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Suppose we wanted to integrate this around a semicircular path above above the for $\Im \left ( z \right )\geq 0$ and $\left | z \right |\leq e$. The integrand has an essential singularity at $z=0$, and so we have to make a little indent inwards(or outwards) at that point on the contour as to avoid crossing the singularity. Let the indent be of radius $\epsilon $, we then have $\oint \frac{z+\bar{z}}{\left | z \right |}dz=0$, according to the Cauchy-Goursat theorem. Now, since we can make the radius of the indent arbitrarily small, this looks all too similar to a principal value we assign to real integrals. Is this indeed the fact? Even stranger is the fact that if we split this integral up into the integral along the x-axis, and one along the arc(ignoring the indent at $z=0$), we somehow overcome the singularity at that point: $$\oint \frac{z+\bar{z}}{\left | z \right |}dz=\int_{-e}^{e}2dx +\int_{arc}^{ } f\left ( z \right )dz\neq 0$$ The fact that this last computation yields a non zero value is a telltale sign that we are dealing with v.p. values here?

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    $\begingroup$ Your integral isn't holomorphic anywhere, and you can't say that it has an essential singularity, and you definitely cannot use Cauchy's theorem. Is this really the integral you want to compute? (For example where does the restriction $|z| \le e$ come from?) But if that really is the question: if you want to integrate over a semi-circle of radius $r$, use that $|z| = r$ and $\bar z = r^2/z$ along the curve. That will help with the computation. $\endgroup$ – mrf Dec 2 '15 at 8:07
  • $\begingroup$ @mrf Another option might be to use Cauchy's theorem for non analytic functions. This was just an exercise in my Complex analysis book. $\endgroup$ – Emir Šemšić Dec 2 '15 at 8:30
  • $\begingroup$ $z=re^{i\phi}$, $\frac{z+\bar z}{|z|}=2\cos\phi$ $\endgroup$ – rych Dec 2 '15 at 9:45
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The integral has the value $\pi r i$ if the radius of the semi-circle is $r$.

The reason: the integral along the real axis from $-r$ to $r$ is $0$. Note that the integrand is ${2x \over |x|}$ on the real axis. The function to be integrated is thus $2$ for $x > 0$ and $-2$ for $x<0$. Hence it is bounded and integrable with the value of the integral $0$.

The integral along the semi-circle is easily done as follows:

Put $z = re^{i\theta}$. Then $dz = rie^{i\theta} d\theta$. One gets two terms, one of which gives $0$ and one gives $\pi ir$.

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