(04/04/2023, 12:23 PM)JmsNxn Wrote: ...

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

(04/04/2023, 12:23 PM)JmsNxn Wrote: 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

(04/04/2023, 09:28 PM)Caleb Wrote:

(04/04/2023, 12:23 PM)JmsNxn Wrote: 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 spent some time thinking about what you've said-- and I think I've come up with some preliminary results. This post is really only some early thoughts, so it probably won't be very well organized. Anyway, first, we need a lemma

Cauchy's Integral theorem for open curves
Cauchy's integral theorem can be derived very directly for circles. In particular, if we have 
\[\int_C z^k dz = i \int_{0}^{2 \pi} r^{k+1} e^{i {k+1} \theta} d\theta = \begin{cases} 2\pi i &  k = -1 \\ 0 & \text{otherwise} \end{cases} \]
Then, by homotopy, this works for all closed loops. However, really, a circle is only special because it loops back on itself when we multiply by any integer. But, for instance, a half circle loops on itself whenever we mutliply by an even integer. So, for a half circle, we will have removed half of every term. In particular, if \(C\) is a half circle, then  
\[\int_C z^k dz = i \int_{-\pi/2}^{\pi/2} r^{k+1} e^{i {k+1} \theta} d\theta = \begin{cases} \pi i & k = -1 \\ 0 & k \text{ odd } \\ r^{k+1}\frac{2 (-1)^{k/2}}{k+1} i & \text{k even} \end{cases}\]
By homotopy again, this means that every open curve with those endpoints evaluates to the same value. A similar sort of thing applies if we take \(C\) to be a third of a circle, in which case every 3rd term gets removed, etc. So, if we compute an integration with open curves, and the endpoint of those open curves lie on a circle and the two endpoints are a rational multiple of pi angle apart, then this weakened version of Cauchy's integral theorem can apply.

Now that we have this out the way, let us return back to the original problem, computing \( \int_{-i \infty}^{i \infty} e^{z^2} \). 

---

Let me share some intuition on where I see this approach going. In my view, \( e^{z^2} \) actually have a residue at \( + \infty \), and a different residue at \( - \infty \). NOTICE, this is not the typical way that people view residues. In complex analysis, normally, one thinks of there as being ONLY ONE INFINITY. In particular, there is only a single infinity on the Riemann sphere, so we can't exactly seperate out \( + \infty\) and \( -\infty \) by disjoint neighborhoods. However, my idea is the following. We think about taking a contour like this

This contour is viewed as enclosing \( + \infty \) but not enclosing \( - \infty \). Then, we map this contour under \( dz \to d\frac{1}{z}\). This is the intuition for the steps I'm trying to achieve, its basically an attempt to seperate out different residues. You can ignore this section if you're not interested in the intution, since this maybe makes no sense at all, but it is nonetheless the background for why I'm doing the stuff I'm doing in the following steps. 

---

Ok, now to the actual stuff. Let's map do the map \( dz \to d \frac{1}{z} \). Now we are dealing with the integration 
\[-\int_C \frac{e^{\frac{1}{z^2}}}{z^2} dz \]
(the extra 1/z^2 comes from the change of variables). The contour I'm looking at looks like this. 

So there is an infinitesimal indent on the side of \( + \infty \). First, observe that 
\[ \frac{e^{1/z^2}}{z^2} = \sum_{n=1}^\infty \frac{1}{z^{2n}(n-1)!}\]
So 
\[\frac{e^{1/z^2}}{z^2}  = \sum_{n=1}^\infty \frac{a_n}{z^n}, a_n = \begin{cases} (n-1)! & \text{n even} \\ 0 & \text{n odd} \end{cases}\]
Now, we apply the Cauchy's open integral theorem on the infinitesimal contour (call it D). It tells us that the value of this contour integral is
\[\int_D  \sum_{n=1}^\infty \frac{a_n}{z^{n}}dz = 2i \sum_{n=1}^\infty \frac{(-1)^n a_{2n}}{\varepsilon^{2n-1}(2n-1)} = \frac{(-1)^n }{(n-1)!}\frac{1}{\varepsilon^{2n-1}(2n-1)} \]
Now, this is obviously terrible to compute as \( \varepsilon \to 0 \). However, notice that we can write this as a contour line integral, in particular, we can write it as
\[ \int_{3/4 - i \infty}^{ 3/4 + i \infty} \frac{1}{2i} \csc(\pi n)\frac{(-1)^n }{(n-1)!}\frac{1}{\varepsilon^{2n-1}(2n-1)}  dn\]
BUT, THEN WE CAN SWITCH THE DIRECTION OF THIS CONTOUR!!! It picks up the pole at \( 2n-1 = 0\), i.e. at \( n= 1/2 \), and then everything else goes away, since the rest of the integral picks up the terms  
\[-2i \sum_{n=1}^\infty \frac{(-1)^n a_{2(-n)}}{\varepsilon^{2(-n)-1}(2(-n)-1)}\]
Which is simply zero as \( \varepsilon \to 0 \). Notice that this means the integral over \( D\) is equal to \( \frac{1}{2i} a_{2n} \) evaluated at \( n = \frac{1}{2} \). This means the residue is equal to \( 2 \pi i \frac{a_{2 \frac{1}{2}}}{2i} = \pi a_{2 \frac{1}{2}} \)

Lets check that this right. We have that \( a_{2n} = \frac{1}{(n-1)!} \). So \( a_{2 \frac{1}{2}} = \frac{1}{ (1/2-1)!} = \frac{1}{\Gamma(1/2)} = \frac{1}{\sqrt{\pi}} \). So we have that the integral is equal to \( \frac{\pi}{\sqrt{\pi}} = \pi \). 

OKAY, THIS IS PRETTY INSANE. It means that we can compute half these residues by looking at the half derivative of the function. It also says something much more interesting. When I see \(a_{2 \frac{1}{2}} \), I'm very strongly tempted to cancel out the 2 and the 1/2. But, in the case of e^z^2, you can do this, and its because there is some kind of incoherence in the derivatives of the function. The odd order derivatives make it look like the function should have derivative 0, but its even order derivatives make it look like it should have a derivative of 1/(n!). I think that in general, whenever fractional integration makes sense, it means that a_n is given by one single form, not a piecewise formula like with the exponential. In that case, we shoudl have \( a_{2 \frac{1}{2}} = a_1\). In that case, everything reduces back down into regular residues, since \( a_1 \) is the residue term. So, I think that we have that this integration method gives the usual residue precisely when the differential integral makes sense and is 'coherent' in the sense that it isn't defined in the kind of piecewise way that e^z^2 is defined. 

Anyway, I'll look at all this stuff later in detail and rewrite this in a coherent way but for now I have to go, hopefully these initial thoughts are still helpful though.

(04/04/2023, 10:34 PM)tommy1729 Wrote: Do you (implicitly) claim to have reduced the problem to fractional differentiation ?

I think you forgot a summation symbol around the third equation.



Am I correct in thinking if there is only 1 pole and it is at the boundary then only the angle of the curve at that point matters ?
Now wait, make that the 2 angles made by the corner , that touches the poles ; the 2 angles are with respect to the cartesian axes.

Also what about singularities and branch cuts at the boundary ?
Is this really only about poles ?


You often talk about doing analytic number theory, but how does this relate ?

Sorry if I am a bit slow.


regards

tommy1729

(04/06/2023, 05:30 PM)JmsNxn Wrote: I AM GOING TO DO SOMETHING I DONT KNOW THE IMPORTANCE OF, JUST FOR CALEB! LISTEN!


We are going to define the Dirichlet \(\eta\) function, as the function:

\[
\eta(s) = \sum_{n=1}^\infty (-1)^{n-1} n^{-s}\\
\]

This function is holomorphic for \(\Re(s) > 0\). Additionally, we know that it is \(O(s)\) bounded for \(\Re(s) > 0\). Which means, there exists a constant \(C_\delta >0\) such that \(|\eta(s)| < C_\delta |s|\) for \(\Re(s) > \delta\) when \(\delta>0\).

Now your first instinct might tell you to "fractionally differentiate" \(\eta\). But that's wrong. We instead take:

\[
\vartheta(w) = \sum_{n=0}^\infty \eta(1+n) \frac{(-w)^n}{n!} = \frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty} \Gamma(s) \eta(1-s) w^{-s}\,ds\\
\]

Where:

\[
\frac{d^{s-1}}{dw^{s-1}}\Big{|}_{w=0} \vartheta(w) = \eta(s)\\
\]

---

Here is where advanced number theory pokes its head. I am just trying to describe the problem, I am not trying to solve it. For you to describe singularities within the function \(\eta(\infty)\); within these fucking crazy beautiful residues your looking at.... You're solving very deep problems that I don't think you are aware of; and that's why I'm doubtful of some of the shenanigans you're doing.

BUT BY GOD IF YOU'RE NOT AT LEAST ON TO SOMETHING!

Dirichlet's function \(\eta(s)\) is meromorphic on \(\mathbb{C}\)......... Let's just move some integrals around and pull out some residues         



This is true and unique and Hilbert Special and fucking JON VON NEUMANNN UNIQUE so long as we assume that these functions are "Ramanujan bounded". Or we subject ourselves to advanced fourier theory. CARLSON WON A FIELDS MEDAL FOR FINDING THE FOURIER RIGOR VERSION OF RAMANUJAN'S RAW.

So I Fucking love this Caleb! But just know, the fucking devil's in the details. And you're going to have to go case by case; because this relates to highly advanced analytic number theory.

Fucking awesome, regards, James

EDIT:

Forgot to mention; all it takes to turn your integrals into my integrals is a variable change. I mean,

\[
\begin{align}
\frac{d^s}{dw^s}\big{|}_{w=0}f(w^2) &= \frac{1}{\Gamma(-s)} \int_{\gamma} f(w^2) w^{-s-1}\,dw\\
&=\frac{1}{2\Gamma(-s)} \int_{\gamma} f(u) u^{-(s+1)/2-1/2}\,du\\
&= \frac{\Gamma(-s/2)}{2\Gamma(-s)} \frac{d^{s/2}}{dw^{s/2}}\big{|}_{w=0}f(w)
\end{align}
\]

I'm still just rapidly posting these things but fuck this is beautiful.