I need to calculate $\int_{\gamma} \frac{1-\sin(z)}{2-\cos (z)}dz$ where $\gamma$ is the upper hemisphere of the circle with center $\pi$ and radius $\pi$, with a positive direction.

The original question was to calculate $\int_{0}^{2\pi}\frac{1-\sin(t)}{2-\cos(t)}dt$ using complex integration.

So I connected the half circle from $2\pi$ to $0$ with the straight line $0$ to $2\pi$ to get a half circle centered around $\pi$ with radius $\pi$. Let's call the curve of this circle $\Gamma$, and the curve from $2\pi$ to $0$ $\gamma$.

We know that $\int_{\Gamma} \frac{1-\sin(z)}{2-\cos (z)}dz=\int_{\gamma}\frac{1-\sin(z)}{2-\cos (z)}dz+\int_{0}^{2\pi}\frac{1-\sin(t)}{2-\cos (t)}dt$

since $2-\cos (z)$ has no roots inside $\Gamma$, it's analytic in $\Gamma$ and so $\int_{\Gamma} \frac{1-\sin(z)}{2-\cos (z)}dz=0$.

All that's left is to find $\int_{\gamma}\frac{1-\sin(z)}{2-\cos (z)}dz$ and I'm having some difficulties.

I can't find an easy parametrization. It's a half circle centered around $\pi$ with radius $\pi$, so the parametrization $z(t)=\pi+\pi e^{it}$ comes to mind, but it's not very pleasant to calculate that integral with that parametrization.

I could use a hand.

The final answer should be $\frac{2\pi}{\sqrt 3}$


As a comment suggested I went on a different approach:

$$\int_{0}^{2\pi} \frac{1-\sin (t)}{2-\cos (t)}dt=\int_{|z|=1}\frac{1-\frac{z-z^{-1}}{2i}}{2-\frac{z+z^{-1}}{2}}dz=\int_{|z|=1}\frac{-z^2+2iz+1}{-iz^2+4iz-i}dz$$

The roots of the denominator are $z_1=2+\sqrt{3}$ and $z_2=2-\sqrt{3}$, the numerator is not zero at those points, these are poles of the first order.



The sum of the residues is $2+4i$, so from residue theorem $$\int_{0}^{2\pi}\frac{1-\sin(t)}{2-\cos(t)}dt=\int_{|z|=1}\frac{1-\frac{z-z^{-1}}{2i}}{2-\frac{z+z^{-1}}{2}}dz=2i\pi(2+4i)=4i\pi-8\pi$$

Is this correct? This doesn't agree with the given answer of $\frac{2\pi}{\sqrt{3}}$

  • 1
    $\begingroup$ Your contour integral is not equal to $\int_0^{2\pi} \frac{1 - \sin t}{2 - \cos t}\, dt$. It will be helpful if you write $\sin t = (e^{it} - e^{-it})/(2i)$ and $\cos t = (e^{it} + e^{-it})/2$. Use the unit circle as the contour, with parametrization $z = e^{it}$, $0 \le t \le 2\pi$. $\endgroup$ – kobe Jan 27 '15 at 19:30

I suggest a different approach. First of all please note that \begin{equation} sin(t)=\frac{e^{jt}-e^{-jt}}{2j} \end{equation} and \begin{equation} cos(t)=\frac{e^{jt}-e^{-jt}}{2} \end{equation} Since $e^{jt}=z$ you can wtite your integral as \begin{equation} -\frac{j(z-j)^2}{1-4 z+z^2} \end{equation} while $dt= dz/(jz)$. Now your integral can be evaluates by adding the residues: \begin{equation} \int{-\frac{(z-j)^2}{z(1-4 z+z^2)}}=2 \pi j (Res(0)+Res(2-\sqrt(3))) \end{equation} The residue in $2+\sqrt(3)$ is outside the circle of radius 1, so it does not contribute to the integral. The results is exactly $\frac{2 \pi}{\sqrt 3}$.


Your contour integral is incorrect. It should be

$$\int_{|z| = 1} \frac{1 - \frac{z - z^{-1}}{2i}}{2 - \frac{z + z^{-1}}{2}}\, \frac{dz}{iz}$$

which simplifies to $$-\int_{|z| = 1} \frac{z^2 - 2iz + 1}{z(z^2 - 4z + 1)}\, dz$$


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