 So, we were here last time we were trying to estimate this sum which we used the Cauchy Maclaurin formula to estimate to be this. This we treat as error term and we estimated the error to be this bounded by this quantity. So, what we can do is therefore, we can write this this sum to be equal to this minus this plus this and this is also sort of error term in the sense that 1 over z plus 1 1 over z plus n this also 1 over z plus 1 1 over z plus n. So, that is like where are we log plus order and that is what we put as error term. Now, going back to this original estimate you remember what we were trying to actually do is to estimate gamma prime over gamma where z is in this interval and we derived that in this interval gamma prime over gamma and absolute value is bounded by absolute value this quantity and this we just now estimated almost we estimated this by that quantity. So, therefore, this is equal this is bounded by order 1 plus order 1 over z plus n plus log of z plus put this in absolute value you agree with this is the first sum inside the absolute sign is bounded by this plus a little bit more that is how that sort of absolute sign that is bounded by this and the last one sum over n of 1 over n is about log of capital N plus this Euler's constant which is order 1 to that goes out it is important to keep the signs here because as we will soon see this is order 1 plus order we just take these two we get log 1 plus z over n. So, therefore, that value of gamma prime z now we can come to this that is limit capital N going to infinity of log of right and what is this as n goes to infinity this vanishes this also vanishes and what we are left with is plus order now z plus 1 in absolute value is certainly bigger than 1 right for the z that we are interested in right now. So, this is all order let me just mention this for z in the interval minus u plus a there we have it a very nice bound on gamma prime over gamma now if you go all the way back what were you trying to do we are trying to calculate the integral along this line and what was inside the integral zeta prime over zeta. So, let us just go back and recall whatever we have left behind and we have already seen that zeta prime over zeta in absolute value is bounded by what quantity we use that functional equation for zeta do you remember I will have to pull that out then plus right plus order 1 this is what we had using the functional equation taking the log differentiating and model taking modulus everywhere you get this. Now, for again z in this interval what we observed was that zeta prime 1 minus z over zeta 1 minus z is order 1 why because 1 minus z is as a very large real part and zeta prime as well as zeta on very z which has a large real part converges to some value which is bounded by a constant. So, that is taken care of what about gamma prime of 1 minus z by 2 or gamma 1 by 1 minus z by 2 again this will have a very large real part, but this is gamma prime over gamma we do not really have an estimate here except that we can just invoke this just obtain this bound right and this part is valid if you recall for all z which are which do not lie on the negative real axis for all other z this equation is valid. So, in particular for 1 minus z by 2 will certainly not be lying on the negative real axis when z real part of z is negative. So, is real actually there on the positive side. So, we can just throw this bound at it and now again real part of z is very large is actually something like u right u plus 1 by 2 actually 1 minus real part of 1 minus z by 2. So, again this will go vanish away this will be order 1. So, actually the same bound applies here we can just go back and is log of 1 plus z by 2 this is what we get from here gamma prime z over gamma z is this and here we have gamma prime z by 2 and then we have gamma prime 1 minus z by 2 which is log of 1 plus 1 minus z by 2 plus order 1. Now, if you recall this log of absolute value of log z is what z is if you write in polar coordinate is like r times e to the i theta varies between minus pi to plus pi is not never equal to for the z of interest is never equal to minus pi somewhere. So, the absolute value of log z would be log r plus log square r plus theta square and the whole school square. Now, theta is a constant and whenever log of we are taking log of z theta is a constant when if we are taking theta of log of h of z for some function h then theta may not be a constant because that really depends on how much rotations does h give to z. But if it is a z or 1 plus z by 2 they do not give any rotation these are linear shifts. So, this is essentially therefore, either in either case we can write this as order log z and then the when we try to bond this we get less than equal to integral zeta prime over zeta is bounded by order log of z x to the z in absolute value is bounded by is actually exactly x to the minus u absolute value of x to the z. And absolute value of z is at least u. So, since it is in denominator we stick that u there this and this is equal to again now this everything is simple. In fact, absolute value of z we can put an upper bound which is square root of u square plus r square and then take everything out the integral is just 2 i r. And so we just get order r log of r square plus u square and this is perfect for us because remember our target was to send u to infinity and when we send u to infinity this vanishes. Remember x is bigger than 1 right x is x was the number of x was that number. So, that we want to count prime numbers less than equal to x is always going to be bigger than 1. So, x to the u sitting in the denominator u sitting in the denominator in the numerator is only log u. So, that is good now this takes care of this integral now we took at this well these 2 are cousins. So, if we take care of 1 we will take care of the other because this really is symmetric. So, we have to take care of this now let me go back and draw this because this is where the fun begins. So, this is the domain of that we have and let us mark out now and this is c. C if you recall was 1 plus 1 by which is less than 2 right. So, the integral here is starting from here it is from c which is a number slightly bigger than 1 going all the way to minus u and stays the imaginary part stays at higher. Now, as we traverse around here and remember keep in mind the what we are integrating we are integrating zeta prime over zeta times x to the z over z. So, as we traverse on this side x to the z over z is very well behaved, but zeta prime over zeta is not at all well behaved because there may be lots of zeros sitting here right. So, we need to not only avoid those zeros, but also we need to worry about even if we can avoid the zeros zeta prime over not the zeros these are poles sorry not only need to avoid the poles. We also need to make sure that the path we take as zeta prime over zeta does not become too big because if it becomes too big then we cannot make the error term small. On the other hand once we are cross this boundary and we are on this side we know there are no poles and actually we know we are far away from any other pole because the remaining poles are here. So, these are just good reason for us to integrate and further can be may be a little unsure about this part because there may be a pole very close by and by the time we have come here we are still not right if we have still not got rid of the effects of that pole. So, what we will do is we will split this integral into two parts one is from here to here C 2 minus 1 and second part is minus 1 to minus this 1 be this will be easy to handle. In fact, I will handle it in 2 minutes using the tools we already developed this will be not so easy to handle. So, we will then want to handle this one. So, what about this one? So, if you look at integral of minus 1 minus i r 2 minus u minus i r zeta prime z over zeta z x to the z by z the absolute value this is bounded by again we take the modulus inside how do you bound zeta prime over zeta again recall the functional equation just go back this holds why does this hold we will be little more careful z is less than equal the real part of z is less than equal to minus 1 that is another reason for you know going minus 1 and below. If real part of z is less than equal to minus 1 what about the real part of 1 minus z we are at this equation real part of 1 minus z is going to be greater than equal to 2. And when it is greater than equal to 2 well zeta prime of 1 minus z over zeta of 1 minus z is bounded that we know anything above 1 is bounded. So, for 2 it is surely bound. So, this goes away. So, we therefore, move in to this part zeta prime over zeta is bounded by this sum this plus this plus what about now what about gamma prime over gamma will be already derived an expression for gamma prime over gamma that is valid here the only invalid part was when negative real axis that is valid here. So, we can just throw this in and we still get order log of absolute values z right. So, doing all this we get this is order log of absolute value of z what is absolute value of x to the z this we cannot say it because it is now the real part absolute value of x to the z is x to the real part that is varying from minus 1 to minus u. So, x is in the denominator therefore, x to the z is in the denominator and we can always substitute with the smallest possible value where the denominator take which is x. So, let us just do that and again absolute value of z the smallest is r again again absolute value of z the largest value of this is square root of r square plus like. So, we just take that in and this is going from minus 1 to minus u I cannot afford to do that. So, I have to 1 over r is fine, but I cannot afford to have replace x to the z by x because then this is going to diverge. So, what I am going to do is keep that x to the real part of z and this is less than equal to log of absolute value of z as I already said it is we can bounded by order log of r square plus u square divided by r and then we have integral. Now, in the integral now the only thing survives is x to the real part of z. So, we are essentially integrating over a real quantity because we are moving from here to only real part changes. So, in these are also you only move the real part. So, basically this is 1 over x to the t dt and what is this integral this is simple to estimate integral of this is going to be still not good enough because this integral is going to be probably is that is e to the minus t log x and integral of that is going to be for minus t equals minus 1 t equals 1 is 1 over x log x and that is what worries me. So, if as u goes to infinity this vanishes what this diverges. So, I made a mistake somewhere this is fine maybe I should not have replace this absolute value of z by r we will never send r to infinity we will at some point, but only for a short while we will actually be setting r to be in square root of x. Now, we do not want to find the integral on the whole line see no if you recall for any value of r we had if this integral related to psi of x with some error term and our target is always going to be to minimize error term. So, we will choose that value of r which minimizes error term and as we will see later on that that value of r will come to about square root of x, but that is a story for future how do we end this story zeta prime over zeta is bounded by order log that seems like a pretty reasonable conclusion. So, how do we take care of this I think clearly my bounds have been oversimplification. So, one needs to find a slightly better way of handling it maybe I should do that instead of in fact, even in log mod z I should just replace it by log of root of r square plus t square and then do the integrals. So, let us just go ahead and do that. So, I do not want to end up here. So, I am going to erase all of this let us go back and here we replace this by take order outside absolute value and now t we are going to send 1 to u what is order log z it is log of r square plus t square in the denominator we get x to the t and for mod z we get r square plus t square. Now, how do you integrate this can one find some easy estimates I think so see what we can say is surely this is order 1 to u and notice that log of this quantity is upper bounded by this quantity to the epsilon for any small epsilon and there is some constant multiplier that gets absorbed in this order. So, we replace or this by this upper bound which then gets absorbed in here. So, we get dt over x to the t r square plus t square half minus epsilon now integrate this how do we integrate this all we can try many things because it is again we just have to worry about some upper bound to this do not have to worry about exactly evaluating this integral. So, what would be a good way suppose we throw away r square here this is still remains an upper bound and then t square becomes t to the t square to the half minus epsilon becomes t to the 1 minus epsilon. So, t to the 1 minus epsilon divide by x to the t if you integrate what do you get let me say or let me keep r square that might be useful later on. So, if you integrate this by parts let us say t square equals u I can do that r square plus u, but then this becomes there is a square root u also which logs in some way and this becomes x to the square root u that is now that is becomes. So, yes of course, throw away t square t is always greater than 1. So, if you throw away t square from this denominator here you are always going to get an upper bound and now this is an integral that is like 1 by log x x to the t r to the 1 minus epsilon when t is 1 you get 1 by x log x r to the 1 minus epsilon and when x is u you get 1 by log x r to the 1 minus epsilon. And so, this is always 1 plus x log x right anyway this will go to 0 and good. So, that gives us a good estimate of this integral this part of the integral which is also same as this part of the integral. So, it is no difference. So, what is left is the messy one how do we get this estimate because in here that whatever we are trying to do using this equation it does not work anymore because the key thing here is that we cannot get rid of this quantity we know how to handle these two these are well behaved in that region as well. So, we can again plug in log z for these two. But this is not well behaved when real z is less than negative 1 real part of this is less than 2 well for a while you can still come up to you can still keep this bounded, but as real z becomes greater than 0 you do not really have any control over. And therefore, we need to we can only express this quantity in terms of some order log z times plus this quantity. But that really gives us nothing because you want to get bound on zeta prime over zeta you get it in terms of another zeta prime over zeta. So, it is not useful. Therefore, we seems that we cannot use this functional equation to bound zeta prime over zeta in the region of interest for us which is unfortunate because we spend a lot of time in trying to understand this. But you have to work with what you are given you cannot use it. So, you cannot use it. So, we have to use something else. So, what is that something else? So, let me first write we cannot use the functional equation. So, as I said we use something else and that something else fortunately is not something totally new we can again reuse the tools we have developed. But we have to approach it slightly differently and the approach we are going to adopt is define another function which we call xi. So, the definition of xi is quite simple if you look familiar to you also if you recall this the last three quantities in this product occur in the functional equation. In fact, xi is motivated from the functional equation. If you further if you recall that that we had pi to the minus z by 2 gamma z by 2 zeta z to be 1 over z times z minus 1 really have to remember what it was plus I mean there is a minus here there was a 1 to infinity t to the z by 2 plus t to the 1 minus z by 2 that is really long term when we were trying to prove that or trying to analytically extend the zeta function over the entire complex plane. So, this is the equation we use to extend the zeta function is this was this equation oh yes d t by t that is something I missed and was assigned plus good. So, this is what we had. So, therefore xi of z is 1 plus. So, let us spend a couple of minutes on understanding xi of z first is it defined over the entire complex plane. Yes, may be except for some may be some poles it is defined over entire complex plane because zeta z is defined over the entire complex plane except for 1 pole gamma z by 2 is defined over the entire complex plane except for those infinitely many poles and this is of course, these are defined over entire. So, xi z is defined over entire complex plane is meromorphic function. So, it may have some poles. So, are there some poles of xi appears where are they is z equals 1 a pole which is a pole of zeta z no why not that z minus 1 takes care of it is any of the negative integers a negative even integers a pole minus 2 is minus 2 a pole for example, no true because that is already taken over care of by 0 of zeta function at minus 2. So, that only leaves out 0 z equals 0 is z equals 0 a pole because that is where this is bounded, but this is unbounded no because that z is taking care of it and these poles are all order 1. So, actually we can therefore, conclude xi is an entire function in fact that was the reason of multiplying out this with z and z minus 1 to take care of the remaining poles xi is an entire function which is very good. And another nice observation which we will be using anywhere, but it is still good to know that xi is perfectly symmetric along z real z equals half line xi z equals xi of 1 minus z that follows because it is really is a functional equation and u z times z minus 1 is also symmetric. So, as I said we are not be bothered about the second observation, but the first one is going to be important because we know about entire functions. What we know about entire functions is that we can write the entire function in terms of their zeros. So, we will have to worry about where the zeros of xi are. So, that is another third part of the observation where are the zeros of xi gamma has no zeros we already know. So, the potential zeros of xi are z equals 0 that is not a 0 because there is a pole which cancel out z equals 1 not a 0 pole cancel out zeros of gamma trivial zeros of gamma sorry trivial zeros of zeta not zeros because they cancel out with the poles of gamma. So, the only zeros of xi are precisely the non-trivial zeros of zeta and that is that and on those any of the non-trivial zeros these are all quantities which are bounded and therefore that remains 0. So, now you see the usefulness of xi it is entire function. So, we can use that product formula to write xi and in that formula all the non-trivial zeros of zeta will occur. These are the zeros that we are worried about because these zeros give rise to those poles of zeta prime over zeta that we need to avoid in order to bound the integral. So, now we will study xi to derive some facts about those non-trivial zeros of zeta. So, that is going to be our next task. So, we are done for the day.