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I tried finding the integral of $\int_{|z|=2}^{}\frac{1}{z^2+1}dz$ but not sure whether it is correct.

$\gamma(t)=2e^{it},t\in[0,2\pi]$

$$\int_{|z|=2}^{}\frac{1}{z^2+1}dz=\int_{0}^{2\pi}\frac{2ie^{it}}{4e^{2it}+1}dt$$

I cannot seem to take it forward from here. Is it possible to use Cauchy's integral formula to get an answer? Thanks

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  • $\begingroup$ Hint: What are the poles of your function? What can you say about the function on and inside the circle of radius two? $\endgroup$
    – Geoff
    Nov 2, 2014 at 6:05
  • $\begingroup$ @Geoff It is analytic except at i and -i $\endgroup$
    – user81883
    Nov 2, 2014 at 6:08

4 Answers 4

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HINT: $\frac{1}{1+z^2}=\frac{1}{(z-i)(z+i)}$. Use partial fractions to expand this and then apply Cauchy Integral Formula. You should get the $0$ (As proved in the other answer).

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  • $\begingroup$ Just what i was looking for can't believe I missed something this simple. Wondered how to use C.I. formula but missed this expansion. Thanks!! $\endgroup$
    – user81883
    Nov 3, 2014 at 17:13
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You can use the residue theorem to solve the problem: $$I:=\int_{\left| z \right| = 2} {\frac{1}{{{z^2} + 1}}dz} = 2\pi \left[ {\operatorname{Residue} \left( {\frac{1}{{{z^2} + 1}};z = i} \right) + \operatorname{Residue} \left( {\frac{1}{{{z^2} + 1}};z = - i} \right)} \right].$$ We have $$\operatorname{Residue} \left( {\frac{1}{{{z^2} + 1}};z = i} \right) = \mathop {\lim }\limits_{z \to i} \left( {z - i} \right)\frac{1}{{{z^2} + 1}} = \mathop {\lim }\limits_{z \to i} \frac{1}{{z + i}} = \frac{1}{{2i}},$$ $$\operatorname{Residue} \left( {\frac{1}{{{z^2} + 1}};z = - i} \right) = \mathop {\lim }\limits_{z \to - i} \left( {z + i} \right)\frac{1}{{{z^2} + 1}} = \mathop {\lim }\limits_{z \to - i} \frac{1}{{z - i}} = - \frac{1}{{2i}}.$$ So $I=0$.

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  • $\begingroup$ Shouldn't that by $I = 2 \pi \color{red}{i} \left[ \ldots \right]$? $\endgroup$ Jun 22, 2020 at 23:56
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This integral is easier if you note that your integrand:

$$f(t)=\frac{2ie^{it}}{4e^{2it}+1}$$ has the property that $f(t+\pi)=-f(t)$. So $$\int_{0}^{2\pi} f(t)\,dt = 0,$$ since the two ranges $[0,\pi]$ and $[\pi,2\pi]$ cancel.


More generally, if $g(z)$ is a continuous function (or otherwise nice, integrable) defined on $|z|=r$ such that $g(z)=g(-z)$, then $$\int_{|z|=r} g(z)\,dz= 0$$

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Enforcing the change of variables $t= \pi+s$ gives

$$\begin{align}I&= \int_{-\pi}^{\pi}\frac{2ie^{it}}{4e^{2it}+1}\:dt=\int_{0}^{2\pi}\frac{2ie^{-is}e^{-i\pi}}{4e^{-2is}+1}\:ds\\&=-\int_{0}^{2\pi}\frac{2ie^{-is}}{4e^{-2is}+1}\:ds \\&= -\int_{0}^{\pi}\frac{2ie^{-is}}{4e^{-2is}+1}\:ds-\int_{\pi}^{2\pi}\frac{2ie^{-is}}{4e^{-2is}+1}\:ds \\&\overset{\color{red}{u=s-2\pi}}{=}-\int_{0}^{\pi}\frac{2ie^{-is}}{4e^{-2is}+1}\:ds-\int_{-\pi}^{0}\frac{2ie^{-iu}}{4e^{-2iu}+1}\:du \\&=-\int_{-\pi}^{\pi}\frac{2ie^{-is}}{4e^{-2is}+1}\:ds \\&=-I \end{align}$$

Hence $$\color{red}{I= \int_{-\pi}^{\pi}\frac{2ie^{it}}{4e^{2it}+1}\:dt=0}$$

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