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Evaluate $$\int_{\Gamma} \frac{1}{z^2+1}$$ where $\Gamma=\{z:|z|=2\}$

One way is to use partial fractions and then cauchy integral formula.

The other way is to look at:

$$\int_{\Gamma} \frac{1}{z^2+1}=\int_{\gamma_1}\frac{\frac{1}{z-i}}{z+i}+\int_{\gamma_2}\frac{\frac{1}{z+i}}{z-i}$$

Where $\gamma_1$ is a curve around $i$ and $\gamma_2$ is a curve around $-i$ but in which direction? clockwise? as $\Gamma$ is anti clockwise?

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No, it is not a sum. Usually, the meaning of $\int_\Gamma$ with $\Gamma=\{z\in\mathbb{C}\,:\,|z|=r\}$ means that we are integrating along the loop $\Gamma\colon[0,2\pi]\longrightarrow\mathbb C$ defined by $\Gamma(t)=re^{it}$.

In this specific case, the value of the integral will be $2\pi i$ times the sum of the residues of $\frac1{z^2+1}$ at $i$ and $-i$. The first residue is equal to $-\frac i2$ and the second one is equal to $\frac i2$. Therefore, the integral is equal to $0$.

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You mean $$\int_\Gamma\frac{dz}{z^2+1}.$$ You can replace the contour $\Gamma$ with the circle of centre $0$ and radius $r$ for any $r>1$. Letting $r\to\infty$ the integral tends to zero, so it must have been zero in the first place.

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  • $\begingroup$ Can we say that there is antiderivative when where $\theta-\frac{\pi}{2} < Arg(z)\leq \theta -\frac{\pi}{2} $ $\endgroup$ – gbox Jun 18 '17 at 13:46

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