 So, this is where we were last time, which is the big or the most important target for our course, to show that this connection between psi of x and the Riemann hypothesis. And how the Riemann hypothesis says something very interesting about the distribution of primes. Now, there are still some loose ends here, one is that we really want to know what pi x, remember what pi x was? Number of primes less than equal to x, but after that definition of pi x, I switched over to psi x and everything since then has been done in terms of psi x, which is good, but still ideally would like to extract out some information about pi of x. So, that is one loose end and the second one is something actually not really loose end, but still it is a very interesting observation that and I mentioned it last time that Riemann hypothesis implies that psi x equal to this. So, that is an implication and since it took us so much time and effort to prove this implication, it might appear that it is a one way implication, but as it turns out that is not true. If psi x equals this, it implies Riemann hypothesis. So, that is what I am going to show you first, because it is very easy proof. So, let us state psi x equals x plus order. So, actually we can prove something stronger here that if psi x is x plus order x to the half plus epsilon for any epsilon greater than 0. I do not even have to assume it is x plus order square root x log square x, I can allow any power of log x or any x to the epsilon as long for any every epsilon greater than 0, then the Riemann hypothesis holds. So, how do we prove this? To prove this, we just go almost all the way back and right in the beginning if you remember, we derived this expression for zeta prime over zeta. Do you remember what that was? Let me just refresh your memory. We know that zeta z for real z greater than 1, this is equal to this product m p 1 over 1 minus 1 over p to the z, this is correct. Now, take log and derivative, this gives you zeta prime z over zeta z. This is of course, for real z greater than 1, this is equal to zeta log, we will convert this into sum and then derivative will bring down sum over prime p as minus here, then differentiate this. You get 1 over this, then minus log p and this is equal to in fact, this is precisely the expression we used in that integral. To replace this sum by zeta prime over zeta and that is how we got psi related to zeta prime over zeta. Now, what is lambda n? Lambda n equals log of p, n is p to the m, 0 otherwise. That is the definition of lambda n and what is psi x? psi x equals n less than equal to x summation of lambda n. In other words, I can write lambda n as psi of n triple and so let us plug this value in for lambda in the above expression. The next step is to convert this into an integral. Now, rearrange this, collect psi n terms together. So, what is the coefficient or multiplier to psi n? You get 1 over n to the z here, you get 1 more. What is that? 1 by n to the z minus 1 by n plus 1 to the z. Anything missing? n equals 1, you have psi 0, psi 0 is trivial. So, this is what you get. Now, look at this. So, this is the integral of plus 1 psi t by t to the z plus 1 d t times z. See, in interval n to n plus 1 psi is constant. So, the psi comes out. So, you get d t over t to the z plus 1, its integral is minus t to the z over z. So, that z z cancels out. So, you get 1 minus 1 over t to the z and then n plus 1 to n. So, this minus goes away. It does not go away. Now, this is good because we can take this z out and make this from 1 to infinity. Now, keep in mind that real part of z should be greater than 1. Now, since we have psi t equals t plus order t to the half plus epsilon, we get zeta prime z over zeta z to be z minus 1 to infinity t plus order z plus 1 d t. So, the first term just gives you minus z 1 to infinity d t over t to the z and the second term. Now, what does first term say? See, real z is greater than 1. So, first term will converge. What does it converge to? It converges to… So, this is basically minus z. This is z. 1 over z minus 1 t to the minus z t to the… When t is infinity and real z is greater than 1, that vanishes. This part vanishes. When t is 1, then you get negative, negative. t is 1. So, you get 0. Now, let us look at zeta prime over zeta and this expression. The expression on right hand side is analytic as long as real z is greater than half. Actually, greater than half plus epsilon because then this part will become real part of this will become more than 1 and then this integral converges. This is analytic except for pole at z equals 1, real z equals 1. So, that is right hand side. Left hand side is zeta prime over zeta. We know that by whole analysis for real z greater than 1, left hand side equals right hand side. Now, by again analytic continuation that zeta prime over zeta is equal to right hand side for real z greater than half plus epsilon. Now, the right hand side, except for pole at quality, I am saying this, except for pole at z equals 1 is analytic everywhere on real z greater than half plus epsilon. This is true for any epsilon because my assumption was for any epsilon psi of x is order x to the half plus epsilon. What does it mean? It means that zeta prime over zeta is also analytic for any z such that real z zeta prime over zeta is analytic for at z equals 1. We already know it as pole at z equals 1, but if the Riemann hypothesis was not true and there was which means that there is a 0 of zeta at real z more than half. Then zeta prime over zeta will have a pole there which will contradict this statement. Therefore, it is extremely simple. It is a 10 minute proof. I got lost somewhere otherwise I would have finished it more quick. The other direction of course, took forever and both sort of arise from the same idea that is this expression for zeta prime over zeta. You have this expression and for psi you have this sum psi of x is partial sum of these. So, that is these two have a very nice entire relation. In fact, later on this course which is our next step once I am done with all of this in the remaining lecture what I will do is two things. One is to generalize this. So, there is this I mean if you look at it from a higher level psi x is a partial sum of some quantities. Then zeta prime over zeta in our case, but in general some other function zeta prime over zeta is this infinite sum with same quantities in an numerator and n to the z in the denominator. This is zeta prime over zeta lambda n over n to the z for all n psi x is sum n less than equal to x lambda n. There is this relationship. So, one is a complex plane object, one is a number theoretic object and there is a we did all this work to establish this nice relationship between this. Now, it turns out that this we can do not only for lambda n, but many other numbers and there is entire theory which is been developed on this and there is really some real remarkable result which I will only be able to give you a very brief glimpse. I do not either have time nor full understanding to explain all of that to you, but this is essentially the starting point for theory of modular forms. So, this some of the objects called modular forms which I will define at some point which are functions of the kind of the properties like zeta function and there is a whole beautiful theory around this and not just theory we one can use this different forms to derive different number theoretic results just as we did for zeta function. That is the one direction. The second thing I would like you to show hopefully I will have time to do that is at least one domain in which we can prove Riemann hypothesis. We cannot prove Riemann hypothesis over for this complex plane of this kind, but as I said we can do now that we after abstracting out the basic ideas we can take this relationship of number theoretic functions and complex analytic functions form of different kind and form the relationship with them and you know pose the same similar hypothesis make the similar hypothesis about that those complex analytic function about where the zeros lie and relate them to the property of this number theoretic functions. So, we can come up with various versions of Riemann hypothesis. Now, most of this versions remain unproven conjectured for unproven, but some versions have been proven. So, I will give you one example which also not in its own very interesting example which is of elliptic curves. So, again certain number theoretic objects the elliptic curves are thought of as some number theoretic objects and then we can associate a corresponding Riemann hypothesis with these objects or rather some certain numbers associated with these objects and then prove that Riemann hypothesis. But before we do all of this I still have to tie a few more things here one is the psi x versus pi x that business and the second thing I want to do before we move forward is to prove a basic version of not quite Riemann hypothesis, but something let us say the starting point towards prove which is what I will prove is that on the line real z equals 1 there are no zeroes of zeta function and that is enough to prove prime number theorem. Because it means that error is in psi x is psi of x is x plus order x to the 1 minus delta for some tiny delta essentially that error term cannot cancel out x and that is that proves the prime number theorem. Good and as it turns out all of these are very easy so I can do that quickly. So, let us investigate this psi and pi psi we know what it is pi we know what it is how do you write one in terms of another see what how does psi x get calculated. You go through all the numbers in sequence whenever you detect a prime power you add log p pi x get constructed whenever you see a prime you add 1. So, the calculation of psi x I can divide it as follows I can split it into stages stage 1 only consider primes whenever you see a prime add a log p stage 2 consider prime powers whenever you see a prime power p square add a log p and that log p therefore is I can instead of talking of log p I will talk in terms of the number that you see whenever you see n to be prime add log n whenever you see n to be a prime power add half of log n log of root n which is half log n whenever you third stages whenever see n to be a prime cube add one third of log n. Now, how many such things you add I mean how many primes you will see you will see exactly pi x primes for each one of those primes you will add log n how many prime squares will you see pi of root x exactly you will see exactly pi of root x prime squares. Similarly, you will see exactly pi of third root of x prime cubes and for each prime square you have to add half of log n. So, this thing I can now let us just do it this way. So, you add for each prime and psi you add log n, but if I derive divide psi so I am varying psi and whenever I see a change in psi it is because either I have seen a prime there or I have seen a prime square there or a prime cube there or prime fourth power. So, at that point I have appropriate logarithmic term. So, instead of I can be alternately stated the way whenever I see a change in psi and divide at t if I see a change in psi and I divide that by log t then either I could in stage one I will be counting all primes. In stage two I will be counting half of pi of square root x one third of pi of x to the one third. So, this is equal to I have to go through numbers in sequence t starting from one and notice change in psi t at whichever t there is a change in psi t that point I have to divide by log t. So, this is captured by this integral because I have to now define this little more formally because psi is not a continuous function. So, psi t is just well it is continuous, but it is not differentiable it has the step like quality, but we can define the this d of psi x as measuring the delta value. So, between psi of d of psi t is well as t varies let me write for d psi t is 0 most of the time and it becomes lambda t at t equals to t at integral point there may be a jump inside. So, at that point of course, this is an instantaneous jump. So, I will just fuzz around this meaning of this integral or this differential d of psi t to mean that at around integer d of this is measuring that jump amount of the jump and the jump is exactly by lambda of t and everywhere says it is 0. So, this is not quite the original meaning of differential this is called Stiglitz integral there is a mathematician was a mathematician called Stiglitz lots of i j's in that spelling who define this first and actually formally analyze them and proved in what situations is this is sensible definition and then because this allows you to work with a larger class of functions even those which are this step wise functions as long as it is continuous not necessarily differentiable we can work with this. So, that is the meaning will assume here and one of the things that he showed was that we can given the reasonably decent condition on psi of x psi of t we can assume this to be functioning almost exactly as a normal differential and again psi of t does satisfy those conditions or fairly mild condition you should not be sort of jumping up should not be function which will jumps up and down it is a monotonically going up function. So, some simple properties which as long as they are satisfy you get want a new page no insult we centered here what do I do about this select all and before we we run into similar problems let us insert some more pages. So, now you see this integral and think of this as a normal integral all the more. So, because now I have a nice expression for psi t psi t is t plus order somewhere. So, let psi t t plus order square root t and then put another function of t as a delta of t then let us just first look at this integral going from 1 to x d psi t over log t this is equal to now d psi t I can with this expression for psi t I can write it as psi prime d t just divide and multiply by d t. So, now you take differential of psi t with respect to t you get 1 plus order what happens here you get t to the minus half delta t plus t to the half delta prime t now look at the first one what is that d t over log t integral from 1 to x that is x over log x all right is very close to being x over log x how do you show this should be simple enough d t over log t integral does not this have a close formula integral of 1 over log t we will tackle that later let us look at the error part what happens here. So, this is you may be just to make life easier let me stick a just keep it delta t and let us assume for the moment delta t is log square t which comes out of the Riemann hypothesis then your 1 to x t to the minus half delta t this equals delta t is log square t. So, t to the minus half log t delta prime t is 2 log t by t. So, this becomes the integration by parts because there is a 1 sitting here when there is where log t becomes 0 it diverges it becomes little funny integral to handle, but there is some I am very simple trick which allows us to do that what happens here. So, this delta prime as I said was 2 log t over t. So, this becomes log t. So, that is 2 t to the minus half delta prime was 2 log t over t. So, this becomes log t. So, this 2 t to the minus half clearly the first is bigger than the second one and since we are bounding the error we can ignore the second one and then what is 1 to x log t over t square root t d t what is this we are again in similar situation may be not this we do integration by parts because here log t is up integrates 1 over square root t. So, what you get is root t by half you get 2 root t log t 1 to x same thing and log t differentiate when you differentiate you get 1 over t. So, you get 2 by root t d t right. Now, you see that this is of course, again if you integrate this you get the bar root t which is again dominated by this and this is basic all of it is order x to the half. So, that is what the error and the main part it should turn out this is basically where is it this is x by log x plus some small order term which gets absorbed into the error. So, it should actually what I want to see here is basically x by log x plus order square root x log x that is what should come out. Model of that integral like maybe I can leave that as an assignment to you and yes. What is it? So, because see it is saying it is less than equal to this right. So, actually the other integral by parts actually works it gives you half x y log x because the second part after the minus sign you will get back exactly the same integral it will be y by y. You get the same thing. It is twice the same integral equals. Mind you should get half x y log x should be x by log x, but your point is good yes why have I replace psi by this order? It is actually not true and we cannot use yes psi is this t plus order this, but I cannot then therefore write psi prime is 1 plus order this derivative. Because all this is saying is that psi is in this band grows like t and then there are fluctuations within this given by this error those fluctuations may make the derivative of psi very large depending on how those fluctuations happen good point. So, I think this whole derivation is wrong. I will need to think about it then I thought I can do it in this simple way, but whatever eventual derivation is going to throw up this x by log x plus order square root x log x and that is equal to this left hand side which is pi x plus half of pi square root x plus one third of pi x to the one third. Now, what we certainly know is that pi what is pi of square root x it is at most square root x pi of x to the one third is at most x to the one third. So, whatever the pi x is the rest of the sum just gets absorbed into the error term order square root x times something 3 times square root x 4 times square root x that is all just get sucked into the. So, left hand side is pi x plus order square root x right hand side is x by log x plus order square root x log log x. So, this is equal to pi x plus order square root x the right hand side this will be equal to x by log x this is assuming Riemann hypothesis and in general if you do not assume Riemann hypothesis you will get delta x. So, whatever is the assumption about psi x for you that it is order x to the half times delta x you just divide by log x this is what the error term you get and just put these two together you will see that pi x equals x by log x plus order square root x delta x by log x. Now, delta x is log square x as given by Riemann hypothesis then you get square root x log x. So, that is the relationship between these two. So, the only thing I need to now sort out is how do you show this for the right hand side.