$$\int_0^{\pi} \frac{2 d\theta}{k - cos\theta}$$
What is k here?
I'm not sure if I understand the question.
I am told from my book the answer is $\frac{2 \pi}{\sqrt{k^2 - 1}}$ but not sure how we got here.
Asked
Active
Viewed 29 times
-1
-
For $|k|>1$ we have $\int_0^\pi \frac2{k-\cos(\theta)},d\theta=\int_{-\pi}^\pi \frac1{k-\cos(\theta)},d\theta$. Then, enforcing the transformation $z=e^{i\theta}$ we have $$\int_0^\pi \frac2{k-\cos(\theta)},d\theta=\oint_{|z|=1}\frac1{k-\frac12(z+z^{-1})},\frac{1}{iz},dz$$Now, apply the residue theorem. – Mark Viola Dec 03 '20 at 23:55
1 Answers
0
Let $I := \int_{0}^{\pi} \frac{2\, d\theta}{k-\cos(\theta)}$.
Then \begin{align*} I &= \frac{1}{2} \int_{0}^{2\pi} \frac{4}{2\left (k-\frac{1}{2}e^{i\theta}-\frac{1}{2}e^{-i\theta} \right )}\, d\theta \\ &= -2i \int_{0}^{2\pi} \frac{1}{2ke^{i\theta}-e^{2i\theta}-1}\, (ie^{i\theta}\, d\theta) \\ &= -2i \int_{\{|z|=1\}} \frac{1}{-z^{2}+2k-1}\, dz \\ &= 4\pi \sum_{\sigma \in \{|z|<1 \}} \mathrm{Res}(\sigma). \end{align*}
Now $z^{2}-2k+1$ has roots $k \pm \sqrt{k^2-1}$. The sole residue inside $\{|z| < 1 \}$ is $k - \sqrt{k^2-1}$ with residue $\frac{-1}{(k-\sqrt{k^2-1})-k-\sqrt{k^2-1}} = \frac{1}{2\sqrt{k^2-1}}$.
Your integral should be $$I = 4\pi \cdot \frac{1}{2\sqrt{k^2-1}} = \frac{2\pi}{\sqrt{k^{2}-1}}.$$
Christopher K
- 3,383