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Calculate $$\int_\gamma \frac{1+\sin(z)}{z} dz$$

where $\gamma$ is the circle of centre 0 and radius $\log(\sqrt2)$ oriented counter clockwise.

Well there is a singularity at 0. The fact that the radius is $\log(\sqrt2)$ has no real relevance as this function fails to be analytic at any circle centered at 0.

I used the Cauchy integral formula to calculate the integral and got $2\pi i$. Does that look correct?

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    $\begingroup$ yes your are right. $\endgroup$ – noname1014 Apr 18 '12 at 0:24
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$$ \int_\gamma \frac 1 z\,dz + \int_\gamma\frac {\sin z}{z}\,dz. $$ The second integral is zero because the singularity at $z=0$ is removable. You can see that by noticing that the power series for $\sin z$ has no constant term, so you can divide the whole thing by $z$ without getting a $1/z$ term.

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