MHB Contour integral representation of Kronecker delta

[ognik]

I'm rather impressed with complex analysis, but clearly I have a lot to learn.

I'm told $ \frac{1}{2\pi i} \oint {z}^{m-n-1} dz $ is a rep. of the kronecker delta function, so I tried to work through that. I used $z = re^{i\theta}$ and got to $ \frac{1}{2\pi} [\frac{{r}^{m-n}}{i(m-n)}e^{i\theta(m-n)} ]^{2\pi}_0 $

I have since found this on the web, so it seems right - all I need do is evaluate it for m=n and m \NE n ...but I'm stuck. Is there one of those nice tricks to use as the next step?

 

[Euge]

More care is needed in your analysis. You need a specified contour in order to perform contour integration. Also, if $m = n$, the expression $\frac{1}{m - n}$ is undefined. Based of your choice of parametrization, I assume that the contour of integration is the circle $|z| = r$. If $m\neq n$, then your calculations are so far correct (again, assuming that the contour is the circle $|z| = r$), and the result is zero (since $e^{i2\pi(m - n)} - e^{i0(m-n)} = 1 - 1 = 0$). If $m = n$, then $$\oint z^{m-n-1}\, dz = \oint z^{-1} dz = \int_0^{2\pi} (re^{i\theta})^{-1}\cdot ire^{i\theta}\, d\theta = \int_0^{2\pi} i\, d\theta = 2\pi i.$$

Hence $\oint z^{m-n-1}\, dz = 2\pi i\delta_{m,n}$, or

$$\frac{1}{2\pi i} \oint z^{m-n-1}\, dz = \delta_{m,n}.$$

 

[ognik]

Sorry, yes that's the contour (I thought a small circle around 0 was a default contour).

I somehow missed both of those, they are straight forward (I might have been a little tired (Doh)) - thanks as usual for the clarity.