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I'm pretty sure I've gotten an integral properly set up, but the integral looks very ugly to compute. Could someone tell me whether I've made a mistake in my setting up of this problem?

Compute $\int_\epsilon$Log(z)dz where $\epsilon$ is the segment connecting 1 to i

Parametrize the path: $z=1-t(1-i) : 0\leq t\leq1$

Compute the differential: $dz=(i-1)dt$

Plug into original function and integrate:

$$\int_\epsilon Log(z)dz = \int_0^1 \left[ \frac{1}{2}Log\left( (1-t)^2 +t^2 \right) + i*arctan\left( \frac{t}{1-t} \right)\right](i-1) dt$$

Did I mess up anywhere in my setup? Or is this just an ugly integral I've got to compute?

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Have you tried parametrizing by a segment on the unit circle?

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  • $\begingroup$ AH! It's path independent! This is because Log(z) is continuous. Is this reasoning correct? $\endgroup$ – druckermanly Feb 9 '14 at 3:26
  • $\begingroup$ Contour integrals are path-independent for continuous functions as long as the area defined by the full path (i.e. original path and new path) doesn't include poles which is $z=0$ in this case. $\endgroup$ – wfh Feb 9 '14 at 3:35
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    $\begingroup$ continuous is not the correct keyword here, but holomorphic. Contour integrals of continuous functions are in general not path-independent. $\endgroup$ – mrf Feb 9 '14 at 9:31
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Another possibility is to find an antiderivative: you can check that $F(z) = z\operatorname{Log} z - z$ will do ($F$ is holomorphic on a neighbourhood of the curve, and $F'(z) = \operatorname{Log} z$.) Hence the value of the integral is just $F(i) - F(1)$.

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