About the integral $\oint \frac{z+\bar{z}}{\left | z \right |}dz$

Question

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?

Answer 1

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 $\pi ir$.