 So, let us estimate the error term first thing today we had this error term. So, some n going from 1 to infinity. So, here we have a pretty good handle on these values this lambda n is bounded by log n at all ways this x to the c r n to the c these are all very clear this is the slightly problematic one log x by n and the reason is that as n goes from 1 to infinity when n comes close to x then this can become very very small and therefore, the whole thing become very large. So, we have to handle that aspect little carefully otherwise bounding this is pretty straightforward. So, in order to achieve that let us split this sum into 3 components n equal to 1 to x by 2. So, everything reasonably below x plus 2 3 x by 2 n plus n greater than 3 x. So, the first and third are pretty straight forward to estimate just let us see that. So, here since n is always reasonably smaller than x the largest value that log sorry the smallest value that log x by n will take is a constant when n is exactly and x by 2 this will be log 2 which is a constant. So, denominator we would like to give make as small as possible. So, this is certainly order n going from 1 to x by 2 lambda n x to the c divided by r n to the c and this we can just write as x to the c by r summation n going from 1 to x by 2. Well, lambda n we can replace by log n at the most or make it even simpler let us take out log x lambda n always less than equal to log x. So, take that out also 1 by n to the c and this sum is what well c remember is greater than 1. So, when c is greater than 1 this sum actually converges actually n going from 1 to infinity 1 by n to the c converges to some constant value. So, we just this similarly if you look at this. So, here again since n is substantially bigger than x always. So, log x by n would be substantially as n increases log x by n will become a larger and larger negative number but larger and larger and here we are taking with absolute values we need not worry about positive negative. So, again it will attain its minimum when x is 3 by 2 x and again it would be a constant. So, therefore, we can again write this as and now here 1 has to 1 cannot substitute lambda n with log x when you keep on increasing to infinity. This series the convergence will depend on exactly what is the value of c is that right lambda n as most log n. So, when would this converge log n that is what I am saying is smaller than any n to the epsilon. So, this is going to converge anyway in does not matter which way you take as long as c is greater than 1 this is going to be order x to the c by r for any c bigger than 1 that is fine. So, that leaves out only the middle term which is this sum and this we have to handle slightly more carefully. Now, here the log x by n sort of is within a small range of values but the problem is that it can become very close to 0. It may become very close to 0 and then there is that is not so nice. So, to handle that let us do the following replacement. It is again simple trick and will easily you know can write this in a more sensible form. If I just say x by n to be let us see what I want this goes between x by n varies between 3 by 2 to half right. So, what I want here is 1 minus z or let us z I am using elsewhere. So, let us 1 minus s. So, what we get is minus half is less than s less than equal to plus half as x n goes between x by 2 to 3 by. So, that is x by n or x by n varies from 2 by 3 to 2 make n by x. It does not matter actually here just by changing the sign you can replace log x by n but will by log n by x. Now, also for this one this once you do this log x by n is same as log 1 minus s and log what is log 1 minus s that is a standard power series for this which is s plus s square by 2 plus s cube by 3 plus and so on right. And now we use the fact that s is in absolute value at most half. So, this is equal to s plus let us just say this sum the of the rest of it is not going to be very large actually. Let us just write it as s by 2 plus square by 3 plus so on. And since s absolute value at most half this will be at most first term will be at most 1 quarter 1 by 3 1 by 12 1 quarter plus 1 by 12 if you take absolute sum which is going to converge to some constant in fact very small constant also I guess it will be less than maybe one actually not I guess I am sure it is going to be less than 1 because this is actually worse than the geometric series. So, this is less than equal to 2 s oh is that what I want no I want I want the other way I want I want to lower bound on this. So, this would be greater than equal to well if you just take s and plus this much the rest cannot reduce it. So, this would be so take the first two terms fine this is square so it is always positive. So, s plus s square by 2 that is positive and the rest can suppose it is all negative and try to subtract out as much as you can you cannot probably you cannot subtract out more than half s from there because this sum of all of these would be at most half s. So, I am leaving out some details for you to work out whether they are pretty straight forward. So, basically what you get is an s is at most half. So, this would be this s could be 0 also so let us say at most half s at least half s. So, now with this we can write this as order summation take this out x to the c by r summation x by 2 less than n less than equal to 3 by 2 x lambda n by n to the c and log x by n is replaced by half of s which is half goes away. So, s is an s is x minus n divided by x this is not quite what I wanted y is an x coming there but anyway seems I cannot of course that is right. Now, n is at least half x. So, this is n to the c I can replace by x by 2 to the c c is a constant 2 to the c goes away. So, it is x to the c below and c is greater than 1 which cancels out with the x above. So, this is at most and this is log x lambda n since n varies between x by 2 3 x y 2 this is log x by 2 x minus n and as x minus as n varies from x by 2 to 3 by 2 x what happens to x minus n. Yeah let us assume x is half integral this is the it is never 0. So, what happens to what x by x minus n will go from minus x by 2 to plus x by 2 right or so we can fold up. So, and it will take half integral values there which is again we can take it to make the interval a little larger and make it take integral values because you can always multiply out by constants and absorb it in it which means it is something like a sum of 1 plus 1 by 2 plus 1 by 3 up to some 1 over x y 2 over 1 over x and that sum is. So, if you just go back to this it is order x to the c by r the log x here and this sum itself is log x. So, that becomes log square. So, just adding all three what you get is now this error is a bit can be a bit too large if c is c is 2 or 3 then the error actually is much bigger than your main term recall what was the main term psi this is the main term right whatever it is psi x what is and this estimate psi x psi x is this plus the error term psi x would be how much at the most x log x can not be more than x log x in the most silly counting you will give you x log x. So, if your c is a little more than say 1.1 then the error is already going through of course, assuming r is fixed number. So, that we do not want. So, let us fix once and for all the value of c which is going to be very close to 1 I cannot make it 1 unfortunately I cannot make it 0.1 kind make it 0.0001 because any of this will just be too much. So, I have to make it a function of x it has to because it has to be bigger than 1 and yet not be a constant. So, this is what I am going to fix 1 plus 1 over log x and then what you get let us look at this psi x is 1 over 2 pi i now you have integral from c minus i r to c plus i r minus zeta prime z by zeta z x to the z by z d z plus order what is x to the c for this value of c that is pretty simple what is x to the c. Last error term there would not be x to the power c because we have cancelled power c by denominator and the minimum value of n can take is x by 2. So, that will cancel out I got it yes you are right. So, it is quite does not matter I am saying even if you take s to x c because I am going to choose a c to be 1 plus 1 over log x. So, what does this make x to the c equal to is e times x which is just a order x. So, x to the c is order x. So, basically you get x log square x and this is you know a pretty good starting point that I have managed to write psi x which is kind of a cousin of the prime counting function as a complex integral plus an error term which is well which is still too big if r is small of course, no the nice thing is that r is in the denominator. So, if I can always send r to infinity and make this vanish completely and then the error term does not matter also. In fact, then c the choice of c is also does not matter can be any constant, but as we will see later on it is more beneficial to not send r to infinity instead set r to about square root of x, but that we will come to later, but for now let us just keep in mind that we will not be actually sending r to infinity. However, if we do send r to infinity what happens then error vanishes and psi x is. So, this is certainly a corollary of this that psi x equals 1 over 2 pi i is precisely this integral along this line minus infinity. But as I said this is only for fun we will not really be using it because in any case as I said we do not expect to sort of precisely compute a value for psi x a formula for psi x because that is although we will compute a precise formula for psi x later on, but that is not going to be very useful in terms of giving us a good estimate. To get a good estimate we will actually use a finite value of r. So, that is the setup we have psi x we have a complex integral which tells us what psi x. So, the next question is how to be evaluate this complex integral and here I will revert back to C plus i r and C minus i r because this evaluation is what is going to determine the psi x for us. Now, the nice thing is that we know how to evaluate this integral at least we know how to go about evaluating this integral we already done this we already have tools for this. What we need to do is make a contour or make a domain whose boundary one of the boundary is this C plus C minus i r to C plus i r close that make a contour integral around that boundary and count the poles inside that region that is the value of the integral and then estimate the value of the integral on the remaining boundaries and somehow set it up. So, that those integrals are very small so that at the worst they add up to some error term not and not substantially and then we will have our value of psi x. So, let us attempt that. So, here is the let me write the integral also C minus i r to C plus i r minus eta prime this is the integral and we have one here and very close to one we have C we already chosen that minus r plus r and we want to integrate from here to here this integrate. So, what is the contour that or what is the domain that we should choose for this? We can extend it this side we can extend it this side we have seen examples of both actually, but if you recall that example that there was in that example x to the z by z was the integrant and depending on the value of x we either went left or we went right and the reason was that if x is less than 1 then you need to go on this side positive side because when you want to bound later on those integrals on the remaining boundaries. If x is greater than or x is less than 1 then on those boundaries you can make z to be very large positive and large. So, x to the z will become very small right and as you you know expand those boundaries that will vanish. On the other hand if x is more than 1 then this strategy will not work because as you go further and further on the positive direction x to the z will keep blowing blow up. So, when x is positive you want to go on the negative side because when and then large then when z takes large negative values again x to the z will become very very small and again you can try to bound it away bound it to 0 or small. So, what this tells us is that we cannot hope to go on the right of this line to integrate this we should try to go to the left of this line. So, the moment you conclude this there is a problem what is the problem. So, let us define this domain. Again possibly the most simple domain that is possible exactly as we did in case of the delta function there is a rectangle going all the way up to minus u here coming down and this is how you integrate along this counter clockwise direction. So, that is the domain D, but as I said we immediately run into the problem what is the problem look at the integrant what is the integrant zeta prime over zeta x to the z by z is the integrant defined on the domain. You want the integrant to be analytic on the domain otherwise nothing works right that is an essential requirement. Now forget the being analytic is it defined on the domain worse. Is that the part of the problem? Yes that is certainly true, but it is even worse x to the z is defined in everywhere 1 by z is defined everywhere except for the pole at z equals 0 that is pretty well understood. What about zeta z? zeta z is this infinite sum where does it converge only for absolute value of z greater than 1 otherwise this infinite sum does not even converge. So, which means this is a line z equals 1 to the left of it including the on the line zeta z is not even defined. So, nearly on there is just this tiny strip to the right of it where the zeta z is defined everywhere else is not defined. So, there is really serious problem there. By the way do you know or do you believe that zeta z actually converges for absolute values at greater than 1? The real z greater than 1 not absolute value greater than 1 real z. So, at least we need to know or we need to ensure that it converges on the right side of this line. So, are you convinced that it does? We did oh then it is perfect. Anyway it is pretty simple just take the absolute value of every term and then you say it comes out to be bounded by a simple geometric integral and then the integral converges. So, zeta z is defined everywhere here, but there are problems on this side. Going further is fine zeta z is defined on the right hand side, but is it analytic on the right hand side on every point? How do you prove it to be analytic? This is an assignment. It is a reasonably straight forward proof just use the tools that you have learned. How do you prove something analytic and apply those tools? So, now we will believe that zeta z is analytic on this half plane which is on the real z greater than 1, but it does not even exist seemingly at least on the other half. So, this approach appears doomed except that we know that an analytic function can be extended beyond its domain. We have seen some examples of this and we know that it can be extended in a very nice way unique way actually. So, maybe zeta has that property there may be it can be extended it is not defined through this infinite series, but maybe there is a way to extend it on the other half. Let us explore this. There is no harm. So, let us write zeta in a slightly different form. This I can write as let us see let us make another attempt. Do it look at a partial sum. This I want to write as integral. So, the key thing is to translate it into integral because that is easier to manipulate and just trying to work out what is should be the form of the integral plug in this integral. So, how far away is this sum from this integral? This integral I can write as equal to sin and regular series is all together. This is equal to summation n going from 1 to x integral y going from n to n plus 1 and n going from 1 to x minus 1 actually. And here you will have n divide by y to the z plus 1 d y and this is equal to what is the integral value? This is dy over y to the z plus 1 is equal to what dy over y to the z plus 1 integral of that is y 1 over z y to the z and a negative sign somewhere and this n to n plus 1. And this is equal to what? Now, what is this sum equal to? We want to expand this. This is equal to n equals 1. What happens? 1 by 2 power z minus plus n equal to 2 2 by 3 power z minus 1 by 2 power z minus 1. Is that right? This does not seem very good. No, these are not very good. Maybe it is not too bad. Now, look at this and this. You get this gives you 1 by 1 power z minus, of course. What does these two give you? What about these two? What about these two? Which is exactly what you want? Except that at the end, the last one, this guy will survive. So, what I get is minus 1 by z and here minus can go away because that will make all the rest of it positive. So, sum n going from 1 to x 1 by n to the z, I think x minus 1 or so. And then the last guy survives which is n which n which is n equals x minus 1. So, that is x minus 1 divided by x y z. So, now, let us get back to this. Good. So, now, we can write using this. What I can say is as equal to x minus 1 over x to the z plus z times this integral y going from 1 to x. And now, sent x to infinity. I should not have chosen x here because x I am using for something else. This is a different x not to be confused with x parameter x of psi. So, send x to infinity here. What happens? The left hand side becomes the zeta function. What happens is the right hand side. The first term, see absolute value of z or real part of z is greater than 1. And because of that, this the first term goes to 0, x as x goes to infinity. So, what I get is zeta function equals this integral. That is nice to know although it does not solve our problem yet, but it is nice to know that there is a not so nice integral that I can write zeta function as because it is not so nice because there is a this floor of y sitting out as integral. But anyway, we are not really integrated interested in getting a value for zeta z. We only want to see if it can be extended beyond its defining domain. And that we can use this integral for. Well, how do we do that? Let us just write this 1 to infinity y minus bracket y, where bracket y is just the fractional part of y. Now, the first integral can be evaluated. That is equal to minus 1 over z minus 1, 1 over y to the z minus 1, y goes taking value 1 to infinity. Now, when y is 1 and infinity, of course, it goes away. The infinity at in pi equals infinity, this is 0 because z is z minus 1 has a real part, which is bigger than 1, bigger than 0. And y equals 1, this becomes of course, all of this is for real z greater than 1. And let me just write zeta z. So, this is also important equation. Now, let us look at the right hand side. Right hand side is an analytic function. Why is it analytic? Well, this is of course, analytic except for z equals 1. And the integrand here is bounded. Well, of course, I have to be careful here. This part, this is analytic for real z greater than 0, z not equal to 1. For z not equal to 1, there is this does not diverge. So, the first term is fine everywhere for real z greater than 0. If you look at this integral, as long as real z is greater than 1, sorry real z greater than 0, y to the z plus 1, the real part of this is bigger than y equally y to the 1 point something. And the fractional y is bounded by 1. So, this integral converges absolutely as long as real z is greater than 0 to some finite value. For every value of z, it converges some finite value. And it is of course, the dependence on z is analytic because the only dependence of z is y to the z plus 1. 1 over y to the z plus 1 is an analytic function in z. And we have seen that when you integrate an analytic function, actually it is both analytic in z as well as in y or for every value of y is analytic. And therefore, its integral is also analytic in z. And it converges for every real z greater than 0. So, the right hand side therefore, is an analytic function which for real z greater than 0 and z not equal to 1. The left hand side is an analytic function for real z greater than 1. And these two coincide on real z greater than 1. Now, just invoke the uniqueness of analytic continuation. This right hand side is the analytic continuation of zeta z. For of course, this is still not fully done. This is 1. And we had zeta z analytic here. And we have the right hand side analytic also in this strip except of course, this a whole at 1. So, it has allowed us to extend the definition of zeta z slightly to the left. Not all the way, but at least slightly to the left. So, that gives us hope that perhaps by doing something more we can extend it all the way to the left everywhere which is what we will do in the next class.