$$ \int_{\gamma}\frac{dz}{z^2+1} = \frac{1}{2i}\int_{\gamma}\frac{dz}{z-i}-\frac{1}{2i}\int_{\gamma}\frac{dz}{z+i} = 2 \pi i \times0$$

By cauchy's integral formula right?

According to my lecture notes this should be $ \pi$

Where $\gamma$ is a closed contour, semicircle of radius $R$. in the upper half plane

  • 1
    $\begingroup$ The semicircle only encloses one singularity, $z=i$, so one of those integrals will be zero and the other will be $\pi$. $\endgroup$ – TorsionSquid Apr 27 '16 at 13:17
  • $\begingroup$ by the Cauchy integral formula, or by direct evaluation of $\int_\gamma \frac{dz}{z+i} = \int_{-R}^R \frac{dt}{t+i} + \int_0^\pi \frac{R i e^{i\theta}}{R e^{i\theta} + i} d\theta$ $\endgroup$ – reuns Apr 27 '16 at 13:40

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