I would like to compute the following integral. This is for a complex analysis course but I managed to around some other integrals using real analysis methodologies. Hopefully one might be able to do for this one too.

$$\int_{0}^{2\pi} \frac{1}{a-\cos(x)}dx, \text{ with } a > 1.$$

Any suggestion will be greatly appreciated.

  • $\begingroup$ Hint: use Euler's formulas and set $z=e^{ix}$. This will turn your integral into an integral along the unit circle. $\endgroup$ – mrf Dec 3 '12 at 21:32

$$\cos x = \frac{e^{i x}+e^{-ix}}{2}$$


$$\int_0^{2\pi} \frac{dx}{a-\frac{e^{i x}+e^{-ix}}{2}} = \oint_C \frac{1}{a-\frac{z+z^{-1}}{2}} \frac{dz}{iz} = - i\oint_C \frac{dz}{az-\frac{z^2+1}{2}} = 2i\oint_C \frac{dz}{z^2-2az+1}$$

where $C$ describes the unit circle $|z|=1$, centred at the origin, parametrized by $e^{iz}$ where $0\le z\le 2\pi$.

Letting $f(z)=\frac{1}{z^2-2az+1}$, we find that the poles of $f$ are at $z=a\pm\sqrt{a^2-1}$. Noting that $a>1$, the only pole in $C$ is the one with the negative sign. Then

$$\operatorname*{Res}_{z = a-\sqrt{a^2-1}}f(z)= \lim_{z\to a-\sqrt{a^2-1}}\frac{z-a+\sqrt{s^2+1}}{z^2-2az+1}= \lim_{z\to a-\sqrt{a^2-1}}\frac{1}{2z-2a}=- \frac{1}{2\sqrt{a^2-1}}$$

And thus we wrap up:

$$\int_0^{2\pi} \frac{dx}{a-\cos x} = 2i\oint_C \frac{dz}{z^2-2az+1} = 2i\left(-\frac{2\pi i }{2\sqrt{a^2-1}}\right) = \frac{2\pi}{\sqrt{a^2-1}}$$



Let $\tan(x/2) = t$. We then get that $$\sec^2(x/2) dx = 2dt \implies dx = \dfrac{2dt}{1+t^2}$$ Also, $\cos(x) = \dfrac{1-\tan^2(x/2)}{1+\tan^2(x/2)} = \dfrac{1-t^2}{1+t^2}$. Hence, \begin{align} \dfrac{dx}{a- \cos(x)} & = \dfrac{2dt}{1+t^2} \dfrac1{a - \dfrac{1-t^2}{1+t^2}}\\ & = \dfrac{2dt}{a(1+t^2) - (1-t^2)}\\ & = \dfrac{2dt}{(a+1) t^2 + (a-1)} \end{align} Hence, $$I = \int \dfrac{2dt}{(a+1) t^2 + (a-1)} = \dfrac2{a+1} \int \dfrac{dt}{t^2 + \dfrac{a-1}{a+1}} = \dfrac2{\sqrt{a^2-1}} \arctan \left(t\sqrt{\dfrac{a+1}{a-1}}\right) + c$$ Writing it in terms of $x$, we get that $$I = \dfrac2{\sqrt{a^2-1}} \arctan \left(\tan(x/2)\sqrt{\dfrac{a+1}{a-1}}\right) + c$$

  • $\begingroup$ Amazingly fast! Thank you very much! $\endgroup$ – user44069 Dec 3 '12 at 21:32
  • $\begingroup$ How is this a complex analysis solution? $\endgroup$ – mrf Dec 3 '12 at 21:33
  • $\begingroup$ @mrf If I correctly interpreted the question, the OP wants to solve some of the integrals without resorting to complex analysis. $\endgroup$ – user17762 Dec 3 '12 at 21:34
  • $\begingroup$ Oh, I read it exactly the other way around, but after reading it again, you're probably right. $\endgroup$ – mrf Dec 3 '12 at 21:37
  • $\begingroup$ But a is greater than 1. It follows that there exists no singularities on the whole complex plane. Then by the Riemann integral theorem the integral result will be 0 directly. Right? $\endgroup$ – Scorpio19891119 Dec 3 '12 at 21:38

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