I want to evaluate the following complex contour integral:

$$ J = \oint_{C}\frac{1}{\cos z+1}dz $$ where $C: |z|=5 $

I began by finding the singularities included inside $C$. With the following limit I find that $z=\pi$ is a 2nd order pole:

$$ \lim_{z\rightarrow\pi}{(z-\pi)^{2}\cdot f(z)} = \lim_{z\rightarrow\pi}\frac{(z-\pi)^{2}}{\cos(z)+1} = \lim_{z\rightarrow\pi}\frac{\frac{\mathrm{d}^{2} }{\mathrm{d} z^{2}}(z-\pi)^{2}}{\frac{\mathrm{d}^{2} }{\mathrm{d} z^{2}}(\cos(z)+1)} = 2 \neq0 $$

I am now trying to evaluate the integral using the residues theorem by which:

$$ J=2\pi iRes(f;z=\pi) $$

The problem I'm encountering is that when trying to evaluate this residue with second order poles, I find this:

$$ Res(f;z=\pi)= \lim_{z\rightarrow\pi}\frac{\mathrm{d} }{\mathrm{d} z}\frac{(z-\pi)^{2}}{(\cos(z)+1)} =\cdots $$

For which I have to apply L'Hospital's rule to the limit about 4 times which is a pain to say the least but seems like it should work. Using Maple I find that the residue is equal to 0 which makes me wonder if I could have seen this all along.

My question is: Is there some theorem or some method that I'm not aware of that can be used to compute this integral? Even though manually evaluating the residues seems to work, I hardly doubt this is the way this problem was intended to be solved given that this could potentially be an exam problem.

  • 2
    $\begingroup$ $\cos \frac{3\pi}{2} = 0$, the only singularities enclosed by the contour are $\pm\pi$. $\endgroup$ Oct 6 '14 at 18:09
  • $\begingroup$ I'm not sure I understand. How is -pi a singularity? It seems to me that it does not cause a problem. Also to me z = 3pi/2 < 5 so it should be included in the contour, what am I not understanding? $\endgroup$
    – user33355
    Oct 6 '14 at 18:14
  • $\begingroup$ $\cos (-\pi) = -1$, so $1 + \cos (-\pi) = 0$. But $1+\cos \frac{3\pi}{2} = 1 \neq 0$. $\endgroup$ Oct 6 '14 at 18:15
  • 1
    $\begingroup$ By the way, the contour certainly is $\lvert z\rvert = 5$? $\endgroup$ Oct 6 '14 at 18:16
  • $\begingroup$ Pardon me, yes it is equal to 5, not lesser than. I see the singularity at -pi now, thank you. For some reason I thought of cos(3pi/2) being -1 but I mistook it for cos(3pi). It is a lot clearer now and I edited the question. I'm still not sure how to compute the residue $\endgroup$
    – user33355
    Oct 6 '14 at 18:20

The singularities enclosed by the contour are $\pm \pi$.

The fastest way to evaluate the integral is symmetry: The integrand is even, the contour is symmetric about $0$, hence the integral is $0$.

The second fastest method is the residue theorem, using a Laurent expansion:

$$1 + \cos (\mp\pi +(z\pm\pi)) = 1 - \cos (z\pm \pi) = \frac{(z\pm\pi)^2}{2} - \frac{(z\pm\pi)^4}{4!} + \dotsc,$$

which gives the Laurent expansion

$$\frac{1}{1+\cos z} = \frac{2}{(z\pm\pi)^2} \frac{1}{1 - \frac{(z\pm\pi)^2}{12} + \dotsc} = \frac{2}{(z\pm\pi)^2}\left(1 + \frac{(z\pm\pi)^2}{12} + O((z\pm\pi)^4)\right)$$

showing that the residue is $0$ with little work.

  • $\begingroup$ Thank you! The Laurent expansion is what I needed I didn't really see it but this makes it much simpler! $\endgroup$
    – user33355
    Oct 6 '14 at 18:28

After taking the derivative, you are left with a limit of $\cdots/(\cos x+1)^2$. Try multiplying and dividing by $(z-\pi)^4$ and using the limit you computed at the start (where you found that $z=\pi$ is a second order pole). Pulling out the factor $(z-\pi)^4/(\cos x+1)^2$, whose limit is $4$, you are now left with a fraction much more amenable to repeated use of L'Hôpital.


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.