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I want to calculate

$$\int_{\gamma} \frac{\sin z}{z^2 + 1}dz$$ where $\gamma$ is the upper-half circle of radius 2 centered at the origin starting at 2. I know that since $\gamma$ is not a closed curve, Cauchy's theorem will be of no use, but I believe that the integrand has a singularity at i, which is between the curve and the real axis. Can I use the residue theorem here?

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1 Answer 1

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Residue theorem is the right way to go. Note that since the integrand is odd on $\mathbb R$,

$$ \int_{\gamma} \frac{\sin z}{z^2 + 1}dz = \int_{\gamma} \frac{\sin z}{z^2 + 1}dz + \int_{-2}^2 \frac{\sin z}{z^2 + 1}dz = 2\pi i \operatorname{Res}_{z=i}\left( \frac{\sin z}{z^2 + 1}\right)$$

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  • $\begingroup$ I think you mean integral from $-2$ to $2$, looking at OP's definition of $\gamma$. $\endgroup$
    – SvanN
    Dec 12, 2018 at 18:42
  • $\begingroup$ @S.vanNigtevecht yeah sorry, already edited $\endgroup$ Dec 12, 2018 at 18:43

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