04/04/2023, 12:23 PM
OMG!
I SEE WHAT YOU ARE DOING.
THIS WAS A HUGE PROBLEM I COULD NEVER SOLVE. I ACTUALLY HAVE AN OLD MO POST ABOUT IT THAT GOT DOWNVOTED TO OBLIVION, LMAO!
Let me spit ball for a moment to remind myself. I'm just going to try to give some context to my understanding of this problem. If I'm way off base let me down easy
Okay; let's refer to a function \(f(z)\) which is holomorphic on the unit disk \(\overline{\mathbb{D}}/1\). This means not only is it holomorphic for \(|z| < 1\) but also holomorphic for \(|z| = 1, z \neq 1\). Holomorphy on a boundary is taken to mean it is analytic on the boundary. You are constructing a contour that looks like:
\[
\int_{|z| = 1} f(z)\,dz\\
\]
This type of contour and integration problem can be related to Mellin transforms (or fourier transforms in general). Let's map \(1^+ \to \infty\) and \(1^- \to -\infty\) (where this is as a path approaching \(1\) along \(\partial{\mathbb{D}}\)). And then additionally we map the unit circle to \(\mathbb{R}\). This is a perfectly legal construction; and looks like:
\[
h(z) = \frac{az + b}{cz+d}\\
\]
I'll find \(a,b,c,d\) but, know first, they exist prima facie. We are mapping a clircle to a clircle and a point to a point (a clircle is just a fancy word for a circle on the Riemann Sphere projection); and therefore this defines a unique LFT. Since \(h(1) = \infty\) the bottom terms are \(c = 1, d = -1\). We now map:
\[
\Im\left(\frac{az + b}{z-1}\right) = 0\,\,\text{if}\,\,|z| =1\\
\]
Which is solvable as \(a = i\) and \(b=i\); and we just end up with the good old fashion automorphism \(h(z) = i\frac{z+1}{z-1}\). We're going to write \(h'(z) = \frac{-2i}{(z-1)^2}\). Additionally; we should know that \(h(h(z)) = z\); so changing variables is pretty standard.
Let's also parameterize the contour \(|z| = 1\) with \(e^{i\theta}\) for \(-\pi \le \theta \le \pi\). Where \(z= e^{i\theta}\) and \(dz = ie^{i\theta}d\theta\). Where then we are at two new expressions for this integral:
\[
\begin{align}
\int_{|z| = 1} f(z)\,dz &= \int_{\mathbb{R}} f(h(x)) h'(x)\,dx = -2i\int_{\mathbb{R}} \frac{f(h(x))}{(x-1)^2}\,dx\\
&= i\int_{-\pi}^\pi f(e^{i\theta})e^{i\theta}\, d\theta\\
\end{align}
\]
You can do more substitutions if you'd like; (to get the mellin transform); but this is enough to talk about Fourier; and I'll stick to that. Let's assume now that \(f(h(x))h'(x) = g(x)\). The function \(g(\xi)\) is holomorphic for \(\Im(\xi) > 0\) by assumption. When we integrate this we get:
\[
\int_{\mathbb{R}} g(x)\,dx = \int_{|z| = 1} f(z)\,dz
\]
Now equally so we have the integral, for \(t > 0\):
\[
\int_{\mathbb{R}} g(x+it)\,dx = \int_{\mathbb{R}} g(x)\,dx
\]
Which follows by Cauchy's theorem; and the assumption that for \(r < 1\):
\[
\int_{|z| = r} f(z)\,dz
\]
Is integrable; which we know to be true........
Now this doesn't look like your problem; not at first glance. But it is your problem. You have a single singularity at a point on the boundary; and you are holomorphic for all intents and purposes away from this singularity on the boundary. I am mapping it to the upper half plane.
Now additionally; I have assumed that \(g(\xi)\) is holomorphic on \(\mathbb{R}/1\). I'm going to take an additional assumption (which was part of my MO question originally, that I probably phrased terribly). Let's assume that \(\mathcal{L}\) is a contour from \(-\infty\) to \(\infty\), where \(\Im(\mathcal{L}) \neq 0\) and \(\Im(\mathcal{L}) < -\delta(x)\), for \(\delta>0\) appropriately small which goes to zero as \(x \to \pm \infty\).
A great example of this technique can be seen in the Newman proof of the prime number theorem. I'm only taking some small notes from there; and not bothering with the rigour yet; just painting a picture. But this allows us to pull out a much cleaner (better bounded, absolute convergence) Laplace Transform (in Newman's case)/Fourier Transform (general case)/Mellin Transform (Ramanujan's raw intuition case).
Then we can also write:
\[
\int_{\mathcal{L}} g(\xi)\,d\xi = \text{Res}(g(\xi),\xi = 1) + \int_{|z| = r} f(z)\,dz\\
\]
What this is saying; is that the "residue at infinity" can be found just like a normal residue; if you change your domains and do some riemann mapping magic. Because; going back to our discussion with \(f\); we've defined:
\[
\text{Res}(f,z=1) = \text{An integral on the real line}\\
\]
But this integral actually equals another integral on the real line that's more manageable. You can pull out the fourier coefficients if you'd like--you're probably getting something pretty ugly; but it's still a nice idea.
A lot of this discussion is mostly just to motivate you that your "residues at infinity" are mappable to normal infinite integrals; and then the application of residue calculus atop that. And it's incredibly fucking advanced. I can't remember my dumb question exactly; but it was essentially the statement that in this restricted scenario we should have some nice mellin transform results--which relate to Ramanujan--which relate to all the beautiful shit you keep seeing
Also, Caleb. Go easy on me. I'm just spit balling here, and trying to give my two cents. I apologize if I'm being stupid, lmao. But if you go backwards from here; you should be able to find a function \(q(z)\) such that:
\[
\int_{|z| = 1} f(z)\,dz = \int_{|z| = r} f(z) q(z)\,dz\\
\]
And it could be more manageable. I apologize I can't be of more help! These are things I would exclusively wrestle with using the Mellin transform; and what you're detailing relates to an arc \(\gamma\) such that \(\gamma(0) = 0\) and \(\gamma(\infty) = \infty\). And similarly \(\gamma*(0) =0\) and \(\gamma*(\infty) = \infty\). Then if a function is differintegrable we have:
\[
\int_\gamma \vartheta(w)\,dw = \int_{\gamma^*}\vartheta (w)\,dw
\]
Which is just a fancy cauchy's integral theorem. At infinity; if this isn't differintegrable, I conjectured that there was an equivalence:
\[
\lim_{R\to \infty} \int_{\gamma_R}\vartheta(w)\,dw - \int_{\gamma^*_R}\vartheta (w)\,dw = 0\\
\]
Then this function was differintegrable... Where \(\gamma_R\) is \(|\gamma| \le R\). This largely went nowhere but would mean fantastic and huge things for determining when a function is differintegrable. Which would again; talk about how we handle fourier transforms, and fourier coefficients of your residues. Have you considered:
\[
\int_{|z| = 1} f(z) z^k\,dz\\
\]
Which are fourier coefficients at their core.
Either way, don't shoot me. I'm mostly just casually spitballing here. I love your work and I just like to talk.
Regards, James
I SEE WHAT YOU ARE DOING.
THIS WAS A HUGE PROBLEM I COULD NEVER SOLVE. I ACTUALLY HAVE AN OLD MO POST ABOUT IT THAT GOT DOWNVOTED TO OBLIVION, LMAO!
Let me spit ball for a moment to remind myself. I'm just going to try to give some context to my understanding of this problem. If I'm way off base let me down easy
Okay; let's refer to a function \(f(z)\) which is holomorphic on the unit disk \(\overline{\mathbb{D}}/1\). This means not only is it holomorphic for \(|z| < 1\) but also holomorphic for \(|z| = 1, z \neq 1\). Holomorphy on a boundary is taken to mean it is analytic on the boundary. You are constructing a contour that looks like:
\[
\int_{|z| = 1} f(z)\,dz\\
\]
This type of contour and integration problem can be related to Mellin transforms (or fourier transforms in general). Let's map \(1^+ \to \infty\) and \(1^- \to -\infty\) (where this is as a path approaching \(1\) along \(\partial{\mathbb{D}}\)). And then additionally we map the unit circle to \(\mathbb{R}\). This is a perfectly legal construction; and looks like:
\[
h(z) = \frac{az + b}{cz+d}\\
\]
I'll find \(a,b,c,d\) but, know first, they exist prima facie. We are mapping a clircle to a clircle and a point to a point (a clircle is just a fancy word for a circle on the Riemann Sphere projection); and therefore this defines a unique LFT. Since \(h(1) = \infty\) the bottom terms are \(c = 1, d = -1\). We now map:
\[
\Im\left(\frac{az + b}{z-1}\right) = 0\,\,\text{if}\,\,|z| =1\\
\]
Which is solvable as \(a = i\) and \(b=i\); and we just end up with the good old fashion automorphism \(h(z) = i\frac{z+1}{z-1}\). We're going to write \(h'(z) = \frac{-2i}{(z-1)^2}\). Additionally; we should know that \(h(h(z)) = z\); so changing variables is pretty standard.
Let's also parameterize the contour \(|z| = 1\) with \(e^{i\theta}\) for \(-\pi \le \theta \le \pi\). Where \(z= e^{i\theta}\) and \(dz = ie^{i\theta}d\theta\). Where then we are at two new expressions for this integral:
\[
\begin{align}
\int_{|z| = 1} f(z)\,dz &= \int_{\mathbb{R}} f(h(x)) h'(x)\,dx = -2i\int_{\mathbb{R}} \frac{f(h(x))}{(x-1)^2}\,dx\\
&= i\int_{-\pi}^\pi f(e^{i\theta})e^{i\theta}\, d\theta\\
\end{align}
\]
You can do more substitutions if you'd like; (to get the mellin transform); but this is enough to talk about Fourier; and I'll stick to that. Let's assume now that \(f(h(x))h'(x) = g(x)\). The function \(g(\xi)\) is holomorphic for \(\Im(\xi) > 0\) by assumption. When we integrate this we get:
\[
\int_{\mathbb{R}} g(x)\,dx = \int_{|z| = 1} f(z)\,dz
\]
Now equally so we have the integral, for \(t > 0\):
\[
\int_{\mathbb{R}} g(x+it)\,dx = \int_{\mathbb{R}} g(x)\,dx
\]
Which follows by Cauchy's theorem; and the assumption that for \(r < 1\):
\[
\int_{|z| = r} f(z)\,dz
\]
Is integrable; which we know to be true........
Now this doesn't look like your problem; not at first glance. But it is your problem. You have a single singularity at a point on the boundary; and you are holomorphic for all intents and purposes away from this singularity on the boundary. I am mapping it to the upper half plane.
Now additionally; I have assumed that \(g(\xi)\) is holomorphic on \(\mathbb{R}/1\). I'm going to take an additional assumption (which was part of my MO question originally, that I probably phrased terribly). Let's assume that \(\mathcal{L}\) is a contour from \(-\infty\) to \(\infty\), where \(\Im(\mathcal{L}) \neq 0\) and \(\Im(\mathcal{L}) < -\delta(x)\), for \(\delta>0\) appropriately small which goes to zero as \(x \to \pm \infty\).
A great example of this technique can be seen in the Newman proof of the prime number theorem. I'm only taking some small notes from there; and not bothering with the rigour yet; just painting a picture. But this allows us to pull out a much cleaner (better bounded, absolute convergence) Laplace Transform (in Newman's case)/Fourier Transform (general case)/Mellin Transform (Ramanujan's raw intuition case).
Then we can also write:
\[
\int_{\mathcal{L}} g(\xi)\,d\xi = \text{Res}(g(\xi),\xi = 1) + \int_{|z| = r} f(z)\,dz\\
\]
What this is saying; is that the "residue at infinity" can be found just like a normal residue; if you change your domains and do some riemann mapping magic. Because; going back to our discussion with \(f\); we've defined:
\[
\text{Res}(f,z=1) = \text{An integral on the real line}\\
\]
But this integral actually equals another integral on the real line that's more manageable. You can pull out the fourier coefficients if you'd like--you're probably getting something pretty ugly; but it's still a nice idea.
A lot of this discussion is mostly just to motivate you that your "residues at infinity" are mappable to normal infinite integrals; and then the application of residue calculus atop that. And it's incredibly fucking advanced. I can't remember my dumb question exactly; but it was essentially the statement that in this restricted scenario we should have some nice mellin transform results--which relate to Ramanujan--which relate to all the beautiful shit you keep seeing
Also, Caleb. Go easy on me. I'm just spit balling here, and trying to give my two cents. I apologize if I'm being stupid, lmao. But if you go backwards from here; you should be able to find a function \(q(z)\) such that:
\[
\int_{|z| = 1} f(z)\,dz = \int_{|z| = r} f(z) q(z)\,dz\\
\]
And it could be more manageable. I apologize I can't be of more help! These are things I would exclusively wrestle with using the Mellin transform; and what you're detailing relates to an arc \(\gamma\) such that \(\gamma(0) = 0\) and \(\gamma(\infty) = \infty\). And similarly \(\gamma*(0) =0\) and \(\gamma*(\infty) = \infty\). Then if a function is differintegrable we have:
\[
\int_\gamma \vartheta(w)\,dw = \int_{\gamma^*}\vartheta (w)\,dw
\]
Which is just a fancy cauchy's integral theorem. At infinity; if this isn't differintegrable, I conjectured that there was an equivalence:
\[
\lim_{R\to \infty} \int_{\gamma_R}\vartheta(w)\,dw - \int_{\gamma^*_R}\vartheta (w)\,dw = 0\\
\]
Then this function was differintegrable... Where \(\gamma_R\) is \(|\gamma| \le R\). This largely went nowhere but would mean fantastic and huge things for determining when a function is differintegrable. Which would again; talk about how we handle fourier transforms, and fourier coefficients of your residues. Have you considered:
\[
\int_{|z| = 1} f(z) z^k\,dz\\
\]
Which are fourier coefficients at their core.
Either way, don't shoot me. I'm mostly just casually spitballing here. I love your work and I just like to talk.
Regards, James