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Given that $$\int_{|z|=1|}\frac{z^2}{2z+1} dz = \frac{i\pi}{4}$$,
show $$\int_{0}^\pi \frac{2\cos 2 \theta + \cos 3\theta}{5+4\cos\theta} = \frac{\pi}{8}$$.

I saw the bounds of the latter integral and thought that I should try and parametrize using $z = e^{2i\theta}$ where $\theta \in [0,\pi]$. This doesn't seem to simplify easily.
i saw this thread: Show that $\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$

and the top answer says:
$$\begin{align} \int_0^\pi \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,d\theta&=\frac12\text{Re}\left(\oint_{|z|=1}\frac{2z^2+z^3}{5+2(z+z^{-1})}\,\frac{1}{iz}\,dz\right)\\\\\end{align}$$
which I don't understand.

How does multiplying a half to the integral with contour $|z|=1$ (parametrized by $z = e^{i\theta}, \theta\in [0,2\pi]$) give the LHS?

I tried looking at it by taking the latter integral and using the substitution $u=\pi + \theta$, in hopes that the integrand simplifies to stay the same but it doesn't, so I can't see why the integral with bounds $0,\pi$ is half the integral that would have bounds $0,2\pi$ (since we would use the parametrization $z=e^{i\theta}$).

Twenty-six colours
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2 Answers2

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With $\theta\to-\theta$ $$ I=\int_0^{-\pi} \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,(-)d\theta=\int_{-\pi}^0 \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,d\theta $$ \begin{align} 2I &=\int_{-\pi}^\pi \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\, d\theta\\ &=\int_{|z|=1} \frac{{\bf Re\,}(2z^2+z^3)}{5+2(z+z^{-1})}\,\dfrac{1}{iz} dz,\\ &={\bf Re\,}\int_{|z|=1} \frac{2z^2+z^3}{5+2(z+z^{-1})}\,\dfrac{1}{iz} dz,\\ &={\bf Re\,}\dfrac{1}{i}\int_{|z|=1} \frac{z^2}{2z+1}\, dz, \end{align} Note that $\overline{z}=\dfrac{1}{z}$ so $\overline{dz}=d\overline{z}=-\dfrac{dz}{z^2}$ and $\dfrac{\overline{dz}}{\overline{iz}}=-\dfrac{dz}{-i\overline{z}z^2}=\dfrac{dz}{iz}$ and also ${\bf Re\,}(z+\overline{z})=z+\overline{z}$. The rest is simple.

Nosrati
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Note that

$$2\int^\pi_0 \cos(n\theta)f(\cos \theta ) d\theta = \int^{\pi}_{-\pi} \cos(n\theta)f(\cos\theta ) d \theta $$

Now you can use that $\gamma(t) = e^{i t}$ where $t \in [-\pi , \pi )$ to prove

$$\Re \oint_{|z|=1} z^n f\left(\frac{z+z^{-1}}{2} \right) \frac{dz}{iz} = 2\int^\pi_0 \cos(n\theta)f(\cos \theta ) d\theta $$

Or any linear combination

$$\Re \oint_{|z|=1} (az^n+bz^m) f\left(\frac{z+z^{-1}}{2} \right) \frac{dz}{iz} = 2\int^\pi_0 (a\cos(n\theta)+b \cos(m\theta))f(\cos \theta ) d\theta $$

So we have by choosing $n=2,m=3,a=2,b=1$

$$\Re \oint_{|z|=1} \frac{z^3+2z^2}{5+2(z+z^{-1})} \frac{dz}{iz} =2\int^{\pi}_{0}\frac{\cos(3t)+2\cos(2t)}{5+4\cos(t)}dt $$

Note that

$$\frac{z^3+2z^2}{5+2(z+z^{-1})} \frac{1}{z} = \frac{z^2(z+2)}{(z+2)(2z+1)} = \frac{z^2}{2z+1}$$

So we deduce that

$$2\int^{\pi}_{0}\frac{\cos(3t)+2\cos(2t)}{5+4\cos(t)}dt =-i \Re \oint_{|z|=1} \frac{z^2}{2z+1} = \frac{\pi}{4}$$

Zaid Alyafeai
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