Amaζing πaϑoloΓical compleξ analyΣiΣ
by math_explorer, Oct 9, 2014, 5:34 AM
[edit: I totally messed up the second most important equation in this post and nobody mentioned it. I guess nobody reads my math content after all.]
Vim doesn't even have a built-in digraph for \vartheta. I shall solemnly propose
Okay, with that out of the way:
[Complex Analysis, Princeton Lectures in Analysis II, Stein & Shakarchi, as usual]
Theorem 2.2. If
then ![\[ \pi^{-s/2}\Gamma(s/2)\zeta(s) = \frac{1}{2} \int_0^\infty u^{(s/2)-1} \left(\vartheta(u) - 1\right)\, du. \]](//latex.artofproblemsolving.com/d/6/1/d613da5e129061157830bc489a0c7d334be803d4.png)
I think this is the point where I last stopped really learning complex analysis and started sort of gaping with my mouth open.
I didn't know what to do, so now, after coming back after X months and still not really knowing what to do, I figure oh well, I guess it might help a little bit to memorize it. As everybody knows, memorizing stuff is easy, but knowing how to rederive it is far more effective. I don't even have the quotient rule for differentiation memorized because I can rederive it. Why am I restating the obvious?
Okay, whatever, on the left side there are three Greek letters with increasing mathematical sophistication and "parameters" that are linear multiples of
decreasing in complexity. First halve, then negate.
On the right side there's an integral. Inside the integral, the most mathematically mundane thing of all,
(it's just a dummy variable), has the most complex exponent application of all, 
. Even operator precedence matters now. It gets multiplied by 
, but at least this and the 1/2 outside the integral are easy to justify as an artifact of different summation bounds --- if you read on, you'll find that it's easier to mentally chunk the 1/2 factor outside the integral next to this term.
There, you now have Theorem 2.2 memorized!
---
The derivation looks like plain dark magic, invoking the equation that appeared out of nowhere:
![\[ \pi^{-s/2}\Gamma(s/2)n^{-s} = \int_0^\infty u^{(s/2)-1} e^{-\pi n^2 u}\, du. \]](//latex.artofproblemsolving.com/6/e/d/6edf71bbb38e64f8880430a4a8cc9e744ba05312.png)
until you realize that the left side is basically using
, the "zeta component" that's figuratively screaming to be summed across 
, and finding a way to absorb it as a fudge factor into the gamma function that results in 
inside the integral, which is the "vartheta component" that's figuratively screaming to be summed across .
In the vartheta expression, you have to subtract 1 and halve because the summing of from 
to 
only gets less than half of what the vartheta component wants, all integers from 
to . For the record,
![\[ \vartheta(t) = \sum_{n=-\infty}^\infty e^{-\pi n^2 t}. \]](//latex.artofproblemsolving.com/f/5/8/f5895575aa2d6d0bf8c33ce8c990206916ab9fb9.png)
Also, if you forgot (I sure did):
![\[ \Gamma(w) = \int_0^\infty x^{w-1}e^{-w}\, dx \]](//latex.artofproblemsolving.com/4/c/1/4c154c0d9a6a1f05c5ac8175fee17039a85aef23.png)
Then you just gratify both expressions' wishes by summing across (the interchange with the integral of which has to be justified, though.)
I'm not going to actually carry out the algebraic work that got swept behind the scenes. Because of all the variable changing, getting the fudge factors to work out is still a bit mind-bending for me. Maybe another time. Furthermore, this "nice" form of the equation is still not uniquely specified from rederiving it --- you could, for example, move the power of
to the other side, and might really want to since it's a "negative" exponent. Alas, things don't work out nicely later if you do that.
The next step is to redefine the πuζζlinΓ jumble on the left side with ξ, because getting to write two leftward curves in a letter is so cool amirite?
![\[ \xi(s) := \pi^{-s/2}\Gamma(s/2)\zeta(s) \]](//latex.artofproblemsolving.com/d/3/8/d3888a7db919ce6efe1bfefbad6fc99b15804a5a.png)
Then we can continue it analytically and magically prove that
.
How? Why? What exactly do we gain from re-expressing the zeta function in terms of the gamma and \vartheta function? Well, it boils down to the functional equation that \vartheta satisfies, which in turn exists because of the Poisson summation formula, which is that weird thing I invoked in that Brilliant CMC thread from last time. And, of course, magic fudge factors.
That's pretty much it. For details, find an actual complex analysis textbook. It does turns out, however, that the difference between the \vartheta and the half-\vartheta sum we're originally using here, which the book allocates yet another Greek letter for
![\[ \psi(t) := \frac{\vartheta(t)-1}{2} = \sum_{n=1}^\infty e^{-\pi n^2t}, \]](//latex.artofproblemsolving.com/e/4/f/e4f640e021aeb4bd38abd0ea91adb84e056cde3a.png)
is more than just an annoying artifact because we have to get rid of the
term in the sum before being able to integrate in .
Hooray, we made it to the bottom of this post!
Vim doesn't even have a built-in digraph for \vartheta. I shall solemnly propose
digraph 9* 977Okay, with that out of the way:
[Complex Analysis, Princeton Lectures in Analysis II, Stein & Shakarchi, as usual]
Theorem 2.2. If
then ![\[ \pi^{-s/2}\Gamma(s/2)\zeta(s) = \frac{1}{2} \int_0^\infty u^{(s/2)-1} \left(\vartheta(u) - 1\right)\, du. \]](http://latex.artofproblemsolving.com/d/6/1/d613da5e129061157830bc489a0c7d334be803d4.png)
I think this is the point where I last stopped really learning complex analysis and started sort of gaping with my mouth open.
I didn't know what to do, so now, after coming back after X months and still not really knowing what to do, I figure oh well, I guess it might help a little bit to memorize it. As everybody knows, memorizing stuff is easy, but knowing how to rederive it is far more effective. I don't even have the quotient rule for differentiation memorized because I can rederive it. Why am I restating the obvious?
Okay, whatever, on the left side there are three Greek letters with increasing mathematical sophistication and "parameters" that are linear multiples of
decreasing in complexity. First halve, then negate.On the right side there's an integral. Inside the integral, the most mathematically mundane thing of all,
(it's just a dummy variable), has the most complex exponent application of all, 
. Even operator precedence matters now. It gets multiplied by 
, but at least this and the 1/2 outside the integral are easy to justify as an artifact of different summation bounds --- if you read on, you'll find that it's easier to mentally chunk the 1/2 factor outside the integral next to this term.There, you now have Theorem 2.2 memorized!
---
The derivation looks like plain dark magic, invoking the equation that appeared out of nowhere:
![\[ \pi^{-s/2}\Gamma(s/2)n^{-s} = \int_0^\infty u^{(s/2)-1} e^{-\pi n^2 u}\, du. \]](http://latex.artofproblemsolving.com/6/e/d/6edf71bbb38e64f8880430a4a8cc9e744ba05312.png)
until you realize that the left side is basically using
, the "zeta component" that's figuratively screaming to be summed across 
, and finding a way to absorb it as a fudge factor into the gamma function that results in 
inside the integral, which is the "vartheta component" that's figuratively screaming to be summed across .In the vartheta expression, you have to subtract 1 and halve because the summing of
from 
to 
only gets less than half of what the vartheta component wants, all integers from 
to . For the record,![\[ \vartheta(t) = \sum_{n=-\infty}^\infty e^{-\pi n^2 t}. \]](http://latex.artofproblemsolving.com/f/5/8/f5895575aa2d6d0bf8c33ce8c990206916ab9fb9.png)
Also, if you forgot (I sure did):
![\[ \Gamma(w) = \int_0^\infty x^{w-1}e^{-w}\, dx \]](http://latex.artofproblemsolving.com/4/c/1/4c154c0d9a6a1f05c5ac8175fee17039a85aef23.png)
Then you just gratify both expressions' wishes by summing across
(the interchange with the integral of which has to be justified, though.)I'm not going to actually carry out the algebraic work that got swept behind the scenes. Because of all the variable changing, getting the fudge factors to work out is still a bit mind-bending for me. Maybe another time. Furthermore, this "nice" form of the equation is still not uniquely specified from rederiving it --- you could, for example, move the power of
to the other side, and might really want to since it's a "negative" exponent. Alas, things don't work out nicely later if you do that.The next step is to redefine the πuζζlinΓ jumble on the left side with ξ, because getting to write two leftward curves in a letter is so cool amirite?
![\[ \xi(s) := \pi^{-s/2}\Gamma(s/2)\zeta(s) \]](http://latex.artofproblemsolving.com/d/3/8/d3888a7db919ce6efe1bfefbad6fc99b15804a5a.png)
Then we can continue it analytically and magically prove that
.How? Why? What exactly do we gain from re-expressing the zeta function in terms of the gamma and \vartheta function? Well, it boils down to the functional equation that \vartheta satisfies, which in turn exists because of the Poisson summation formula, which is that weird thing I invoked in that Brilliant CMC thread from last time. And, of course, magic fudge factors.
That's pretty much it. For details, find an actual complex analysis textbook. It does turns out, however, that the difference between the \vartheta and the half-\vartheta sum we're originally using here, which the book allocates yet another Greek letter for
![\[ \psi(t) := \frac{\vartheta(t)-1}{2} = \sum_{n=1}^\infty e^{-\pi n^2t}, \]](http://latex.artofproblemsolving.com/e/4/f/e4f640e021aeb4bd38abd0ea91adb84e056cde3a.png)
is more than just an annoying artifact because we have to get rid of the
term in the sum before being able to integrate in .Hooray, we made it to the bottom of this post!
This post has been edited 2 times. Last edited by math_explorer, Apr 26, 2015, 10:45 AM
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