# Fractional Calculus and Residues

1. Oct 27, 2013

### Harrisonized

According to Fractional Calculus, the power rule can be written as

(dm/dzm) zn = n!/(n-m)! zn-m

For example,

(d1/2/dz1/2) z1/2 = (1/2)!/(1/2-1/2)! z0 = (1/2)√π

To find the residue of f(z) = f(z)/(z-z0)m at z→z0, the formula is Res(z→z0) f(z) = 1/(m-1)! dm-1/dzm-1 f(z).

For example, when I use the formula for the residue of 1/(z-1)1/2 at z→1, I get the following:

1/(1/2-1)! d1/2-1/dz1/2-1 1
= 1/(-1/2)! (2/√π) z1/2 = 2/π

Actually, the formula for any power residue can also be computed.

Res(z→z0) f(z)
= 1/(m/2-1)! (dm/2-1/dzm/2-1) z0 = (1/(m/2-1)!)(0!/(1-m/2)!) z1-m/2
= 1/[(m/2-1)!(1-m/2)!] z1-m/2, where m is an odd integer.

Using this formula, I can construct a table:

m=1, Res = 2/π √z0
m=3, Res = 2/π 1/√z0
m=5, Res = -2/(3π) 1/z03/2
m=7, Res = 2/(5π) 1/z05/2
m=9, Res = -2/(7π) 1/z07/2
m=11, Res = 2/(9π) 1/z09/2
.
.
.
m, Res = (-1)(m+1)/2 2/[(m-2)π] 1/zm/2-1

http://www.wolframalpha.com/input/?i=1/[(m/2-1)!+(1-m/2)!]+z^(1-m/2),+m=[1,15]

It's interesting to note that for m=2, Res=1, and every other even integer above that gives a 0 value. This is consistent with the correct results.

What is the meaning of such a residue? Every source I can find on the internet says there shouldn't be a residue, and Wolfram says the residue is 0.

Last edited: Oct 27, 2013