{2}[\theta]$), not the imaginary part (which is $2\sin[\theta]\cos[\theta]$ instead). – user32240 Sep 2 '12 at 6:17
1  
@Michael: We have $\cos[\theta]=\frac{1}{2}(\cos[\theta]+i\sin[\theta])+(\cos[\theta]-i\sin[\theta‌​])$, you missed a factor. – user32240 Sep 2 '12 at 6:27
@user32240 : Fixed now, I hope. – Michael Hardy Sep 2 '12 at 18:34
up vote 1 down vote

Hint: put $z={\rm e}^{i\theta}$ and change the integral to the form $ \int_{|z|=1}f(z)\,dz \,,$ then use residue theorem. Exploit the identity $$ \cos\theta = \frac{1}{2}\left({\rm e}^{i \theta} + {\rm e}^{- i \theta}\right)\,. $$

share|improve this answer
Are you suggesting doing an integral of the form $\int_{|z|=1}\frac{z^{2}}{a+bz}$? But though it has an obvious pole at $z=-\frac{a}{b}$(which is outside of the contour), I do not see how to get the original integral from $\cos[\theta]=\frac{z+\overline{z}}{2}$. But thanks for the hint. – user32240 Sep 1 '12 at 8:37
yeah ... kinda something like that. But instead of squaring ... you can take Real part of $z^2$ which makes the problem very simple. Kinda confused on that – hasExams Sep 1 '12 at 8:38
I am not sure how to deal with the bottom part, $(a+b\cos[\theta])$ is not easily treatable. – user32240 Sep 1 '12 at 8:45
up vote 1 down vote

We should have

$$\frac{\cos^{2}\theta}{a+b\cos\theta}=\frac{1}{b}\cos\theta-\frac{\frac{a}{b}\cos\theta}{a+b\cos\theta}.$$

The first part of the definite integral is easily evaluated. The second part is the same as evaluating

$$\int \frac{\cos\theta}{a+b\cos\theta}d\theta=\int\frac{1}{b}\left(1-\frac{a}{a+b\cos\theta}\right)d\theta.$$

So we only need to evaluate

$$\int\frac{1}{a+b\cos\theta}d\theta.$$

This can be done by various trignometry identities such as using $\tan(\theta/2)$. A detailed step by step proof can be found here (click show steps).

share|improve this answer
thanks for idea ... I think i can evaluate $\frac{1}{a+b\cos[\theta]}d\theta$ very easily – hasExams Sep 1 '12 at 8:25
well, I am not sure if this is the "fastest possible" way to do it. I guess you are expecting some contour integral or manipulations involving $\Gamma$ functions. – user32240 Sep 1 '12 at 8:28
I'm supposed to use contour integral ... the $z^2$ part is awful – hasExams Sep 1 '12 at 8:40

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.