9
$\begingroup$

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?

$\endgroup$
  • 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
13
$\begingroup$

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 $\arctan(+\infty)$:

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
3
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.