The integral is$$\int_0^{2\pi}\frac{\mathrm dθ}{2-\cosθ}.$$Just to skip time, the answer of the indefinite integral is $\dfrac2{\sqrt{3}}\tan^{-1}\left(\sqrt3\tan\left(\dfracθ2\right)\right)$.

Evaluating it from $0$ to $ 2 \pi$ yields$$\frac2{\sqrt3}\tan^{-1}(\sqrt3 \tanπ)-\frac2{\sqrt3}\tan^{-1}(\sqrt3 \tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2i\int_C\frac{\mathrm dz}{z^2-4z+1}=2i\int_C\frac{\mathrm dz}{(z-2+\sqrt3)(z-2-\sqrt3)},$$ where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-\sqrt3$ is inside the domain of the region bounded by $C$, then: $$2i\int_C\frac{\mathrm dz}{(z-2+\sqrt3)(z-2-\sqrt3)}=2πi\frac{2i}{2-\sqrt3-2-\sqrt3}=2πi\frac{2i}{-2\sqrt3}=\frac{2π}{\sqrt3}.$$

Using real analysis I get $0$, using complex analysis I get $\dfrac{2π}{\sqrt3}$. What is wrong?

  • 2
    $\begingroup$ OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0. $\endgroup$ – Ben Millwood Jan 2 at 12:56

The problem with the real approach is that you make the change of variable $t=\tan\left(\dfrac{\theta}{2}\right)$ for $0 < \theta < 2 \pi$.

This is problematic since your substitution need to be defined and continuous for all $\theta$, but you have a problem when $\theta=\pi$.

Edit: Note that if you split the integral into $\int_0^\pi+\int_\pi^{2 \pi}$, you are going to get the right answer, as for one integral you are going to get $\arctan(- \infty)$ and for the other $\arctan(+\infty)$:

$$\int_0^{2 \pi} \frac{\mathrm{d}θ}{2-\cos \theta}=\int_0^\pi \frac{\mathrm{d}θ}{2-\cos \theta}+\int_\pi ^{2 \pi} \frac{\mathrm{d}θ}{2-\cos \theta}\\ = \lim_{r \to \pi_-} \int_0^r \frac{\mathrm{d}θ}{2-\cos \theta}+ \lim_{w \to \pi_+} \int_w^{2 \pi} \frac{\mathrm{d}θ}{2-\cos \theta}\\= \lim_{r \to \pi_-} \left(\frac{2\tan^-1( \sqrt{3} \tan( \frac{ r}{2}))}{ \sqrt{3}}-0\right)+ \lim_{w \to \pi_+}\left(0- \frac{2\tan^-1( \sqrt{3} \tan( \frac{ r}{2}))}{ \sqrt{3}}\right).$$

  • $\begingroup$ Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else? $\endgroup$ – khaled014z Jan 2 at 0:26
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    $\begingroup$ @khaled014z See the edit. Let me know if you want more details. $\endgroup$ – N. S. Jan 2 at 0:27
  • $\begingroup$ Brilliant, that was kind of a tricky substitution, thank you $\endgroup$ – khaled014z Jan 2 at 0:31
  • $\begingroup$ When this is next edited, you want tan^{-1} twice in the last line. $\endgroup$ – Teepeemm Jan 2 at 14:12

Note that that tangent function, $\tan(x)$, is discontinuous when $x=\pi/2+n\pi$. So, the antiderivative $\frac2{\sqrt{3}} \arctan\left(\sqrt 3 \tan(\theta/2)\right)$ is not valid over the interval $[0,2\pi]$.

Instead, we have

$$\int_0^{2\pi}\frac{1}{2-\cos(\theta)}\,d\theta=2\int_0^\pi\frac{1}{2-\cos(\theta)}\,d\theta=\frac{4}{\sqrt3}\left.\left(\arctan\left(\sqrt 3 \tan(\theta/2)\right)\right)\right|_0^\pi=\frac{2\pi}{\sqrt3}$$

  • $\begingroup$ Don't you mean $x = \frac{\pi}{2} + n\pi$? $\endgroup$ – user150203 Jan 2 at 3:18
  • $\begingroup$ Hi Mark ! Happy New Year ! $\endgroup$ – Claude Leibovici Jan 2 at 3:33
  • $\begingroup$ @DavidG Yes, of course. Thank you for the comment. $\endgroup$ – Mark Viola Jan 2 at 14:36
  • $\begingroup$ @ClaudeLeibovici Hi Claude! Happy New Year to you too! $\endgroup$ – Mark Viola Jan 2 at 14:36

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