[MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n))

[Caleb]

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

Fair point-- I didn't explain the shape of my contours well enough. Look closely again at the shape of the contours I am choosing, for instance, the one for \( e^{\frac{1}{z^2}} \) or \( e^\frac{1}{z^10} \) are easy ones to see. Also, to be specific, the contour I have is drawn in white, I'm not sure if I mentioned that before. 

All the contours are going THROUGH THE NON-ANALYTIC POINT AT 0! That means you can't apply Cauchy's theorem for residues! The integrals you have shown are regular circles-- I know how to deal with these cases. For instance, the contour I'm looking at for \( e^\frac{1}{z^2} \) is over the contour C where C is the path
\[ -i \to 0 \to i \to i+1 \to -i + 1 \to -i \]
Actually, if we want to be precise, we need to remove \(0\) from this path, since \( e^\frac{1}{z^2} \) is not defined there.


Okay, so I've introduced a weird object, why do I care about it? I care about such weird path as these because they allow me to pick up, for isntace, the residue at \( + \infty \) without picking up the residue at at other angles. For instance, I might wonder-- what is the integral \( \int_{-\infty}^\infty e^{-z^2} dz\) equal to? I could draw this as a contour over the imaginary line of \( e^{z^2} \), it looks like this
   

BUT, notice that I want to pick up the residue at \( + \infty \) WITHOUT ALSO PICKING UP THE RESIDUE AT \( -\infty \). In particular, I make the change \(dz \to d\frac{1}{z}\) and now I'm looking at \( \frac{1}{z^2} e^{\frac{1}{z^2}} \). Then, the contour integral around the full circle is zero-- but this isn't what we want, the integral is clearly now zero so thats not helpful. The contour we actually want is this one
   


In particular, notice that 
\[\int_{-\infty}^\infty e^{-z^2} dz = \sqrt{\pi}\]
But 
\[\int_{|r| = 1} \frac{1}{z^2} e^{\frac{1}{z^2}}dz = 0 \]
Whereas (where C is the contour I drew in the picture)
\[\frac{1}{i} \int_{C} \frac{1}{z^2} e^{\frac{1}{z^2}}dz =\sqrt{\pi}  \]
As expected