 So, this is where we were last time we were trying to evaluate this integral which is not too difficult to fix now. So, let us continue here see what happens to the first integral that is basically half of V to the minus half minus z by 2. So, this would be V to the half minus z by 2 divided by half 1 to infinity right. Similarly, what happens to the second integral something similar V to the minus z by 2 and then here we get 1 to infinity this of course, we do not integrate we will just write it as. So, this is just a slight rewrite of the same thing and now if we look at the actual values here what do we get well here of course, while evaluating this integral we have to assume that for example, real part of z is bigger than 1 because otherwise if real part of z is less than 1 then when you set V to infinity it diverges which you do not want. So, under the assumption that real part of z is bigger than 1 when you set V to infinity this is 0 when you set V to 1 then what do what you get is 1 over z minus 1 fine. Similarly, here we need to assume that real part of z is bigger than 1 sorry bigger than 0 which is anyway fine because we have already assumed real z is bigger than 1 when V equals infinity this goes to 0 when V equals 1 then this becomes lots of minuses here this should be right and plus this integral. Now, let us switch back I guess several of these before. So, this is what the expression for zeta was zeta z pi to the minus z by 2 gamma z by 2 is this integral which we have split into these two parts and this is the part that we just managed to rewrite as that long integral this one fine. So, let us just plug that value in here and see what we get equals stick that in which is 1 over z minus 1 minus 1 over z this is actually also equal to 1 over z z then plus this 1 to infinity V to the 1 minus z by 2 and in addition this part which we have not touched and we will just copy this part as it is which is again changing the variable name from u to v we get v to the z by 2 w v. So, this is the result of all this hard work through Fourier analysis and what not, but what we get out of this is something very interesting see the right hand side first thing to notice in the right hand side is not obvious, but once you realize it becomes obvious is that the right hand side is invariant under the substitution z by 1 minus z the first term is 1 over z z minus 1 you replace z by 1 minus z z minus 1 becomes minus z and z becomes 1 minus z. So, that is again same what if you do the same thing in this integral v to the 1 minus z of become v to the z by 2 nothing no z here which and v to the z by 2 become v to the 1 minus z by 2. So, these two integral just swap with each other. So, this implies that so if right hand side is invariant of course, the left hand side is also invariant under the substitution and which means and what does this show us well we know the definition of zeta z for all z with real part greater than 0 real part greater than 1 it is defined by that infinite sum we did little bit of playing around with it to define it also between the strip 0 and 1 and now the question that we started with was how to what definition do we give to zeta z 1 z has negative real component. Now, stick a negative real component to z what do we get say zeta of minus half well pi to the minus. So, let us say z is just take an example suppose z is minus half plus i t from complex number with real part minus half then zeta of minus half plus i t pi to the minus i t divided by 2 this equals zeta of 1 minus z 1 minus z is what of minus i t. Now, this is well defined pi to the whatever power is well defined gamma of minus quarter plus i t by 2 is well defined if the gamma is well defined on any non integral negative number. So, this is well defined gamma of half minus i t is well defined pi to this minus is well defined and gamma again gamma of quarter minus half i t by 2. So, that gives me a definition of or a value to gamma minus half plus i t. So, in fact if you just think about it for a moment this is 0 and you take the line let us say this is 1 this is half this relationship which I just derived this relationship can be used to define the value of zeta at any point here using the value of zeta at this point which is just a reflection around the line z equals plus half real z equals plus half right because gamma z or sorry zeta z is defined in terms of zeta 1 minus z which is a that is a symmetric quantity with respect to z equals half line. And since we know what the value of zeta is on this domain we just extend the same thing to get the value of zeta on this domain. So, that is that is that is it I mean it is although this relationship if you think back that this integral we said when they evaluating this integral and sending this is 0 that real z must be greater than 1 which means that this relationship holds for real z greater than 1. So, there is a question here why when this relationship only holds for real z greater than 1 how can we conclude this equation that is because analytic continuity look at the right hand side it is easy to see that this is defined everywhere the right hand side except for 2 points at z equals 0 there is a pole at z equals 1 there is a pole because of this right. So, this gives poles at z equals 0 and 1 at all other points this term is defined what about the integrals see if you recall the definition of w v w v was sum n going from 1 to infinity e to the minus pi n square v. So, this infinite sum converges for every v and in fact it is easy to see that this infinite sum is going to decay very fast something like e to the minus v at least as badly as e to the minus v also because every single term there is e to the minus first term is e to the minus v second term is e to the minus 4 v e to the minus 9 v e to the minus 16 v when you add all of them up you probably do not get anything better than maybe e to the minus 2 v actually we will get something even less. So, this decays really fast sorry now e to the minus 2 v e to the minus v by 2. So, this decays really fast and this is a only a polynomial the key difference from the earlier case was that when the integral is from 0 to infinity and z real z is a big bigger than 1 here for example, this becomes 1 over v to the sum positive power and when v is 0 this diverges. So, this integral now is from 1 to infinity this part never diverges this behaves at most a polynomial either as in the numerator a polynomial or in the denominator a polynomial which does not matter I mean this is decays. So, fast this will converge for every value of z similarly for every value of z this also converge. These both are clear. So, these two converge on all z these two are converge on all z except for 2 poles at 0 and 1. So, right hand side is defined for is basically a meromorphic function x with 2 poles at 0 and 1 and it right hand side agrees with the left hand side on the half plane real is greater than 1. Therefore, by analytic continuity and its uniqueness we can analytically continue the left hand side and make it equal to the right hand side on the entire complex plane which allows me to extend the definition of zeta z all over because I know pi to these a minus z by 2 and gamma z by 2 are defined everywhere except for some poles. So, this equation itself guarantees that zeta z exists on all points except for some few poles which we will see in a moment and then we can use this relationship to derive this relationship between zeta z and zeta 1 minus z which gives us a nice way of calculating zeta z on the negative side using zeta values on the positive side. Therefore, have this called the functional for zeta and this is as I already said this is very important equation which will allow us to talk about zeta values in different context and we will keep coming back to this equation. It also very nicely relates zeta function with the gamma function. Now, coming back to this well the same equation or in fact if you go back to the previous one which I should write as like this and I already observed that this has exactly 2 poles at z equals 0 and z equals 1. So, let us using this let us find out all the poles of the zeta function over the entire complex plane because this allows me a definition of zeta function over the entire complex plane. We already know zeta z has a pole at z equals 1 and in fact that is a simple pole of order 1 and on the right hand side also we see a pole of order 1 at z equals 1. So, this seems to be fine you know that gamma z by 2 that is gamma half does not have a pole is not that does not diverge. So, it is everything is perfect what about z equals 0 the right hand side of the pole at z equals 0 does zeta have a pole at z equals 0 zeta 0 should be zeta 1. That is only for gamma function yeah exactly see gamma has a simple pole at z equals 0 and that matches with the simple pole on the right hand side therefore, zeta does not have a pole at z equals 0. If I cannot even have a 0 at z equals 0 because if it had a 0 at z equals 0 that will cancel out the pole at of gamma on the left hand side and therefore, you will not have a pole on the right hand side. So, this is all the conclusions from this are zeta z does not have a 0 or a pole let us move back these are the 2 poles on the right hand side which cause the poles or 0 on the left hand side poles actually on the left the right hand side has no other poles. So, which means whatever pole zeta has if at all it has they must cancel out with 0's of gamma, but does gamma have a 0 anywhere on the negative complex plane. In fact, if you recall that functional equation for gamma gamma z is equal to gamma 1 plus z over z. So, if gamma z is 0 then gamma 1 plus z must also be 0 which means gamma 1 2 plus z must be 0 3 plus z. So, that all points must be 0 all the way up, but at some point the real part of k plus z will become positive and then we know that gamma of that quantity is not 0. So, that actually said also that the gamma function is never 0 on the entire plane. So, if gamma function is never 0 on the entire plane it surely is not 0 on the negative complex plane and therefore, zeta has no poles on the negative. So, which means zeta has exactly 1 pole at z equals 1. Well, we can observe something more also here because gamma has some poles on the negative side. In fact, at all negative all negative integers it has a pole. It also has a pole at 0 you are right it has a pole at 0, but that is fine that pole already is taken care of, but it has a pole at minus 1 minus 2 minus 3 which means to get that pole in this equation we have to set z equals minus 2 minus 4 minus 6. So, all every negative even integer will cause a pole on the left hand side, but right hand side has no poles at those points which means zeta must have a 0 at those points to cancel out those poles and the order of what is the order of those poles of gamma say go to the order of pole of gamma at minus 1 it is order is 1. In fact, all of these poles have order 1 because the gamma minus 1 is gamma 0 over minus 1. So, it is inherits the order of the pole at gamma 0 and every subsequent one. So, all these poles of gamma are of order 1. Therefore, zeta must have a 0 at every negative even integer and these 0 must be of order 1. So, a simple 0 is same as a pole 0 of order 1 minus oh because there is z by 2 here is there any other 0 of zeta function on the let us say negative complex plane use what yes. So, use the functional equation there if there is a 0 on the say at z which is on somewhere on the negative complex plane then there will be a 0 corresponding. Now, gamma at those points is non 0 whether gamma z by 2 or gamma 1 minus z by 2 because we know gamma has no 0s and no poles also except for 0 minus 1 minus 2 which we have already taken care of at all other points at any other point if zeta z has a 0 then zeta 1 minus z also is 0. If that is negative so, that means if there is minus 1 here then this will become zeta 2 that zeta 2 is 0 and zeta 2 is 0 which actually pi square by 6. In fact, zeta z for any value of z with real part greater than 1 will not be a 0 because again that it is a convergent series and numbers is get added up except well there are amplitudes complex amplitudes which can subtract also. But the series sum one can argue a bit to see that it is never 0 on the side of real z greater than 1 which in turn means that it will have no 0 when real z is less than 0 is symmetric. Symmetric around real z equals half. So, if there is no 0 on real z greater than 1 there is no 0 on real z less than 0. In fact, that may give that as an assignment. So, by the symmetry well symmetries of values certainly holds for the 0s. If zeta z is half then zeta 1 minus z need not be half. But if zeta z is 0 then zeta 1 minus z must be 0 because of the way the equation is set up. So, therefore, since there is no 0 here 0 free region let us say that this is also 0 free except of course these minus 2 minus 4 minus x minus 8 and so on where the 0s are caused by the gamma function. So, apart from this 0s of zeta function and minus 2 minus 4 minus 6 minus 8 all other 0s if at all they exist must be in this region because gamma has a pole at those points. So, 0 and pole cancel each other out on the left one side and other side. So, the remaining 0s are in this trip between 0 and 1 and this is called critical strip of zeta function because the behavior of zeta function in this strip is something that we would like to understand why I will explain very soon. But that is where we have some very interesting behavior of zeta function. These 0s which are at negative even integers these are called trivial 0s of zeta function trivial because these are basically caused by poles of gamma function and they are not really of much interest to us. While they are of interest, but they do not really change the things in a any significant way as we will see later. So, with all of this we have at least achieved our one of the targets which was to define zeta function over the entire complex planes. So, that we can take a contour now I do not know if you remember the origin of all of this. We had psi x equals 1 over 2 pi i integral c minus i r to c plus i r zeta prime z over zeta z x to the z by z d z plus 1 over 2 pi i by z. So, we have to order x log square and this is how we started and we said let us evaluate this integral this is the error term to evaluate this integral we use this familiar technique of defining a domain and using the residue theorem Cauchy's integral formula essentially. But in order to use that one thing that became clear is that we cannot go on the right side because if you go on the right side since x is bigger than 1 x to the z will diverge and we will not be able to bound the integral this side. So, we have to come on the left hand side and look at this region and c was if you recall 1 plus 1 by log x just a little more than 1. Now, in order to do this domain integral we needed the definition of zeta function in this domain which we have now managed to achieve once zeta is defined in this domain and zeta is analytic also by this analytic continuation make sure that zeta is analytic everywhere except for the pole at z equals 1 and so zeta prime z is perfectly well defined everywhere. So, this function is defined well defined and we can now go ahead and do the integral. Now, in order to do the integral we will of course, invoke this theorem of Cauchy integral formula that integration along the boundary of a domain equals the or the residue theorem equals the sum of the residues at all the poles right inside the domain. So, what are the residues of this function this whole function that we have what are its residues in this region or forget the residues poles let us forget let us find out poles in this region one pole is very clear. Z equals 0 pole right here, but that and that takes here of this guy no other pole because of this the remaining poles are because of zeta prime over zeta what poles are caused by zeta prime over zeta will it have pole at z equals 1 zeta has a pole, but there is 1 over zeta prime over zeta think about this see zeta has a pole at z equals 1 it is a simple pole. So, if you look at the Lorentz expansion around z equals 1 for zeta you will get something like 1 over z minus 1 plus higher degree term differentiate this what do you get you get minus 1 over z minus 1 whole square plus higher degree terms. In fact, the higher degree terms will only be constant and higher there will be no term like 1 over z 1 over z minus and then you look at zeta prime over zeta. So, in the numerator you have 1 over 1 z minus 1 whole square in the denominator you have 1 over 1 minus z plus higher degree term you multiply both by 1 minus z. So, the denominator becomes a polynomial or a power series numerator still has 1 over z minus 1 plus a power series now at z equals 1 what happens pole same order pole order 1 pole at z equals 0 1 upon z is not there zeta zeta is not 0 or does not have a pole or a 0 at z equals 0. In fact, this character zeta prime zeta is a pretty bad character because whenever there is a pole of zeta or there is a 0 of zeta this becomes a pole. One simple way of seeing this is that zeta prime over zeta is no no it does. So, let us look at so it is very forget about zeta take any function just consider let us say some function m z with pole of order k at z is 0. So, let us will just assume this as a pole at 0. So, this means f z is 1 over z to the k not 1, but it is some constant c c over z to the k plus higher degree terms what is derivative of f it is minus c k over z to the k plus 1 plus higher degree terms. So, f prime z over f is minus c k over z to the k plus 1 plus higher degree term c over z to the k plus higher degree term multiply and divide by z to the k we get minus c k by z plus higher degree term c plus higher degree term. Now, this at z equals 0 is a pole, but is a pole of order what is the residue at this pole residue is calculated by multiplying it with z and taking the limit as z goes to 0. You multiply the whole thing by z take the limit as z goes to 0 you get minus c k over c. So, you get residue minus k let us instead suppose f k as a 0 of order k then you can write f k as sorry f k as a 0 of order f z f z as z to the k some c times z to the k plus higher degree term. So, what happens to f prime z c k z to the k minus 1 plus higher degree term. So, what happens to f prime over f c k z to the k minus 1 plus higher degree term over c z to the k minus k plus higher degree term z to the k minus 1 can be cancelled out from numerator and denominator you get c k plus higher degree term over c z plus higher degree term. So, this as a pole at z equals 0 of order 1 and what is the residue. So, f prime over f for any function f has this very nice or strange property if you will that whenever f has a 0 or a pole f prime over f has a pole of order 1 and the residue is precisely the order of the 0 or pole where the order of pole is taken in the negative sense. So, that is the knowledge we have now coming back to zeta prime over zeta which is one instantiation of f prime over f we have to now worry about every single pole and 0 of zeta function. So, well surely we know that here is a pole at z equals 1. So, that is going to be a pole of zeta prime over zeta, but not only that minus 2 m minus 4 m minus minus 2 minus 4 minus 6 these are all 0 of zeta function therefore, poles of zeta prime over zeta, but not only that all the 0's wherever they are in this critical strip and we have no idea how many of where they are and how many of them are they are also poles of zeta. So, in order to be able to estimate what this integral is we will need to do a lot of work now this part is easy minus 2 minus 4 minus 6 very clean we know it is a we know the residue, residue is because these are poles no these are 0's of order 1. Therefore, residue at each one of them is 1 simple this is again 1 also we know it is a pole of order 1. So, therefore, residue is minus 1 these are the characters which we really need to understand these 0's here we need to get a handle on where they are how many of them are there. So, that we can calculate the residues the right number of residues appropriately the residues we can only we can calculate easily by just knowing how many of them are there, but there is there is a reason why we need to know exactly where these guys are and the reason is the following that as you take the integral see that strategy is going to be that we take this contour integral anticlockwise and we will argue that this integrals are very small and therefore, can be absorbed in the error and. So, this contour integral is roughly same as the integral this one this integral and therefore, we get psi x not to argue that these contour integrals are very small or this path integrals are very small we will have to be careful as we move along say from here to here we may be coming close to a 0 of zeta function that is where it will be because here we know exactly where are the 0's we will can avoid them by going far away from there and also come and come down of course, we cannot avoid them by too much, but we can certainly split them right in the middle. So, distance 1 this side distance 1 this side so reasonably far apart, but here we have to have some control over taking this path otherwise if this comes too close to a 0 what happens this person diverges or it becomes very large as we move the integral becomes large and then we cannot argue that the integral along this line is small. So, that is why it is important to know the locations of these 0's also. So, that we can avoid the 0's as we move from right to left. So, we need to know the numbers we need to know the locations and then we can say something useful hopefully about the contour integral and therefore, we can conclude something useful about psi function. So, that is the task I wrote us any questions or zeta prime over zeta has a pole at that point. So, as if you this path goes very close to a pole or a 0 of zeta function which is a pole of zeta prime over zeta then the integral will shoot up very close. So, we have to define what very close is it becomes essentially we want to avoid a situation where when this function takes a very high value along some point on the path integral path right. Other takes a very high value then of course, we cannot argue that the whole integral is small. So, that is the situation we have to avoid. So, we cannot just blindly take a no pick a point here and just move to the left. We might even be hitting a pole if you as you move along and then the whole integral is identified. So, that is as I said is the task I had of us and we are now it sort of tells us that we have to analyze this distribution of zeros in order to get a handle. But let us assume wherever those zeros are. So, let us say we can at least write down this suppose. So, let us use a symbol rho which I will uniformly use to represent zeros of zeta function in that critical strip. So, then what is this contour integral if you recall that this is the domain d there is a negative sign also. So, this is the negative what is the value of this integral by residue calculus it equals the sum of the residues at all the poles. So, what are the poles here poles are of course, at one order of zeros I am not sure I think we can say something, but that is not very important for us because zeta prime or zeta will always have a order 1 pole only the value of the residue will change value of residue will change. But what we can say is we just let rho run over all zeros if the order is k then let we can sort of pretend that they are k 0 sitting at the same point. So, as rho runs over all zeros it runs over k times over that. So, this equals the sum of residues at all the poles. So, first let us take care of the poles that we know well what is the residue as the pole when z equals 1 zeta prime over zeta takes the value minus 1. So, minus zeta prime over zeta will take the value plus 1 and, but notice that the residues of the whole function not only this is this entire function we are considering at its poles. So, what is the residue of this at z equals 1 x then moving back z equals 0 is pole of this. Now, the residue of this therefore, is minus 1 because of this being a pole and this will then be zeta prime 0 over zeta 0 as the pole residue. Then moving further back at minus 2 minus 4 minus 6 what is going to be the residue. This will give always a minus why would there be a minus the pole is pole will give a residue of minus 1 no going back to minus 2 for example, minus 2 is a 0 of zeta. So, its residue will be plus 1 zeta prime over zeta and since there is a minus here there will be a minus 1 here and this would be minus x square by 2 minus x to the 4 by 4 and so on. So, these are the residues corresponding to the 0s at the trivial 0s and then there would be residues corresponding to non-trivial 0s within the critical strip and what about there. So, there will be 0. So, its residue will be plus 1 it is minus 1 here and this would be x to the. So, let us just write this as you are right and there will be a minus here. So, that is the formula we get we will do it next time.