How to Solve this Integral $\int_{\mid x\mid =r} xdz$?

Question

This is NOT a duplicate question.

I need some help computing the integral

$\int_{\mid z\mid =r} xdz$

But I really want to use the fact that

$x=\frac{1}{2}(z + \frac {r^{2}}{{z}})$ when the circle is oriented counter-clockwise in the integral.

Should I just plug that into the integral? I have seen the similar question to this and understand how to use parameters to get that the integral equals $r^{2}i\frac{\pi}{2}$. However, I am interested in using the fact mentioned above, which requires a different method than the solutions already discussed on this site, so it is not a duplicate. How would I go about this?

Answer 1

$|x|=r$ is $x=-r\lor x=r$ and the integral can be translated to

$$\int_{|x|=r}x\,dz=-\int_{z=-r-i\infty}^{-r+i\infty}r\,dz+\int_{z=r-i\infty}^{r+i\infty}r\,dz,$$ assuming "counterclockwise traversal".

I guess that this can be said to equal zero in the Cauchy sense.

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Assuming a typo in the problem statement,

$$\int_{|\color{red}z|=r}x\,dz=\int_{t=0}^{2\pi}r\cos(t)\,d(re^{it})=r^2\int_{t=0}^{2\pi}\cos(t)ie^{it}dt=\\r^2\int_{t=0}^{2\pi}(i\cos^2(t)-\cos(t)\sin(t))\,dt=i\pi r^2$$ by the average values of $\cos^2(t)$ and $\cos(t)\sin(t)$. The sign is positive for a counterclockwise traversal.

Answer 2

The solution depends on the tools you have. Using this fact makes me suspect that you can use the residue theorem:

$\int_{|z|=r} \Re z dz = {1\over2}\int (z+r^2/z) dz = {1\over2}(\int z dz + r^2\int z^{-1} dz) = {1\over2}(0+r^22\pi i)$

To just go ahead and just plug it into the integral by using $\Re z = {z + r^2/z\over2}$ doesn't make much sense:

$\int\Re z dz = \int{z + r^2/z\over2}dz = \int_0^{2\pi} {re^{i\varphi}+r^2/(re^{i\varphi})\over2}ire^{i\varphi}d\varphi = \int_0^{2\pi}r{e^{i\varphi}+e^{-i\varphi}\over2}ire^{i\varphi}d\varphi = \int_0^{2\pi}r\cos(\varphi) ire^{i\varphi}d\varphi$

that is a complicated way to arrive at the same integral as if we didn't use that fact and just went ahead and evaluated the integral (and instead using the fact that $\Re re^{i\varphi} = r\cos\varphi$):

$\int\Re z dz = \int_0^{2\pi} r\cos(\varphi) ire^{i\varphi}d\varphi$