I am doing contour integration, and I'm trying to solve a couple problems. I am not sure if I am doing them correctly, so if anyone can explain the steps to me, it would be much appreciated:

$(a)\int_{|z| = 1}\frac{\sin z}{z}dz$ , counterclockwise

$(b) \int_{|z| = 2}\frac{1}{z^2-1}dz$, counterclockwise

For (a), I think I should use the Cauchy formula, $f(z_0) = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z - z_0}dz$, so that I could set $z_0 = 0$, which gives me $f(0)2\pi i = \int_\gamma \frac{\sin z}{z-0}dz = \sin(0)2\pi i = 0$. I think that another way to solve it is to parameterize, but I have not gotten past $\int_{|z| = 1}\frac{\sin z}{z}dz = \int_0^{2\pi} \frac{\sin(e^{i\theta})}{e^{i\theta}}\frac{ie^{i\theta}\cos(e^{i\theta}) - \sin(e^{i\theta})}{e^{i\theta}}d\theta$.

For (b), I separated out the denominator to get $\int_{|z| = 2} \frac{1}{(z-1)(z+1)}dz$ and was not sure where to go from there.

As a quick side question, when is it most appropriate to use the Cauchy formula or to parameterize? And when do I need to worry about the points where the function does not exist (i.e. in (a), z = 0 and in (b), z = 1, -1)? And are there other ways to compute these integrals? All help is appreciated.


Your computation of the first integral, through Cauchy's integral formula, is just fine. It's not amazing that you cannot compute it from the definition of contour integral.

Concerning the second one, now you use the fact that$$\frac1{(z-1)(z+1)}=\frac{1/2}{z-1}-\frac{1/2}{z+1}.$$

  • $\begingroup$ By your second comment, do you mean that I should do the algebra, or that I should use the fact that $e^{i\theta}$ is never 0? There is a theorem that states the integral of a closed contour over an analytic function is 0, I think. Is this correct? If so, is it what I should be using? $\endgroup$ – johndisk Oct 9 '19 at 17:19
  • $\begingroup$ What I was suggestiong is that you do\begin{align}\int_{\lvert z\rvert=2}\frac{\mathrm dz}{(z-1)(z+1)}&=\frac12\int_{\lvert z\rvert=2}\frac{\mathrm dz}{z-1}-\frac12\int_{\lvert z\rvert=2}\frac{\mathrm dz}{z+1}\\&=\frac122\pi i-\frac122\pi i\\&=0.\end{align} $\endgroup$ – José Carlos Santos Oct 9 '19 at 17:26
  • $\begingroup$ I was referring to "It's not amazing that you cannot compute it from the definition of countour integral." -what do you mean by using the definition of the contour integral for (a)? $\endgroup$ – johndisk Oct 9 '19 at 17:31
  • $\begingroup$ What you tried to do was to parametrize your closed path and to use that parametrization to compute the integral That approach, which is logically correct, almost always leads to integrals which are very hard to compute directly. $\endgroup$ – José Carlos Santos Oct 9 '19 at 17:33

for (a), may you write

$$\begin{aligned} \int_\gamma {\frac{\sin z}{z} \mathrm{d}z} & = \int_{0}^{2\pi} {\frac{\sin(e^{i\theta})}{e^{i\theta}} \mathrm{d}(e^{i\theta})} = i\int_{0}^{2\pi} {\sin(e^{i\theta}) \>\mathrm{d}\theta}\\ & = i\int_{0}^{\pi} {\sin(e^{i\theta}) \>\mathrm{d}\theta} + i\int_{\pi}^{2\pi} {\sin(e^{i\theta}) \>\mathrm{d}\theta}\\ & = i\int_{0}^{\pi} {(\sin(e^{i\theta}) + \sin(e^{i(2\pi-\theta)})) \>\mathrm{d}\theta}\\ & = i\int_{0}^{\pi} {(\sin(e^{i\theta}) + \sin(e^{-i\theta)})) \>\mathrm{d}\theta}\\ & = 2i\int_{0}^{\pi} {\left(\sin\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right) \cos\left(\frac{e^{i\theta}-e^{-i\theta}}{2}\right)\right) \mathrm{d}\theta}\\ & = 2i\int_{0}^{\pi} {(\sin(\cos\theta) \cos(i\sin\theta)) \>\mathrm{d}\theta}\\ & = 2i\int_{0}^{\pi/2} {(\sin(\cos\theta) \cos(i\sin\theta)) \>\mathrm{d}\theta} + 2i\int_{\pi/2}^{\pi} {(\sin(\cos\theta) \cos(i\sin\theta)) \>\mathrm{d}\theta}\\ & = 2i\int_{0}^{\pi/2} {(\sin(\cos\theta) \cos(i\sin\theta) + (\sin(\cos(\pi-\theta)) \cos(i\sin(\pi-\theta))) \>\mathrm{d}\theta}\\ & = 0 \end{aligned}$$


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