complex-analysisresidue-calculus
How do you evaluate the following integral? Here we take $a>1$.
$$\int^{2\pi}_{0} \frac{1}{(a+\cos\theta)^{2}} d \theta=\frac{2\pi a}{(a^{2}-1)^{\frac{3}{2}}}.$$
I know I have to use the Residue Theorem, however, I am stuck on which contour to use, and also how to find the pole of the function. Any hints are greatly appreciated.
This seems quite silly to use complex analysis for.
$$\begin{aligned}\int_{-\infty}^{\infty}\frac{x-1}{x^3-1}\,dx &=\int_{-\infty}^\infty \frac{dx}{1+x+x^2}\\ &= \frac{2}{\sqrt{3}}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)\mid_{-\infty}^{\infty}\\ &= \frac{2\pi}{\sqrt{3}}\end{aligned}$$
If you're hell-bent on using the Residue Theorem though, let $\Gamma_R$ be the semi-cicular contour in the upper half-plane which is centered at $0$ and of radius $R$. Then, you know that
$$\int_{\Gamma_R}\frac{1}{z^2+z+1}\,dz = \int_{-R}^{R}\frac{dx}{1+x+x^2}+\int_{A_R}\frac{dz}{1+z+z^2}$$
where $A_R$ is the upper circular arc of $\Gamma_R$. Now, $\displaystyle \frac{1}{1+z+z^2}$ has only one pole in the interior of $\Gamma_R$, at $\displaystyle z_0=\frac{-1}{2}+i\frac{\sqrt{3}}{2}$.
Thus, you see that for every $R$ the Residue Theorem gives
$$\int_{\Gamma_R}\frac{dz}{1+z+z^2}=2\pi i\, \text{Res}\left(\frac{1}{1+z+z^2},z_0\right)$$
Thus, if you can show that
$$\displaystyle \int_{A_R}\frac{dz}{z^2+z+1}\xrightarrow{R\to\infty}0$\quad\mathbf{(1)}$$
You will have that
$$\int_{-\infty}^{\infty}\frac{dx}{1+x+x^2}=2\pi i\, \text{Res}\left(\frac{1}{1+z+z^2},z_0\right)$$
The residue, as I've said, is up to you. To prove $\mathbf{(1)}$ let me give you a hint: Jordan's Lemma.
Consider a contour that looks like this:
in the limit as $R$, the radius of $C_R$, goes to $\infty$. The contour should be constructed so that the point $z=a$ is inside the contour. The residue theorem then tells you $$\oint_C\frac{dz}{(z-a)\sqrt{1-z^2}}=-\frac{2\pi i}{\sqrt{1-a^2}}=\frac{2\pi}{\sqrt{a^2-1}}$$ where $\sqrt{a^2-1}$ is to be evaluated on branch $\arg(a-1),\arg(a+1)\in [0,2\pi)$. You can show that the integral over $C_R$ goes to zero as $R\rightarrow \infty$. The integral over the segments of the contour parallel to the imaginary axis obviously cancel each other. And you stated that you have already shown that the integral over $C_\rho$ goes to zero as $\rho\rightarrow 0$.
So now we must deal with the straight segments on opposite sides of the branch cut. We selected the branch with $\arg(1+z)\in [0,2\pi)$, $\arg(1-z)\in [-\pi,\pi)$. The discontinuity over the branch cut is captured by $\arg(1+z)$. So parametrize the upper segment using $z = \xi - 1$ and the lower segment by $z = \xi e^{2\pi i}-1$. Then we have:
\begin{align} \oint_C\frac{dz}{(z-a)\sqrt{1-z^2}}&=\int_2^0\frac{d\xi}{(\xi -a-1)\sqrt{2-\xi}\sqrt{\xi}}+\int_0^2\frac{e^{2\pi i}d\xi}{(\xi e^{2\pi i} -a-1)\sqrt{2-\xi e^{2\pi i}}\sqrt{\xi e^{2\pi i}}}\\ &=\int_2^0\frac{d\xi}{(\xi -a-1)\sqrt{2-\xi}\sqrt{\xi}}-\int_0^2\frac{d\xi}{(\xi -a-1)\sqrt{2-\xi}\sqrt{\xi}} \\ &=\int_1^{-1}\frac{dx}{(x-a)\sqrt{1-x^2}}-\int_{-1}^1\frac{dx}{(x-a)\sqrt{1-x^2}} \qquad \text{(subst. }x=\xi-1\text{)} \\ &= -2\int_{-1}^1\frac{dx}{(x-a)\sqrt{1-x^2}} \\ &= \frac{2\pi}{\sqrt{a^2-1}} \end{align} From this you can conclude that $$\int_{-1}^1\frac{dx}{(x-a)\sqrt{1-x^2}} = -\frac{\pi}{\sqrt{a^2-1}}$$ remembering that $\sqrt{a^2-1}$ is to be evaluated on branch $\arg(a+1),\arg(a-1) \in [0,2\pi)$.
Best Answer
Use the contour $|z| = 1$
First change the cosines into exponential forms.
$$\large\int \frac {1}{(a+\frac{e^{it}}{2} + \frac{e^{-it}}{2})^2} \ d\theta$$
$$z = e^{i\theta}\\ d\theta = \frac {1}{iz} dz$$
$$\large \oint_{|z| = 1} \frac {1}{iz(a+\frac{z}{2} + \frac{z^{-1}}{2})^2} \ dz$$
Which simplifies to:
$$\large \oint_{|z| = 1} \frac {4z}{i(z^2 + 2az+ 1)^2} \ dz\\ \oint_{|z| = 1} \frac {4z}{i(z+ a + \sqrt {a^2-1})^2(z+ a - \sqrt {a^2-1})^2} \ dz$$
Has one pole inside the contour.
The residual at $z = -a+\sqrt {a^2 - 1} = 2\pi i \frac {d}{dz} \frac {4z}{i(z+ a + \sqrt {a^2-1})^2}$ evaluated at $z = -a+\sqrt {a^2 - 1}$