Solving the integral $\int_{0}^{2\pi}\frac{dt}{(a+b\cos(t))^2}$

Question

For solving the iuntegral: $\int_{0}^{2\pi}\frac{dt}{(a+b\cos(t))^2}$ I will make it a complex one using the substitusion: $e^{it}=z$ So the integral becomes: $$\frac{4}{i}\oint_V \frac{zdz}{(bz^2+2az+1)^2}=\frac{4}{i}\oint_V \frac{zdz}{(z-z_1)^2(z-z_2)^2}$$ Where: $$z_1=-\frac{a+\sqrt{a^2-b^2}}{b} ; z_2=-\frac{a-\sqrt{a^2-b^2}}{b}$$ We also have the condition: $a>b>0$ So only $z_2$ is inside the contour. I tried to find the only residue by the formula: $$\text{Res}(f;z_2)=\frac{1}{(m-1)!}\lim_{z\to z_2} \frac{d^{m-1}}{dz^{m-1}}((z-z_2)^mf(z))$$ $$m=2$$ $$\Rightarrow \text{Res}(f;z_2)=\lim_{z\to z_2}\frac{d}{dz}\left(\frac{z}{(z-z_1)^2}\right)$$ However In my textbook it says that: $$\text{Res}(f;z_2)=\lim_{z\to z_2}\frac{d}{dz}\left(\frac{z}{b^2(z-z_1)^2}\right)$$ Where does that $b^2$ come from? I just can't understand.

Answer 1

It is not correct that $$ \frac{4}{i}\oint_V \frac{dz}{(bz^2+2az+1)^2}=\frac{4}{i}\oint_V \frac{dz}{(z-z_1)^2(z-z_2)^2}, $$ because it is not correct that $$ bz^2+2az+1 = (z-z_1) (z - z_2). $$

To see why it is not correct, compare the coefficients of $z^2$: the coefficient of $z^2$ on the left side is $b$, while that on the right side is $1$. They cannot be equal unless $b = 1$.

It is correct that $$ bz^2+2az+1 = \color{red}{b} (z-z_1) (z - z_2). $$

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Well, there are still mistakes in your original post. I will solve the integral below.

Put $z = \mathrm{e}^{\mathrm{i}t}$. We have $$ \cos {t} = \frac{z^2 + 1}{2z} $$ and $$ z = \mathrm{e}^{\mathrm{i}t} \implies \mathrm{d}z = \mathrm{i} z\, \mathrm{d}t \implies \mathrm{d}t = \frac{\mathrm{d}z}{\mathrm{i}z}. $$ Hence $$ a + b \cos {t} = \frac{bz^2 + 2az + \color{red}{b}}{2z} $$ and $$ \frac{1}{(a + b \cos {t})^2} = \frac{4z^2}{(bz^2 + 2az + \color{red}{b})^2} $$ and $$ \begin{aligned} & \int_{0}^{2\pi} \frac{\mathrm{d}t}{(a + b \cos {t})^2} \\ = {} & \oint_{|z| = 1} \frac{\mathrm{d}z}{\mathrm{i}z} \frac{4z^2}{(bz^2 + 2az + \color{red}{b})^2} \\ = {} & \frac{4}{\mathrm{i}\color{red}{b^2}} \oint_{|z| = 1} \frac{\color{red}{z}}{(z^2 + 2kz + 1)^2}\,\mathrm{d}{z} \\ = {} & \frac{4}{\mathrm{i}\color{red}{b^2}} \oint_{|z| = 1} \frac{\color{red}{z}}{(z - z_1)^2 (z - z_2)^2}\,\mathrm{d}{z}, \end{aligned} $$ in which $$ \begin{aligned} & k = \frac{a}{b}, \\ & z_1 = -k - \sqrt{k^2 - 1}, \\ & z_2 = -k + \sqrt{k^2 - 1}. \end{aligned} $$

Since $k > 1$, $$ \begin{aligned} z_2 = {} & -(k - \sqrt{k^2 - 1}) \\ = {} & \frac{-(k - \sqrt{k^2 - 1})(k + \sqrt{k^2 - 1})}{k + \sqrt{k^2 - 1}} \\ = {} & \frac{-1}{k + \sqrt{k^2 - 1}} \end{aligned} $$ is in the interval $(-1, 0)$, and $$ z_1 = -(k + \sqrt{k^2 - 1}) < -1. $$ Hence only $z_2$ is inside the contour.

The residue is $$ \mathrm{Res}(f;z_2)=\frac{1}{(m-1)!}\lim_{z\to z_2} \frac{\mathrm{d}^{m-1}}{\mathrm{d}z^{m-1}}((z-z_2)^mf(z)), $$ in which $m = 2$ and $$ f(z) = \frac{z}{(z - z_1)^2 (z - z_2)^2}. $$ It is not hard to find that $$ \mathrm{Res}(f;z_2) = -\frac{z_2 + z_1}{(z_2 - z_1)^3} = \frac{k}{4 (k^2 - 1)^{3/2}}. $$ Hence $$ \begin{aligned} & \int_{0}^{2\pi} \frac{\mathrm{d}t}{(a + b \cos {t})^2} \\ = {} & \oint_{|z| = 1} \frac{\mathrm{d}z}{\mathrm{i}z} \frac{4z^2}{(bz^2 + 2az + b)^2} \\ = {} & \frac{4}{\mathrm{i}\color{red}{b^2}} \oint_{|z| = 1} \frac{\color{red}{z}}{(z^2 + 2kz + 1)^2}\,\mathrm{d}{z} \\ = {} & \frac{4}{\mathrm{i}\color{red}{b^2}} \oint_{|z| = 1} \frac{\color{red}{z}}{(z - z_1)^2 (z - z_2)^2}\,\mathrm{d}{z} \\ = {} &\frac{4}{\mathrm{i} b^2} \cdot 2\pi\mathrm{i} \cdot \frac{k}{4 (k^2 - 1)^{3/2}} \\ = {} & \frac{2\pi k}{b^2 (k^2 - 1)^{3/2}} \\ = {} & \frac{2\pi kb}{b^3 (k^2 - 1)^{3/2}} \\ = {} & \frac{2\pi a}{(b^2)^{3/2} (k^2 - 1)^{3/2}} \\ = {} & \frac{2\pi a}{(a^2 - b^2)^{3/2}}. \end{aligned} $$

Answer 2

One mistake has been pointed out by @Juliamisto

Another mistake is

$$\int_{0}^{2\pi}\frac{dt}{(a+b\cos(t))^2}=\frac{4}{i}\oint_V \frac{\color{red}z~dz}{(bz^2+2az+1)^2}$$

you miss a factor $z$ on the numerator.