04/02/2023, 05:31 AM
Hmmmmm.
Now I'm confused. I think you may be missing some background knowledge on complex analysis.
\[
\int_{|z| = 1} e^{1/z}\,dz = 2\pi i\\
\]
If I take \(k>1\):
\[
\int_{|z| = 1} e^{1/z^k}\,dz = 0\\
\]
This is not numerically verifiable; but it's mathematically true. Additionally we can substitute \(\xi = 1/z\); and take taylor coefficients at \(\xi\); which relate directly to the above values; but as "residue at infinity" relates to "residue at zero".
When we introduce:
\[
\int_{|z| = 1} e^{1/\sqrt{z}}\,dz\\
\]
It is impossible to take this integral without considering it in an "improper manner". Which was my point.
I apologize if I'm not making any sense. But from that point; could you perhaps elaborate further from this perspective? I'm not trying to be a jack ass, I'd just love to get a better grasp of what you mean???
Again, sorry.
James
Now I'm confused. I think you may be missing some background knowledge on complex analysis.
\[
\int_{|z| = 1} e^{1/z}\,dz = 2\pi i\\
\]
If I take \(k>1\):
\[
\int_{|z| = 1} e^{1/z^k}\,dz = 0\\
\]
This is not numerically verifiable; but it's mathematically true. Additionally we can substitute \(\xi = 1/z\); and take taylor coefficients at \(\xi\); which relate directly to the above values; but as "residue at infinity" relates to "residue at zero".
When we introduce:
\[
\int_{|z| = 1} e^{1/\sqrt{z}}\,dz\\
\]
It is impossible to take this integral without considering it in an "improper manner". Which was my point.
I apologize if I'm not making any sense. But from that point; could you perhaps elaborate further from this perspective? I'm not trying to be a jack ass, I'd just love to get a better grasp of what you mean???
Again, sorry.
James