Zeta Explained #84: Perron's Inversion Formula

[00:00:00]Hello. Welcome to Zeta Explained. Today we're talking about Perron's inversion formula.

[00:00:06]Now we're going to start with the definition of the Riemann zeta function.

[00:00:08]Um it's defined as the sum from n equals 1 to infinity of 1 over n to the s, and we can generalize this to any function in the numerator, not just the 1 function. So in general, a Dirichlet series is of the form we have some function like capital S, which is the sum from n equals 1 to infinity of little f of n over n to the s.

[00:00:26]And in this video series I usually write the same letter to denote the correspondence or the mapping um of the Dirichlet series. So we have capital F versus little case f.

[00:00:36]And this is a table of various Dirichlet series discussed so far in the video series. This is not an exhaustive list, um but you can see that these Riemann zeta function is just one example of a Dirichlet series where the little case function is 1 and the capital function is zeta. But there's all these other ones as well.

[00:00:52]Um and in particular, we're going to talk about this one a little bit towards the later part of the video. So just remember like this one.

[00:01:00]Now one question is what is the point of deriving all of these mappings um from like a little case function to the big function? So one thing is it allows us to use complex analysis to say things about the arithmetic functions.

[00:01:13]So in other words, you can actually like we've seen some examples in prior episodes where to figure out something about this function like the Euler totient function, which on the surface looks like just a function about divisors and prime numbers, you can actually translate it into zeta of s minus 1 over zeta, do some complex analysis to derive some results about this function, and then translate the results back um to the Euler totient function.

[00:01:37]So we're going to go through the main tool to do this.

[00:01:42]Um in previous episodes we've kind of used this tool, but we haven't really talked very um in-depth about it. So that's going to be the point of today's episode of like what exactly is this tool that allows us to um say things about this function.

[00:01:56]So the main tool is called Perron's inversion formula. So, suppose f of s and little f of n satisfy this condition, which is just that they're a Dirichlet series, um where the series is absolutely convergent in some region, then Perron's inversion formula, so the idea of inversion formulas in general is you have like one function written in terms of another function, but you often want to reverse the process. You want to write the other function in terms of the first function.

[00:02:25]So, Perron's inversion formula says, if this is true, then the sum of n less than or equal to x uh with this tick mark on it, which I'll explain in a second, of f of n is equal to 1 over 2 pi i times an integral of f of s times x to the s over s ds.

[00:02:41]So, you can see how like this part now, it's not quite f of n on its own, but it's at least a much simpler form of f of n than in this infinite sum over n to the s.

[00:02:51]Um and on the right side now we have this capital F function, which was by itself, and now it's inside some complicated-looking integral.

[00:02:58]And the sum with a tick mark means you only add 1/2 of the last term um n if x is equal to n's, and I'll get to what that means in a second.

[00:03:06]Okay, so um the way this video is going to work is I'm not going to actually prove this theorem right away. We'll actually save that towards the end. Um but we're going to go through a bunch of examples um and like how to use the theorem first. So, first of all, let's actually go through like what is this uh sum with a tick mark mean?

[00:03:23]Okay, so if you consider this very simple sum of n less than or equal to x with a tick mark of 1, so this just means every time you get to a new number, you add 1.

[00:03:34]Um so, if x is 0.99, then the sum is equal to 0 because there are no natural numbers less than or equal to 0.99.

[00:03:42]And when you get to 1.01, it's saying like how many natural numbers are there up to 1.01? Well, it's just the number 1, so at 0.99, this gives you 0, at 1.01, this gives you one. But the tick mark is special about what happens if it's exactly equal to one.

[00:03:58]So if you just said um like the normal sum of n less than or equal to x of one, then if x is equal one, then this should be one. But with this tick mark, it's that it's actually equal to 1/2.

[00:04:10]Uh you just like think about this like definition right here.

[00:04:15]Um and you can see like how it kind of like it's just like in right in between like the zero and one. Uh like right when it crosses this threshold, it takes the half point value or the midpoint value. And you can see like when it crosses to two, it goes to 1.5 first before it jumping from one to two.

[00:04:35]So that that's really all this sum is saying.

[00:04:39]All right. So now let's actually go through how to use the integral. So I'm sure like this integral looks very complicated and unless you've seen complex analysis before, you might have like no idea how to actually evaluate this integral.

[00:04:51]So the first thing we're going to do is turn the integral from a an infinite integral into a finite integral. So notice here we have a minus I infinity and a plus I infinity.

[00:05:00]And we're going to convert it to a minus IT and a plus IT and then take the limit as T goes to infinity.

[00:05:06]Okay, so pretty standard stuff so far.

[00:05:10]And it's still too hard to do this line integral from C minus IT to C plus IT.

[00:05:15]So what we're instead going to do is we're going to consider a uh the integral around a entire rectangle.

[00:05:24]So the integral from C minus IT to C plus IT, if you visualize it here, and I've just drawn some like kind of arbitrary dots on it for now just to illustrate the point later.

[00:05:34]Um the integral that we care about is this blue section right here going from C minus I T to C plus IT.

[00:05:41]But I'm going to claim it's actually easier to consider the rectangle contour like this. So you're going to go from C - IT to C + IT then to -U - IT then to -U - IT and then to C - back to C - IT.

[00:05:59]And basically by the definition of integrals like the integral over C like I'm going to call C the entire like rectangle contour is equal to the integral over the four segments added together like this.

[00:06:14]And because integration is a linear thing like we can actually just like move this part of the like this part back all the way to the left. So, this integral is really what we care about.

[00:06:25]We only really care about C - IT to C + IT. We don't really care about the other three legs of this integral other [snorts] than to as like used to compute this first part. So, I'm just going to rearrange this equation to move this to the left and this uh integral over the entire rectangle to the right.

[00:06:40]And you get that this integral is equal to the entire integral minus the other three legs.

[00:06:46]Um an intuitive way of thinking about this integral is just this statement right here. So, the right side integral like the one we care about is equal to the total integral minus the sum of the three other legs of the integral.

[00:06:58]Okay. So, once again, maybe this makes sense, but maybe you're wondering like how does this make anything any easier to calculate? Like we just turned one integral into four integrals.

[00:07:07]Okay. So, the idea is uh using complex analysis, we can actually straight up compute what this total integral is.

[00:07:14]And then the hope is that the other three integrals uh might be either easy to bound or possibly can show that they go to zero and therefore you just get that the desired integral that you want care really care about is just the thing that you can actually calculate, the total integral.

[00:07:31]All right. So, the way you calculate the total integral is by something called the residue theorem.

[00:07:36]So, in complex analysis when you take a an integral um in a closed loop like this then the value of the integral depends on what's known as the residues of all the poles inside the integral.

[00:07:50]Like inside the region enclosed.

[00:07:53]So, if you know what are all the poles are, then you can just compute all of the residues at those poles and add them together and that's the value of this uh integral.

[00:08:04]Okay, so what I'm saying is uh let's use Yeah, let's I think this is fine.

[00:08:10]I'm saying these two things are equal.

[00:08:13]So, 1 over 2 pi i times this integral over C, that's the entire rectangle, is equal to the sum over each pole s naught of the residue of this integrand at s naught.

[00:08:25]And then I'm keeping everything else the same. And this thing you can actually usually like when you actually use Perron's inversion formula, this part is not too hard to kind of show like to calculate usual usually. And usually this term dominates the other three terms unless you've chosen something where the integrals like are very badly paved.

[00:08:45]All right, so again, the standard use of Perron's inversion formula is you first locate the poles, then you compute the residues, and then you show that the three remainder integrals are either bounded or possibly like in the best case scenario they actually go to zero, although you don't always have that as we'll see in the examples.

[00:09:01]So, therefore this integral that you actually care about, this integral right here, can be written as a sum of residues minus these other three remainders. And again, hopefully these remainders are small relative to the residues that you actually care about.

[00:09:16]Okay, so I'm sure this was very abstract. Um, I'm going to walk through three examples. The first two examples I would say are relatively trivial, and the third example is going to be more interesting, but I want to go through the trivial examples anyways to build intuition about like how to actually like what this formula is actually saying and like how to think about and how to like in practice just use the formula and do the calculations.

[00:09:38]Okay, so the first example is we're just going to let f of s equal the zeta function. Um and therefore the little case of f of n function is just the one function.

[00:09:49]Okay, so um using this like definition of the zeta function over here, then if you apply Perron's inversion formula, this is the general Perron's inversion formula, and this is the specific case using the zeta function, then it says that the sum of n less than equal to x with a tick mark of one is equal to this integral with where the f of s is now zeta of s.

[00:10:13]All right, and in this example, when you um think about the poles of this integrand, so the question is where when does this integrand um like blow up to infinity is essentially the idea of like where the poles are.

[00:10:28]And there's two poles. One is when the the zeta function itself has a single pole at s equals one.

[00:10:35]And if you look at the other term, this x to the s power over s, well, that has a pole at zero because if you plug in s equals zero, then like x to the zero is one, but then you get divided by zero, so that's another pole.

[00:10:47]So, there's two poles as at one and at zero.

[00:10:52]Now, we know from episode 11 uh when we calculate the Laurent series of the zeta function that the zeta has this pole at s equals one with a residue of one.

[00:11:03]And the pole at like if you isolate out the one divided by s, you don't even think about the x to the s part, you just look at one over s, then one over s has a pole at s equals zero, and this trivially has a residue of one.

[00:11:19]So, therefore you can compute the residue of the entire integrand by just evaluating the rest of the integrand at these two values of like these two pole locations of one and zero.

[00:11:29]And then we're going to need the fact that zeta of zero is equal to minus a half. So, again, the idea is this pole right here is going to lead to this term x to the one over one.

[00:11:40]And this pole over here is going to lead to zeta of zero over or uh times x to the zero.

[00:11:47]Okay, so just to kind of walk through the calculation one time. I'm not going to do this for all of the examples, but just for this example, what I'm doing is this pole is right at s equals one.

[00:11:57]So, the zeta function is equal to one.

[00:11:58]So, I'm going to cross out the zeta function because we already know the residue of zeta is one.

[00:12:02]Then you plug in one to the rest of the equation. So, x to the one over one. So, that's why you get x to the one over one.

[00:12:10]For the other pole, so for the s equals zero, um we just take zeta of zero times x to the zero power because we've already crossed out this uh term over here. Um and then you get zeta of zero times x to zero.

[00:12:23]And if you compute this out, you get x minus one half.

[00:12:28]And therefore, the original integral that you care about, just the right side of the integral, is equal to the total integral, which we said is now just the residue, so x minus one half, and then minus the integral over the other three edges.

[00:12:42]Okay, but before doing any more math, let's think about like what this means.

[00:12:45]Like is there any intuition for what is this What is this x to the minus or x minus one half mean?

[00:12:51]Well, what we already know is that the x minus one half is supposed to be the principal term or the kind of the main part of this sum right here, the sum of n less than or equal to x with a tick mark of one.

[00:13:03]So, let's first compare that before we're doing any more math.

[00:13:08]Okay, so we've plotted these two things, these two functions. So, the red function is the sum with a tick mark of n less than or equal to x of one.

[00:13:17]And this is just saying that whenever you encounter whenever x goes up and crosses a new natural number, then it jumps up by one, but at exactly that number, it takes the half point, the midway point.

[00:13:29]Um so, that's kind of what the red function looks like.

[00:13:33]And the blue is the thing that we calculated from the residues of x - 1/2, and you can see that it's just looks like a pretty good approximation of the red step function.

[00:13:43]And the interpretation now is that we just went through a very difficult and roundabout way, an overly powered way of saying that a step function that goes up by one at every integer, grows roughly like x.

[00:13:56]And you can probably see that if you if you didn't do any complex analysis at all, and you just looked at this function, you say, "Okay, like what does this grow like? Is it linear? Is it quadratic?" You'd probably say that this is linear. So, that's basically all we've proved so far.

[00:14:11]Um and this is a bit anti-climactic, so we should try something less obvious.

[00:14:17]Um and I'll also mention that we kind of skipped the step where we have to bound those three remainder integrals.

[00:14:23]Um the math is quite tedious, I'm not going to go through it. Um but you can see that the error, which is the difference between this red function and the blue function here, is a sawtooth function.

[00:14:34]And this implies that the other three edges of that contour integral are not going to go to zero, uh but they are going to stay bounded, because this sawtooth is bounded.

[00:14:42]Um but I am going to skip the bounding math and go to some more interesting examples. So, let's go to example two.

[00:14:48]So, let capital F of s be equal to zeta of 2s, and let this interesting function I sub square of n be the indicator function.

[00:14:57]So, an indicator function is just always equal to one or zero, depending on n, um if n meets a certain condition, and it's going to be equal to one if n is a square number, and zero otherwise.

[00:15:09]And what you see there is if you try to write out zeta of 2s, you write it as one the sum of one over n to the 2s, which looks like this.

[00:15:18]And then if you do two to the 2s, that's just four to the s, and three to the 2s is 9 to the S.

[00:15:23]So, you can see that this is just the sum from n equals 1 to infinity of I sub square of n over n to the S power.

[00:15:31]Remember, this is just the indicator of is n a square number. If yes, then one.

[00:15:35]If no, then zero.

[00:15:38]Then Perron's in form inversion formula on this um expression or this equation says that the sum of I sub square of n is equal to this integral of zeta of 2S.

[00:15:51]So, this is the same integral as before, but there's a 2S inside the zeta instead of an S.

[00:15:56]And this sum is now, instead of going over one, it's going through the indicator function of square numbers.

[00:16:03]And you should really think of this as just roughly the number of square numbers up to X.

[00:16:09]All right, let's continue with the math.

[00:16:11]So, in this example, there are once again only two poles. Um but now, there's not a pole at one, but there's a pole at 1/2.

[00:16:18]And this is because the zeta function itself kind of has a pole at one, but you don't have zeta of S, you have zeta of 2S. So, for 2S to be equal to one, then you need S to equal 1/2. So, that's why there's now a pole at 1/2.

[00:16:31]And there's still a pole um at zero like before. This X to the S divided by S causes the pole at zero.

[00:16:39]And if you do the same calculation as before, um in this case, this thing actually has a residue of 1/2, so you actually have there's an extra 1/2 as compared to the calculation on the last page. Um but you'll get that this simplifies to square root of X minus 1/2.

[00:16:53]And now, I'm going to show you the graph once again of like what's the geometric interpretation like of this result here.

[00:17:00]And how does it stack up against this indicator function?

[00:17:04]Well, um so, the red is the indicator function, or at least the sum of the indicator functions.

[00:17:09]And [snorts] the really the interpretation of this function is just that it's a function that steps up by one every time you get to a square number. So, at 1, 4, and 9, and 16, and 25, and so on, it just steps up by 1.

[00:17:22]And, you know, with that technicality, at exactly that integer that's a square number, it only jumps up by 1/2, but then it kind of finishes jumping by the rest of the 1/2 right after that.

[00:17:33]And the blue line is just square root of X minus 1/2, and it's just an approximation of this indicator like this sum of indicators function.

[00:17:41]And you can see that this is probably not the most impressive result either, because if you kind of just think really hard about like what is the number of square numbers below a given number X, then it's you would probably conclude that it is roughly square root of X, and you didn't need to do complex analysis uh to figure this out.

[00:17:58]Okay, so we what we did is we just once again went through a very difficult and roundabout way of saying that if you have a step function that goes up by 1 in every square number, then this grows roughly like square root of X. So, again, I think it's interesting that it gives a this it gives you the correct result, um but it's kind of obvious that it's true.

[00:18:18]Um but I will say it's interesting that you can use complex analysis to like derive this result. Um but furthermore, I'm going to point out the following heuristic.

[00:18:28]The further to the right the pole is, the higher the power of X is in the formula.

[00:18:33]So, in the first example, I'm going to actually show you a graph of this in a second, but in example 1, we had a pole at 1, and therefore the approximation function grew like X instead of square root of X, so X to the 1.

[00:18:45]Whereas in this example, there was a pole at 1/2, and this led to like square root of X instead of like linear term of X.

[00:18:54]And then also this the pole at 0, the black dot, um leads to the constant term this like minus 1/2 here.

[00:19:03]So, the idea is just that we generally care about the rightmost poles uh when we do this Perron analysis.

[00:19:10]Um okay, so let's go to this example.

[00:19:13]So, um, so this is I'm I'm trying to illustrate the point here. So, the further to the right the pole is, the higher the power of X it generates.

[00:19:21]Um, and visually you can see that in example one, there's a pole at one, and the when you compute the residues, you get X as your first term.

[00:19:30]And X obviously looks like a linear function. And in the second example, the pole was at 1/2.

[00:19:37]And when you plug into this residue formula, you get cuz S is now 1/2 here, um, you get X to the 1/2, and therefore your function grows like X to the 1/2.

[00:19:47]And I find this very, very interesting, um, that the power of this function is dependent on where the pole is in this complex plane.

[00:19:58]Um, and, um, we're going to see more applications of this like intuition. Um, but I think if there's like one takeaway from this video, I think it's like this sentence right here. It's that the location of the poles, specifically the rightmost poles, uh, determine how fast this function grows.

[00:20:16]All right, so now let's get to a interesting example. So, I claim the first two examples are not very interesting. We're just saying approximately how many integers are there up to X, which is kind of obviously X.

[00:20:25]And then how many square numbers are there up to X, which is perhaps obviously the square root of X. But now let's go to something much, much less obvious involving prime numbers.

[00:20:35]So, we're going to let f of s equal negative zeta prime of s over zeta of s, zeta prime of s being the derivative of the zeta function.

[00:20:42]And we know from episode six that this course is a Dirichlet series corresponding to the lambda function, or the von Mangoldt function. Um, this function is equal to log of p if n is a prime power of the prime p.

[00:20:56]And it's equal to zero otherwise. And if you apply the Perron inversion formula to this Dirichlet series, then you get that the sum of n less than or equal to X, with the tick mark, of course, of lambda of n is equal to this contour integral where this thing is now negative zeta prime of s over zeta of s.

[00:21:12]And another way of writing the sum is actually this psi of x which is known as the Chebyshev psi function. We've actually talked about this function many times previously in the video series, but like for now like this is like what we're going to define it as. So this is a function that is counting up whenever you encounter a prime number or a prime power. That's like roughly the way to think about it right now.

[00:21:37]Because like this function is zero if you're not at a prime power. So this sum of this thing it you're it's like an accumulation function that's like going up by some amount whenever you get to a prime or a like a prime power like this.

[00:21:52]Now in this example unlike the past two examples, so the previous two examples there were only two poles, but now we actually have an infinite number of poles.

[00:22:01]I I guess especially as you try to expand the interval of the bounds of the interval to infinity.

[00:22:07]So you have a pole still at s equals one because zeta prime of s has a singularity at s equals one s equals one. You still have this black dot the pole at s equals zero because this thing is inducing a pole.

[00:22:21]But now you have zeta on the denominator.

[00:22:27]So this function this whole integrand will blow up to infinity if you're dividing by zero.

[00:22:34]So anytime zeta of s is equal to zero, then this thing will be a pole.

[00:22:39]And the zeta function is known to have trivial zeros and non-trivial zeros. So the trivial zeros are these green dots and the non-trivial zeros are the red dots, but I'm going to claim from our intuition earlier that remember the only the rightmost like poles like contribute the most to the like growth rate. So this is kind of why I like when you like read literature about the Riemann zeta function, no one really cares about the trivial poles or the trivial zeros, cuz as you can see in this diagram, like they don't really contribute much. They're like way, way, way further to the left than all of these things.

[00:23:10]All right. And if you follow the math in episode 47, um and like I guess just using the residue computations, if you try to sum all the residues of all these points, you get the following. So, let me draw some arrows to correspond this. So, this pole is going to give you the X term right here.

[00:23:28]Then the red dots are going to give you this sum over here.

[00:23:32]The green dots are going to give you this sum. And finally, the black dot gives you this term here.

[00:23:39]Um so, that's just the sum of the residues.

[00:23:45]And from our intuition earlier, the this term, like the further to the right they are, the more they matter, like the higher the power of X.

[00:23:53]Um so, you can see that this term is a linear term of X because this has real part one.

[00:23:57]Then you get the non-trivial zeros. And the ones we know of have a real part of one half, but no one knows for sure where they all are. That's This is like something about the Riemann hypothesis, which we'll talk about in a second.

[00:24:09]And then I've kind of like changed the order of these two things a little bit.

[00:24:12]The next thing that matters is actually this black term, the minus log of 2 pi.

[00:24:16]And this is because we know that zeta prime of zero over zeta of zero is equal to log of 2 pi.

[00:24:20]Uh I think we we derived this result back in episode 46. And then finally, you have the trivial zeros, which are like this. Um but if you think about these functions as X increases, this thing actually decays to zero.

[00:24:34]So, again, like it's not even a constant term. It's actually growing slower than a constant. It's decaying to zero.

[00:24:41]All right. So, let's draw a plot of this thing. So, I'm sure this might sound very abstract. So, what are we doing here?

[00:24:48]Um now I'm going to plot the sum of n less than or equal to X of lambda of n, which is the Chebyshev psi function, versus this approximation of X minus log of 2 pi. So, what I'm doing is I'm just not including the trivial zeros or the non-trivial zeros. I'm just saying, let's approximate by this X term and the minus log of 2 pi term right here.

[00:25:08]And you can see that this looks like a pretty good approximation. So, um what the red function is doing, let's ignore the blue line for now. The red function is when you get to a prime number or a prime power. So, if you get to number two, you jump up by log of two.

[00:25:22]When you get to three, you jump up by log of three.

[00:25:24]And when you get to the number four, you jump up by 1/2 of a log of four.

[00:25:29]In other words, you jump up by a log of two if you think about the exponents and logs.

[00:25:34]And that log five, you jump by log five.

[00:25:36]Now, six is a composite number that's not a prime power, so you don't jump at six. And at seven, you jump up by log of seven, and this keeps going. So, at eight, you jump by 1/3 log of eight, which is the same as jumping up by log of two.

[00:25:51]And interestingly enough, this blue line that this thing is roughly a linear function.

[00:25:59]So, we've kind of stated a like kind of non-obvious statement about prime numbers and prime powers, that the waiting of If you think about this, this is really a statement about the waiting of prime numbers or the distribution of prime numbers. And we said that if you wait the prime numbers and prime powers in this certain way, then it's roughly a linear function of X.

[00:26:19]That's kind of interesting.

[00:26:21]Although, we can actually make this approximation better um if we add all the non-trivial zeros and the trivial zeros as well. But, remember the non-trivial zeros matter more.

[00:26:30]So, let's figure out like let's see like what exactly is this thing? Like, how do we add this?

[00:26:38]Now, this is a result um this is actually a formula known as von Mangoldt's explicit formula. So, you can actually say that the psi function, this sum, is actually exactly equal to X minus the sum over the non-trivial zeros, minus the sum over the trivial zeros, and then minus this log of 2 pi.

[00:26:55]And recall that the full formula should have some integral terms as well, but they actually go to zero in this situation. So, this would be the full formula if we want to rigorously state this formula.

[00:27:05]But it turns out we kind of already did the math in episode 47 that these three intervals, so remember it's equal to the sum of the residues, minus the three like remaining edges of the interval, but this thing actually goes to zero.

[00:27:17]So, therefore it's just the sum of the residues, which gives you like this thing.

[00:27:22]Uh and I think I actually messed up here. So, this uh note that this here these are two infinite sums, whereas I think I wrote two finite sums here. Pretend these are both infinite sums. So, this should psi of x should be equal to these things with an infinite sum, not a finite sum.

[00:27:35]So, I apologize here.

[00:27:38]Now, the Taylor series, so one way to write this trivial zeros term um explicitly is note that the Taylor series for log of 1 minus e can be used to show the following.

[00:27:52]So, this whole thing right here is just equal to 1/2 log of 1 minus x squared.

[00:27:59]And if you want you can think about expand this out as a Taylor series and see where that goes.

[00:28:05]Now, for the red term, so that's how you deal with the green term. Now, this is how you deal with the red term.

[00:28:10]If you let If you write the non-trivial zeros rho as 1/2 plus i gamma and the conjugate uh of rho is equal to 1/2 minus i gamma.

[00:28:22]And note that this sum is kind of a weird sum. It only converges if you uh sum them uh when they're paired with their conjugate zeros. Um I guess one thing I didn't mention is the zeta function, anytime you have a non-trivial zero, there must be a conjugate zero like reflected across the real axis.

[00:28:41]And this sum will not converge if you like just like only went in one direction. You kind of have to pair the zeros like two at a time like this.

[00:28:51]Now, if you do some math about these exponentials like X to the row power, you can see that these are actually just sine cosine waves and sine waves.

[00:29:00]And if you add X to the row over row with the conjugate version X to the conjugate row over conjugate of row, then you get that this is equal to two root square root of X over 0.25 plus gamma squared of a cosine wave and a sine wave.

[00:29:16]And therefore, because these terms go to zero, this approximation actually gets more and more accurate as you add more and more non-trivial zeros.

[00:29:23]Um so, this is a graph that we actually showed back in episode 45, I believe. Um but the idea is that you have this base function like square function, which is kind of like a straight line almost. At the beginning, it's not quite a straight line because of this logs thing, but it kind of converges into roughly a straight line.

[00:29:42]And the idea is if you start adding these non-trivial zeros, you start approximating the actual kind of like step function.

[00:29:50]Remember, this function is jumping up by like log of two when you get to two, jumps up by log of three when you get to three, jumps up by log of two when you get to four, jumps up by log of five when you get to log log when you get to five, does not jump on log of six or it does not jump up on six and so forth.

[00:30:05]And if you just like add all those cosine and sine waves, then um you can see that the red line here like this line is like those kind of looks like it approximates the step function.

[00:30:18]Um although as you go to the right, you can see the red line is actually like wiggling around a bunch.

[00:30:22]So, that's the general idea. Um so, you can actually use this formula to get the exact like prime counting function basically of psi of X, uh which only jumps on primes and prime powers.

[00:30:35]Now, let's actually revisit this contour for a second um to get like an alternate interpretation of the prime number theorem.

[00:30:40]Um So, the rightmost pole um of this thing of our interval was at real of s equals one.

[00:30:49]And that's this thing right here.

[00:30:50]Actually, let me use a different color.

[00:30:52]So, this thing um generates the main term of just x.

[00:30:57]Like x to the one is just equal to x.

[00:31:00]And the error term comes mainly from the non-trivial zeros, these red dots here, and the growth rate um depends entirely on the real part of the zero. So, if you think about the like when you take x to the row power, then any imaginary part of this doesn't really matter. It's the real part really that matters as far as the growth rate is concerned.

[00:31:20]Um and we're going to ignore the black dot, and we're going to ignore all the like green dots because they're further to the left.

[00:31:29]And we have proven before um like you can look at episode 22 if you want that the real part of these non-trivial zeros must be between zero and one.

[00:31:40]But, we actually don't know exactly where they are. So, technically some like we don't know if the Riemann hypothesis is true. So, if the Riemann hypothesis is false, then in theory there could be a zero sitting very very close to this right edge of the critical strip, like near like real part equals one. So, you could have a zero with a real part of 0.999999, and therefore the that zero would contribute a term like it'll be x you know, like psi of x um is equal to x, and then this term right here would correspond to like some constant like, you know, a * x to the 0.999, which grows almost as fast as x itself.

[00:32:24]Now, the prime number theorem is equivalent to proving that there are no zeros on the line of real part equals one.

[00:32:31]And this kind of proves that so, if there were a zero with real part one, then it would kind of compete with this term right here.

[00:32:38]Um so there'd be a term growing at like one and this other term also grows like one.

[00:32:44]But because there are no zeros on this line, then all these zeros contribution is going to grow less than X um asymptotically.

[00:32:52]So therefore as X gets very very very large, then psi of X will approach X um or at least the ratio of psi of X divided by X approaches one uh because all the zeros are to the left of this line.

[00:33:04]Now, this is not a very rigorous statement because there's you have to deal with infinity, so like in theory there can be this there can be infinite like sequence of zeros getting close to the line um and you have to deal with that case. Um so I'm going to claim that if you try to rigorously turn the statement that there are no zeros on the line real part equals one um into the prime number theorem, it's actually still quite hard to do. Um so we spend episodes 47 and 48 like rigorously proving kind of the statement.

[00:33:31]Now, if the Riemann hypothesis is true, then all of the non-trivial zeros, all these red dots, are exactly on this vertical line real part of S equals 1/2.

[00:33:40]And using our intuition from before, um this means that the error term is bounded by roughly X to the 1/2.

[00:33:46]Although there's a plus epsilon here because you have to deal with the fact that there's an infinite number of them.

[00:33:51]And this is actually gives the tightest possible bound um for this error term of the prime number theorem or, you know, a lot of people phrase it as the tightest possible bound for the distribution of prime numbers.

[00:34:01]And in particular what I mean by the tightest possible bound is the following. So in theory, if the Riemann hypothesis is false and there's this zero here with a real part of 0.999999, then you would get that psi of X uh from our contour integral is equal to X from this term plus something that's big O of X to the 0.999999 plus epsilon from this term over here.

[00:34:27]Whereas if the Riemann hypothesis is true, then all of the zeros are on this line with real part equals 1/2, um and using our intuition from earlier that if, you know, you have a zero on some region, then the growth is going to depend on the real part of that thing, then it's going to grow roughly like x to the plus O of x to the 1/2 plus epsilon, 0.5 plus epsilon. So, you can see how this statement right here is much stronger than the statement over here.

[00:34:51]It's saying that the growth, the error term actually grows much slower than the error term here.

[00:34:55]Um and that's kind of like one implication like of the Riemann hypothesis is it kind of bounds like how like wildly this the primes are uh distributed.

[00:35:08]Okay, so that's probably a lot, so I'm going to kind of wrap up, um start wrapping up the video now. Um so, using our three examples of Dirichlet series, so we went through these three examples and the Perron inversion formula and a lot of complex analysis, we derived two trivial-sounding statements and one very non-trivial statements.

[00:35:24]Um so, we said that the count of positive integers up to x is roughly x.

[00:35:28]Sounds pretty trivial.

[00:35:30]Then we got that this count of square numbers up to x is roughly square root of x, and that might sound very trivial as well. But then we got that the weighted count of prime powers up to x is roughly x plus an oscillating contribution like a sine and cosine waves from non-trivial zeros of the zeta function.

[00:35:47]And I think like this is kind of the power of like the Perron inversion formula.

[00:35:53]All right, and then I'll end with an actual proof, um but I'm going to claim we already did the like the hard part of the proof back in episode 47.

[00:36:01]So, we've used Perron's inversion formula a bunch, but how do we know that it's true?

[00:36:05]Well, in episode 47, we proved this simple version of Perron's inversion formula. So, we already know this, so let I of y be an indicator function which is equal to zero if y is between zero and one, equal to 1/2 if y is exactly equal to one, and it's equal to one if y is greater than one.

[00:36:23]Then this thing we proved in episode 47 says that if y and c are greater than 0, then I of Y is equal to 1 over 2 pi I times this integral from C minus infinity to C plus infinity of Y to the S over S DS.

[00:36:36]And if you kind of look at this, this kind of looks like the main version of the formula we want.

[00:36:42]And I'll mention two like asides here that aren't really useful in what we're about to prove, but these are just like interesting things to think about. Um if you take this integral from 0 to infinity of I of Y times Y to the minus S minus 1 dy, um I haven't really explained like like the motivation behind this, but let's just think about this integral for a second.

[00:37:03]Then because the way I of Y is defined, um I of Y is only non-zero if Y is at least 1.

[00:37:09]So you can actually rewrite this as the integral from 1 to infinity of just Y to the minus S minus 1 because like Y is just this thing is just 1 if Y is greater than 1.

[00:37:18]Now you can ignore this 1/2 thing because it's on a set of measure 0 here.

[00:37:25]Okay, so you get this integral from 1 to infinity of Y to the minus S minus 1 dy, and this is just an elementary integral from like I'm sure you've taken a calculus course to solve this integral.

[00:37:34]You this is just Y to the minus S over minus S, um this improper integral here, and this just evaluates to 1 over S.

[00:37:45]And this is um if you can look at this, this is essentially an inversion formula um of this function right here.

[00:37:55]So uh right here, the lowercase of F doesn't mean a Dirichlet series, it's more about uh Fourier transforms and Mellin transforms. So this is actually just a Mellin transform and an inverse Mellin transform.

[00:38:07]So you have a little case function f of x, then the Mellin transform capital F of S is the integral from 0 to infinity of f of x times x to the minus S minus 1 dx.

[00:38:16]And inverse Mellin transform lets you kind of put pull the x out and put the x in um as little f of x equals 1 over 2 pi i times the integral from c minus infinity to c plus infinity of capital F of s x to the s ds.

[00:38:31]So, you can see that this um statement right here is just the statement of this form like this.

[00:38:40]Where the capital F of s is 1 divided by s.

[00:38:42]And the little case function is this I of y or I of x uh indicator function.

[00:38:48]And then Perron's this basic version Perron's inversion formula, inversion formula, excuse me, is just the inverse Mellin transform like this function right here. So, the indicator function is equal to this integral uh where f of s, remember, is equal to 1 over s.

[00:39:02]>> [snorts] >> And that's exactly what we're saying right here.

[00:39:08]But again, the way we proved this back in episode 47 was not by just saying it's a inverse Mellin transform. It's we actually like did all the bounding of this thing.

[00:39:17]And I'll mention that the Mellin transform and inverse Mellin transform can be themselves thought of as Fourier transforms, but again, I don't want to derail the topic too much. All right, so let's use this lemma to prove Perron's inversion formula.

[00:39:29]And I claim we're almost all the way there already. So, suppose you have a Dirichlet series where you write capital F of s in terms of the little lower case f of n.

[00:39:38]Then you take this you start with the integral already written out.

[00:39:45]And you replace this f of s with a Dirichlet series.

[00:39:48]So, it's the sum from n equals 1 to infinity of little f of n over n to the s.

[00:39:53]And because we're assuming, so we assume that the c is large enough such that the series is absolutely convergent, then you can interchange the sum and the integral.

[00:40:02]So, in other words, I'm going to pull this sum of f of n outside of the integral.

[00:40:07]And so, it's the sum from n equals 1 to infinity of f of n times this integral.

[00:40:12]And this integral here, if you think think the x divided by n, um it this is actually exactly the simple version that we already proved the lemma one.

[00:40:23]So remember lemma one says this that the this integral right here um is equal to I of X divided by N where I of X over N has these definitions right here.

[00:40:35]And therefore this whole thing is just equal to I of X over N this indicator function.

[00:40:44]And finally if you kind of interpret this sum, so the sum from N equals one to infinity of F of N times I of X divided by N well think about which terms you're actually adding. So if N is less than X you're adding one.

[00:40:57]If N is equal to X you're adding one half.

[00:41:01]And if N is greater than X then you're adding zero.

[00:41:04]So this is actually not an an an sorry excuse me not an infinite sum it's a finite sum up to X. And if X is exactly um equal to an integer N then this only adds half of the last value. So therefore this exactly matches the definition of the sum with a tick mark.

[00:41:21]So this thing is just equal to the sum with a tick mark of N less than or equal to X of F of N. And this proves Perron's inversion formula since we started with this thing and ended at this thing.

[00:41:31]Cool.

[00:41:32]Um so we went through a lot of interesting examples. Hopefully um it's now a little bit more clear like what Perron's inversion formula means um and like why the right most poles like matter the most. Um so thanks for watching. We'll see you next time.