 rotation to speak in this wonderful seminar. So I will talk about some joint work with Hang Buyi about negative moments of the Riemann-Zeta function. So I want to, maybe not everyone here is doing analytic number theory, so I want to start by giving some introduction about the Riemann-Zeta function. So how do we define it? Well, when the real part of S is bigger than 1, then the Riemann-Zeta function is given by Dirichlet series, one over the Dirichlet series of 1 over n to dS. And it also has an infinite Euler product, so an infinite product over primes of 1 minus B to the minus S inverse. So when the real part of S is bigger than 1, things are relatively easy and well understood. Now it's more interesting what happens to the left of 1. So it turns out that the Riemann-Zeta function has a meromorphic continuation to the whole complex plane, and it has a simple pole at S is equal to 1. And it satisfies a functional equation so we can relate zeta of S to zeta of 1 minus S. So because of the functional equation, it turns out that the most interesting place where we can, where we want to study the Riemann-Zeta function is the so-called critical strip. So we're in between the lines with real part of S0 and real part of S1. Fundamental question about the Riemann-Zeta function is the question about the location of its zeros. It turns out that understanding zeros tells us a lot of information about, for example, distribution of primes. So we know that the Riemann-Zeta function has so-called trivial zeros, that's negative even integers. And the Riemann hypothesis says that the non-trivial zeros of zeta should all lie on the line with real part of S equal to 1 half. Another big open question is the Lindelof hypothesis, which is actually implied by the Riemann hypothesis. So the Lindelof hypothesis gives a strong upper bound on the sides of the Riemann-Zeta function on the critical line. So the critical line is the line with real part of S equal to 1 half. So the Lindelof hypothesis says that zeta of half plus i t is bounded by t to the epsilon for any small epsilon. So here you think of t as being large. It grows, and then zeta of half plus i t is never bigger than t to the epsilon. So the main focus of the talk would be moments of the Riemann-Zeta function. So let's talk a little bit about how moments were introduced. So they have a long history. They were introduced more than 100 years ago by Hardy and Littlewood. So the 2kth moment of zeta is defined by this i k of t. It's the integral from 0 to t of zeta of half plus i t to the 2k, absolute value. The reason why Hardy and Littlewood introduced these moments is that they realized that the Lindelof hypothesis is equivalent to showing that the 2kth moment is bounded by t to the 1 plus epsilon for any small epsilon for all positive integers k. So the fact that the Lindelof hypothesis implies the upper bound is easy, the other way is true as well. Of course, they weren't successful in showing that the 2kth moment is bounded by t to the 1 plus epsilon for all k. The Lindelof hypothesis is still open today, but this started off the research into moments of zeta, and not only moments of zeta, but moments of other else functions, of else functions and families. So for a long time, moments were not really well understood, not even conjecturally, but this changed with work of Keating and Snape. So they conjectured that if you look at the 2kth moment, then you have an asymptotic formula with the main term of size t log t to the k squared. So this part was relatively well understood for a long time, and it follows by pretty simple considerations. But then the interesting question is what is the constant in front of this main term of size t log t to the k squared? It turns out that you can essentially write it as a product of two other constants, so one constant ak and another constant gk. The ak, again, it was pretty well understood, even before the work of Keating and Snape. It's given in terms of an infinite Euler product, so it's this infinite product over primes, which converges. But the gk was not really understood until the work of Keating and Snape. So before discussing what gk is, let me just explain the log t to the k squared part and the ak part. So heuristically, if you write the Riemann zeta function in terms of a Dirichlet series, so zeta of half plus it to the k, that's given by the Dirichlet series dk of n over square root of n times n to the minus it. So here dk of n is the case divisor function. So this counts the number of ways in which we can write n as a product of k integers. So if we ignore all questions about convergence and we write the 2k power of zeta half plus it in terms of Dirichlet series, we get this double sum over integers m and n, dk of m, dk of n over square root of mn, and we're integrating this m over n to the it. So you can see if m is equal to n, for example, the integral is very simple. So we expect to get a main term. Now if m and n are far apart, then we expect to get a lot of cancellation in the integral of m over n to the it. So heuristically, we only keep the so-called diagonal terms, which correspond to m is equal to n. And then we get the sum of dk of n squared over n. Say we truncate somewhere, say we truncate at t. And then we get an asymptotic formula with a main term of size log t to the k squared and this arithmetic constant a k. So this is where the a k comes from and where the log t to the k squared comes from. And in general, whenever you have a family of health functions, it's by using this kind of heuristic arguments, you can usually write down the power of log pretty easily and the arithmetic constant as well. But more interesting is what happens with this constant gk. So gk was known in a few cases, in a few small cases. So g1, the second moment of the Riemann data function was computed by Hardy and Ditton within the same 1916 paper. G2, the second or the fourth moment was computed by Ingham. Nothing beyond that is known unconditionally. For g3, there was a conjecture of Convian Gauss and for g4, there was a conjecture of Convian Gonec but nothing beyond g4 was known, not even conjecturally. So then Keating and Snape had the idea of modeling the Riemann's data function using random matrix theory. So they said that the Riemann's data function should behave in some way, should behave in the same way as the characteristic polynomial of unitary matrices. So if you want to compute the two-case moment of data, you should look at the two-case moment of the characteristic polynomial in the unitary ensemble. You compute that random matrix theory integral and you end up with this constant gk. And they conjectured that the gk that you get from random matrix theories, the constant that goes in the moment asymptotic formula for moments of the Riemann's data function. There is another way in which you can write down this, you can arrive at the same conjecture. This is the hybrid model due to Gonec, Hughes and Keating. So before we had to put in this constant ak in some sense by hand, so random matrix theory doesn't know about primes. So we had this correction, this ak factor which was the arithmetic part and then the random matrix theory part and then Gonec, Hughes and Keating had this idea of writing the Riemann's data function as a product between a part which involves primes. So they truncate the Euler product at some point x and then they have a product over zeros. So we write the zeros as half plus gamma n and we look at zeros which are very close to the point d within one over log x. And then if we assume a certain independence conjecture, so if we assume that the two-case moment is given by the two-case moment of the product over the prime primes times the two-case moment of the product over the zeros, then you actually can write down this same Keating-Snath conjecture, but now this has the advantage that you can write the ak and the gk at the same time. So you don't have to put in the ak by hand. Okay, now there is a more refined conjecture. This is work of Conry Farmer, Keating, Rubinstein and Snathes and the methods, the heuristic method they use is called the recipe. So they predict that the two-case moment is asymptotic to t times a polynomial in log t of degree k squared. And then you have some power saving error terms, say t to the one minus delta for some positive delta. pk is this polynomial of degree k squared and of course the leading coefficient of pk so the coefficient of log t to the k squared agrees with this Keating-Snath, random matrix theory conjecture. So this conjecture, the Conry Farmer, Keating, Rubinstein and Snathes one is not based on random matrix theory at all and it holds for integer k. And there is another more recent approach of Conry and Keating. They use long-distance polynomials and results on divisor correlations and they get exactly the same conjecture as the recipe gives you. So I mentioned before that in terms of rigorous results, we can only compute the second and the fourth moment rigorously, work of hard dilute, little woodwork of Ingham. For the fourth moment, Heath Brown later improved the fourth moment asymptotic so Ingham recovered the leading order term, the Keating-Snath conjecture. Heath Brown recovered the entire polynomial of degree k squared, of degree four, this case. Unconditionally, this is all that is known. For the sixth moment, Ingham computed the sixth moment conditional on certain conjectures about ternary additive divisor sums. And more recently, for the 38th moment, Ingham, Shan and Wong used similar results about, so conjectures about quaternary additive divisor sums and they recovered the eighth moment. So although in terms of asymptotic formulas, only the first four moments are known, we do have upper bounds and lower bounds, good upper bounds and lower bounds. So under the Riemann hypothesis, there are lower bounds of the right order of magnitude for all positive k. Well, this is work of Ramachandran Heath Brown. There are unconditional sharp lower bounds for k, which is greater than or equal to one due to rod zeppelin and Sanderarajan. And recently there are sharp upper bounds for k between zero and one, again, unconditional due to heap and Sanderarajan. So we have unconditional lower bounds for all positive k. Now, in terms of upper bounds under the Riemann hypothesis, there are some older results of Ramachandran Heath Brown for k between zero and two. There are sharp upper bounds for k of the form one over n, where n is an integer due to Heath Brown for k of the form one plus one over n. And we have sharp unconditional sharp upper bounds for k between zero and two due to heaps rod zeppelin and Sanderarajan. So what can be said about bigger k? So these results that I mentioned here about upper bounds mostly hold for small k or special values of k. Well, we have upper bounds of the right order of magnitude and this is conditional. So for all positive k, there's work of Sanderarajan, which shows that the two kth moment is bounded by T log T to the k squared plus epsilon. So he almost has sharp upper bound up to this plus epsilon here. And this was later improved by Harper, who under the Riemann hypothesis removed the epsilon. So for all positive k, you have a sharp upper bound of size T log T to the k squared. And these results will be important in what I will talk about next. Any questions before I move on? Okay, so the picture for positive moments is pretty well understood. We have asymptotic formulas in certain cases. We have good upper bounds and lower bounds, either unconditional or conditional. The more difficult question is, what happens for negative moments of data? And these are not as well understood. So there's a conjecture of GONEQ, which says the following. So here we're considering the negative to kth moment. Now, because we are looking at the negative moment, so we have the Riemann zeta function in the denominator, we have to be slightly away from the critical line so that we avoid the zeros on the critical line. So we introduced this small shift delta in the negative to kth moment. So now we're looking at the integral of zeta of half plus delta plus IT to the minus to k. And we want to study this integral. So the question is, how does this behave with delta? So we should expect when delta is closer and closer to the critical line, we're closer and closer to the possible zeros of zeta on the critical line, so things become more complicated. So GONEQ's conjecture states the following, depending on how big the shift is. So if the shift is reasonably big, by reasonably big, I mean that the shift is bigger than one over log T. So the families, we're looking at zeta of half plus delta plus IT, we integrate up to capital T, so we can think of this as being a family of size T. So if delta is bigger than one over log of the size of the family, one over log T, then GONEQ's conjecture says that the negative to kth moment should be asymptotic to one over delta to the k squared. And now if delta is smaller, so if delta is less than one over log T, so here we're really in the range when we're very, very, very close to the critical line, then he conjectures that the negative to kth moment should have different asymptotic formulas depending on the size of k. So if k is small, then we expect the log T to the k squared. If k is bigger than one half, then we have this other asymptotic formula, so we have a bigger power of log T. If k is equal to one half, then he actually predicts that you get a certain logarithmic correction, so you have a term here log of one over delta log T. Okay, so for k is equal to one half, something special happens and you see this logarithmic correction. However, it turns out that if you use random matrix theory again, then this partially contradicts GONEQ's conjecture. So there is work of, so there's a question if GONEQ's conjecture is unconditional. I mean, this is a conjecture, so it's not, I think the way he gets to the conjecture is by assuming the Riemann hypothesis, but everything is heuristic. It's conjecture. There's- Alexandra, what does DT means in the second display? In the second display? Yeah. So you mean the integral or? The integral minus k delta T, then DT. This part here? No, lower. Lower to the left. Lower to the left, yeah. Next one. Oh, sorry, yeah, it's a typo. Yeah, thank you. Yeah, there's no DT. Also, I would like to mention the, just sorry, when you mentioned the fourth-pile moment income and then his Brown asymptotic with power saving, I think the record is power saving T to 2-3rd. Yeah, that's right. Yeah, okay, thank you. Yeah, thank you. So if we are to use random matrix theory against to model the Riemann's data function, so now we're looking at negative moments of characteristic polynomials in the unitary ensemble. This was work of Berry and Keating and Blater, Forrester and Keating for other ensembles of random matrices, which model other families of L functions. Then these computations partially contradict Gonex conjecture in certain ranges. So for example, they suggest extra transition regimes whenever K hits an odd integer over two. So when K is equal to one-half, their conjecture, the random matrix theory computations agree with this Gonex conjecture. So you see this logarithmic correction when K is equal to one-half. However, what Berry and Keating do and Forrester and Keating later on is they suggest that when K is equal to three-halves, K is equal to five-halves and so on, you see extra correction factors. So it turns out that this has to do with a phenomenon of clustering of zeros. So for random matrices, Berry and Keating realize that clusters of eigenvalues actually start contributing when you are looking at higher and higher moments. So if we want to write down the conjecture taking these random matrix theory computations into account, this is what you would get. So here I wrote conjecture in quotes because the Berry, Keating and Forrester, Keating papers are really just random matrix theory computations. So they don't write down what you would expect for the Riemann data function. But if you are to translate their work into this number theoretic context, then this is the conjecture that you would get. So when Delta is big enough, so Delta is bigger than one over log T, then this agrees, the conjecture that you get from random matrix theory agrees with Gonex conjecture. So you'll get the one over Delta to the K squared. And then you can also pretty easily write down the arithmetic constant, a K. So this didn't appear in Gonex conjecture, but it's not hard to write it down. Now when Delta is less than one over log T, then the conjecture that you would get is the following. So here J is an integer, bigger or greater than or equal to one. So when K is equal to J minus one-half, so you have an odd integer over two, then this is what you get. You have this logarithmic correction here. When J is equal to one, so K is equal to one-half, this actually agrees with Gonex conjecture because this term Delta log T to the minus J times J minus one is zero when J is one, is one, sorry, the power is zero. However, when K is between J minus one-half and J plus one-half, then you get a different formula. So this doesn't agree with Gonex conjecture when K is bigger than one-half. Again, the arithmetic factory K is relatively easy to write down. So here new K is the case, the case coefficient of the negative case power of Zeta. Any questions about this? You can also use the same heuristic ideas behind the recipe. So recall that the recipe, the Kory-Farmer Kitting Rubinstein-Sneith conjecture was initially written down for positive K or for integer positive K. If you use the same heuristic ideas for negative integer K, then you actually get something which completely agrees with this random matrix theory prediction. But that conjecture wouldn't tell you about, it wouldn't tell you anything about what happens all the integers over two. So we could miss these correction factors of log of one over delta log P. So what is known about these? In the same paper where Gonex made these conjectures, he also got lower bounds. So these are conditional on the Riemann hypothesis that he got lower bounds consistent with the conjecture for all positive K in this range when delta is big enough. So delta is bigger than one over log T. He also got lower bounds consistent with the conjecture when K is less than one path in this other range when delta, in this small range when delta is close to the critical line. So again, recall that these are the ranges where random matrix theory agrees with Gonex conjecture. So these are the less controversial ranges. Gonex also studied discrete negative moments. So if you're looking, if you're summing over zeros of Zeta with height up to T and you're looking at Zeta prime evaluated at the zeros to the minus two K, then under the Riemann hypothesis and the further assumption that all the zeros are simple, Gonex got a sharp lower bound. So everything that is known so far is a concerns lower bounds and this conditional. So nothing was known about upper bounds. So in recent work with Hang Buoyin, we get upper bounds in certain ranges. So everything is conditional. So under the Riemann hypothesis, if delta, so if the shift is bigger than one over log T to the one over two K, then we get a sharp lower bound. So I write it here as log T to the big O of one. The big O of one can be made completely explicit and it's actually, we can make it to be log T to the K squared. So in fact, we can get sharp bounds here, log T to the K squared, maybe K squared plus epsilon, almost sharp bounds. Okay, so recall that we are expecting, in this range, we are expecting to get the log T to the K squared. Now if delta is smaller than one over log T to the one over two K, we also need, we have this condition that the log of one over delta is at most log log T. So this means that delta can be, might be at least one over log T to some constant. Okay, so we can't have, for example, delta to be one over exponential of T or something like that. Then we get this bound here. So the negative two K's moment is bounded by T to the K times this ratio, log of one over delta over log log T minus one half. So you can think of the log of one over delta over log log T, let's think of it as a constant. So for example, if delta is say one over log T, then this is one. So what does this bound tell us? Well, the trivial bounds. So if you are using just a point-wise bound for one over zeta, then the trivial bound is just T to the K times log of one over delta over log log T. So this part here is the trivial bound. So in this range when delta is small, we're saving power of T to the one half over the trivial bound. So of course we are far from getting sharp bounds in these ranges, but it's still a non-trivial bound. Any questions? And in fact, once you have sharp or almost sharp upper bounds, then you can actually get an asymptotic formula for the negative to case moment. So here I wrote the corollary. Let's assume that delta is bigger than log log T over log T. Then we can actually get an asymptotic formula for the negative to case moment. So we have a main term here, the arithmetic factor A, K, and zeta of one plus two delta to the K squared. So for delta small, this is of size one over delta to the K squared. So it's what the Connex conjecture predicts what's random matrix theory would predict in these ranges. But for these kinds of ranges, so when delta is bigger than log log T over log T, we need K to be less than one half in order to get an asymptotic formula. So when delta is roughly one over log T, you need it to be slightly bigger than one over log T. Then we get asymptotic formulas for small case. And in fact, we can write it down. So here I wrote down the error term is a little low of one, but we can write it down explicitly. It's a T to the one minus two delta times some power of log T. And similar results can be obtained for negative moments, for example, for quadratical functions over function fields. This is some prior work of mine. So this is a different family of L functions. It's a discrete family. So it's not continuous family like the Riemann-Zeta function. Any questions? Okay, so before talking about the, some of the ideas in the proof, I want to talk about some applications and some generalizations of negative, of studying negative moments. So one both application and generalization is the ratios conjecture. So this precursor to the ratios conjecture for Zeta was a conjecture of Palmer from 1993. So now instead of looking at negative moments or positive moments, you're looking at ratios. So you're looking at ratios of Zeta. So you have two Zeta functions in the numerator, two Zeta functions in the denominator. You're integrating on the critical line. So S here is one half plus IT. And you have some small shifts, alpha, beta, gamma, delta. Former originally conjecture that you can take them to be of size one over log T. Let's assume that the real parts of these shifts are positive. Then if you're looking at the ratio of these Zeta functions over Zeta functions, then you predicted a certain arithmetic formula. A certain asymptotic formula. So why is this interesting? Well, it turns out that this conjecture actually implies many interesting results about zeros of Zeta. So for example, it implies the pair correlation conjecture of Montgomery, a very strong result. So this essentially says that the pair correlation of zeros of Zeta is the same as the pair correlation of eigenvalues of matrices from the Gaussian unitary ensemble. So noticing that there is this connection between ratios and zeros of Zeta, studying ratios became something that people were interested in. So it turns out that if you adapt this method, the recipe of corn reformer, Kitting, Rubinstein, and Snays, you can actually write down a conjecture for ratios as well. This was a conjecture of corn reformer and CERN power. So this is a more refined conjecture than farmers. So it predicts some lower order terms and it's more specific. So it includes these arithmetic factors too. So you have this function A of alpha, beta, gamma, delta, which is given in terms of product over primes. You have, again, you're looking at the same quantity. So two Zeta functions over two Zeta functions. You have these shifts alpha, beta, gamma, delta. And the conjecture was originally stated for the real parts of the shifts in the denominator. So we're looking here. So the real parts bigger than one over log T. So if you recall from Gonex conjecture, one over log T was, when you're to the right of one over log T, it's the range which is in general better understood. Okay, so they predicted this asymptotic formula for the real parts of the shifts in the denominator bigger than one over log T. You have certain restrictions on the real parts in terms of the upper bounds for the real parts. These are there just to ensure that the arithmetic factor A of alpha, beta, gamma, delta converges. So they are pretty mild. But the important part is that you have the lower bound of one over log T. And in fact, they say that you can probably take the real parts to be less than one over log T. So let's say one over log T squared, as long as the shifts in the numerator go to zero at the same rate. So they should also be of size roughly one over log T squared. So what are some applications of the ratios conjecture? I should mention if you know something about negative moments, then usually you can also say something about the ratios conjecture, at least some partial results. But now what are some applications of the ratios conjecture? So a recent, very nice application which at the beginning at first sight seems not to have anything to do with Riemann's data function or L functions, is the proof that almost all integers without local obstructions can be written as the sum of three cubes. So this is recent work of Victor Wang and its condition alone, things like the ratios conjecture or negative moments of L functions. There are also applications to, for example, computing lower order terms for the pair correlation of zeros of zeta. So these were obtained previously heuristically by Bogomolny and Keating, using hard deleterate type arguments, but the computations were pretty involved. And it turns out that using the ratios conjecture, you can write down these terms in a much easier way. And you can do many other things. For example, you can compute modified moments of the Riemann's data function or of other L functions. And this can have applications to certain non-vanishing questions. So one application that I like very much is to the question of non-vanishing for L functions associated to Dirichlet characters. So there is a conjecture of Chowla, which says that if you're looking at L one half chi, so here chi is a Dirichlet, any Dirichlet character, then the conjecture says that L one half chi is never equal to zero. So this is a strong conjecture. If you focus on special families of characters, so let's say that you focus on the easiest family of quadratic characters, then Sanderergen showed that more than 87.5% of these L values are non-vanishing. And then conditional on the generalized Riemann hypothesis, Osloek and Snyder showed that more than, roughly 94% of these L values are non-vanishing. And they did that by computing the so-called one level density of zeros. So I'm not going to say exactly what this is. So the one level density of zeros is just a way of looking at low lying zeros in that family of L functions. It turns out that this gives you information about non-vanishing at the critical point. Now, if you assume the ratios conjecture, then you can actually compute the one level density of zeros for this function whose Fourier transforms have any support. And this would imply that 100% of these L values are non-vanishing. So this would give you the strongest results towards Chola's conjecture that we would have available so far. And before talking about some of the ideas in the profile, I also wanted to mention that if you're looking at the ratios conjecturing random matrix theory, so here we're looking at ratios of characteristic polynomials of certain matrices and certain assembles, then most of the times you can actually write down exact formulas. So this is work of many people. I mentioned a few here, Conry-Farmer-Zermbauer, Borodin-Straholtz, Conry-Forrester-Sneith, many others. So I just wrote down an example of such a theorem. It's for the unitary symplectic group. Here, this is the characteristic polynomial and you can write down an exact formula. So the situation in random matrix theory is very different from what we have in number theory where in most cases we don't know what's happening. So not even conditional on all the conjectures that you want. Okay, so now I want to describe some of the ideas in the proof. I hope it won't be too technical. So we're looking at the negative two case moment. Some of the ideas are similar to the work of Sander Erjan and Harper on getting upper bounds or almost sharp upper bounds for positive moments of zeta. But of course we have some extra problems coming from the possible zeros which are close to the critical line. So we start with an inequality for log of one over zeta. So if you think of the Euler product for one over zeta, you take the log of that, you can essentially bound log of one over zeta by the sum over primes of say you truncate somewhere it's size X, you have these coefficients A of P, you can just think of them as roughly one. Okay, so you bound log of one over zeta by one over P to the one half plus delta plus IT and you truncate somewhere at some point X and the point is the very important point is that you have flexibility in how you choose your X. And then you have a second term here coming from the zeros, a log T over log X times log of one over one minus X to the minus delta. So you can see that this term is going to be very big and it's going to cause a lot of problems if delta is small. Okay, so if delta is small, then this term is going to be very big. So now if we're taking the two case power of one over zeta, then we get this inequality here. So we have this huge term here one over one minus X to the minus delta to this two K over log T over log X and then you have an exponential of the sum over the price. Okay, so in order to understand the negative two case moment we have to understand the exponential of the sum over the price. And we have to be careful in how we choose our truncation or point where we stop the X. So now what do we do? Well, this idea goes back to work of Sanderarajan for upper bounds for positive moments. So we divide our primes into several intervals and the point is that primes on different intervals behave differently. So you expect that small primes behave differently from bigger primes. Okay, so we divide our primes into intervals say P is between T to the beta G minus one and T to the beta J. Again, this is a sequence which we have to choose carefully. It's not really, I'm just going to say that beta one is of size log P over log T and then we at each iteration we mod the beta J is like R times beta J minus one. And when do we stop? Well, we stop when beta capital K is a small constant. So in some sense this X the end point is a very small constant. Okay, so let's assume now that we are in this range when delta is bigger than one over log T to the one over 2K. So this is the easier range when the shift is big enough. So for each point T, we have three different possibilities. Okay, so we start with the small primes. So P one, this is the contribution from small primes. It's the contribution from primes less than T to the beta one of one over P to the one half plus delta plus IT. So in general, we don't expect the small primes to have a huge contribution. So we're looking at those points T for which the small primes have a big contribution. So we have to decide what exactly it means to be big. Here I wrote that big means that the contribution from small primes is bigger than beta one to the minus D. So with my choice of beta one, this is roughly log T over log log T to the one minus epsilon. Okay, so if the contribution from small primes is bigger than this quantity, which is maybe not that important for the purpose of this talk, then we have to show that this doesn't happen too often. So this is what we're going to use. We're exploiting the fact that if the contribution from small primes is big, then we actually gain because it doesn't happen. It happens on a set of measures zero. Okay, so we have the integral over the exceptional set is not. So it's usually hard to bound integrals over these very special sets. So we want to bound it by an integral over the whole interval zero to T. So we bound this by the integral from zero to T of zeta. And then we have this quantity here, beta one to the D times the contribution from the small primes raised to some power S naught. And we have flexibility in how we choose the S naught. And we want to choose it in a suitable way so that we actually get this contribution small. Okay, so now we have this quantity, beta one to the D S naught. And this goes to zero. So this is the important part. And then we separate the contribution from the primes with the moment of zeta. So we bound this by the fourth moment using Cauchy Schwartz and then the moment of the sum over the primes. So for the first integral, so now we want to bound each of these integrals. For the first integral we can use, we don't really have upper bounds for negative moments. So there are no results in the literature. So we don't know what to do except for use a very bad pointwise bound for the negative of the zeta function. So for the first integral here, we use just the pointwise bound one over zeta is bounded by this quantity. And this is going to be, it's going to be a very weak bound because we use this strong pointwise bound for each value of zeta. But nevertheless, we're going to win over this trivial bound because we have this quantity here which goes to zero. And then we have to compute moments of the sum over the primes, but we can do that. So this is a fairly classical thing to do, computing moments of sums over the primes. We can do that as long as the sum over the primes is not too big. So this is a sum over primes. We can write it as a Dirichlet series, a sum over n of certain coefficients. Let's call them a n over n to the half plus i t. And here n must be less than t to the two S naught beta naught. So the point is that as long as two S naught beta naught is not too big compared to t, then we can actually compute these moments. Okay, so we will end up, if we put things together, we will end up with getting this contribution from the exceptional set is little o of t. So it's very small. It's negligible. Okay, now we move on. So now if we have a point t for which the contribution from small primes is well behaved, then we're looking at what happens with primes on the second interval. So now we're saying, well, if the contribution from primes on the second intervals, on the second interval is not well behaved, so it's bigger than expected, then we also want to exploit the fact that this doesn't happen too often. If the contribution from primes on the second interval is well behaved, then we move on to the third interval and so on. So at each step, say step j, we're assuming that the contribution from all the primes up to the jth interval is small enough. So here we have these parameters, the beta j's that I defined before. So now we're assuming that the contribution from the primes up to the jth interval, each of these contributions is small, but the contribution on the j plus one interval is big. So it's bigger than this beta j plus one to the minus d. Okay, so let's call this set tj. Then what do we do? Well, recall, remember that we were bounding one over zeta by an exponential of a sum over the primes. Okay, so we had the exponential of the sum over the primes, say p up to x of one over p to the one half plus delta plus it. Okay, so essentially we want to use the fact that we can approximate the exponential by a truncated Taylor series whenever the sum over the primes is small. Okay, so we have this nice inequality here. If say t is less than L over e squared, then we can essentially bound the exponential of t by the truncated Taylor series of the exponential. We call it EL of t. So this is the truncated Taylor series at the point L here. Okay, so this is exactly what we're going to use since the contribution from primes on the say h interval with h less than j. Since this is small, then we can essentially approximate the exponential of the sum over the primes on the h interval. We are approximated by this truncated Taylor series. So we truncate the Taylor series at beta h to the minus d. Okay, so now when we take the integral on the tj interval, we use our key lemma, so our inequality for one over zeta with x to the point where we truncate being t to the beta j. So then we have this exponential here, the exponential of the sum over the primes. And now the primes go up to this t to the beta j. So what do we do? Well, as we did for the exceptional set, we don't really want to have the interval over this special set dj. We want to have an integral over the whole interval zero to t. So we bound this by the integral from zero to t. For primes up to the jth interval, we bound the exponential by these truncated Taylor series. And then we multiply by this quantity, beta j plus one to the d times the contribution from the j plus one interval to sum power is j plus one, which we can choose, we have to choose appropriately. And the point is that this quantity here is always bigger than one, okay? So we bound this by this integral. And now this might look a bit ugly, but it's actually, we can rewrite this whole thing inside the interval as a Dirichlet series. So the first part here, this is a Dirichlet series of a given length. It has length sum of beta h to the one minus d with h less than j, less than or equal to j. Now, if we look at the primes on the j plus one interval raised to the sj plus one, again, we can write this as a Dirichlet series. So it's a Dirichlet series of size t to the s times sj plus one times beta j plus one. So the point is whenever, when we unwrap this big Dirichlet series, as long as we choose our parameters wisely, this is going to be a short Dirichlet series. So we can actually compute the integral from zero to t when this Dirichlet series is not too long, okay? So we can actually show that this contribution is also going to be small. So the point is that we have this quantity here, beta j plus one to the d, which goes to zero, which helps us, okay? So this contribution is also going to be small. And then we get to the final interval where all the contribution of the primes on all the intervals up to k, all those contributions are small. So let's call this set dk, then we just apply our key lemmas and we choose the point where we try to be t to the beta k. So again, we estimate the exponential of the sum over the primes by these truncated Taylor series. And if we choose our parameters well, this is not going to be a long Dirichlet series. So we can actually compute moments, okay? And we end up with this bound of log t to the k squared plus epsilon. So recall that we expect to get the log t to the k squared. And once we have this bound, this upper bound, then we can actually get an asymptotic formula in these ranges. So this is something that happens many times with moments. If you want to get an asymptotic formula, many times you need an a priori upper bounds, a sharp upper bound in order to get some kind of asymptotic formula. So this is what we do later on. We use this upper bounds to get an asymptotic formula. So everything that I've said here applies for big shifts. So when the shift is bigger than one over log t to the one over two k, if the shift is smaller, then the problem is a bit harder because as I said, the smaller the shift is, the closer we are to the critical line. So we do the same thing as we did before, but we have to do some kind of inductive argument. So let's denote by u, this quantity log of one over delta over log log t. Again, let's think of this as a constant. So if delta is say one over log t squared, this is true. Okay, then the point is that we do this argument that I just presented and we get some kind of bound here. So recall that the k u t to the k u is the trivial bound. So now we save a little bit over the trivial bound, but then we can do this argument again. So recall that whenever we were bounding the contribution from the exceptional set, we were using some a priori bound on the negative four key moment. So we keep doing that and that each step, we use the a priori bound that we obtained in the previous step of the induction. So at the second step, for example, we end up with, so this is the previous bound that we had in the first step. And now we save a little bit more over it and we keep going. And at the end, we end up with this upper bound of size k u times one minus one over two k u. Okay, so I think I will stop here. Thank you very much. And I see that there was some question in the chat as well, so I can probably answer later after that.