Fractional Calculus and Residues

[Harrisonized]

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

(d^m/dz^m) zⁿ = n!/(n-m)! z^n-m

For example,

(d^1/2/dz^1/2) z^1/2 = (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)! d^m-1/dz^m-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)! d^1/2-1/dz^1/2-1 1
= 1/(-1/2)! (2/√π) z^1/2 = 2/π

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

Res(z→z₀) f(z)
= 1/(m/2-1)! (d^m/2-1/dz^m/2-1) z⁰ = (1/(m/2-1)!)(0!/(1-m/2)!) z^1-m/2
= 1/[(m/2-1)!(1-m/2)!] z^1-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/z₀^3/2
m=7, Res = 2/(5π) 1/z₀^5/2
m=9, Res = -2/(7π) 1/z₀^7/2
m=11, Res = 2/(9π) 1/z₀^9/2
.
.
.
m, Res = (-1)^(m+1)/2 2/[(m-2)π] 1/z^m/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.

 

[fresh_42]

I don't know fractional calculus, but here are some interesting papers:
https://www.math.ust.hk/~maykwok/courses/ma304/06_07/Complex_6.pdfhttps://www.degruyter.com/downloadpdf/books/9783110409475/9783110409475.3/9783110409475.3.pdfhttps://arxiv.org/ftp/arxiv/papers/1106/1106.3011.pdf

 

[Svein]

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.

Yes - they say the same thing.

For a simple pole the residue of $f(z)$ at $a$ equals the value of the function $(z-a)f(z)$ for $z=a$. Thus, if $a$ is not a pole, there is no residue and the value of $(z-a)f(z)$ is 0.