 So, let us continue our investigation into psi and before I dive into this little more let us first clearly understand what we want out of psi as we know that psi z is z z minus 1 our eventual target is to estimate log no zeta prime over zeta. So, we do this take the log and differentiate we get psi prime over psi equals minus and therefore, if you look at zeta prime z over zeta z in absolute value that is bounded by of course, gamma prime zeta over 2 gamma zeta over 2 which is which we already know is like order log of z again in the range we are interested in which is sort of up there. So, works fine and the first three quantities here are tiny. So, they got all absorbed into order log z plus psi prime over psi the. So, that is the quantity we need to estimate in the range where z varies from a little more than 1 plus i r 2 minus 1 plus i r. So, we already have done this job for gamma function estimating gamma prime over gamma through this first we showed that it is an entire function without any zeros then we looked at 1 over gamma which is an entire function without with zeros of course and then we expressed entire function of order 1 and then we expressed 1 over gamma as a product which gave us an expression for gamma prime over gamma. So, that is the same approach we use for psi also we know that psi is an entire function we know that it is zeros are precisely the non-trivial zeros of zeta function the only thing we need to establish at this point is the order of this psi and psi is an entire function, but what is the order of psi. Now, for this let us recall the definition of psi there are two definitions one is through this product which we already have written here, but this does not give too much recluse because of course gamma we know exactly the order of, but zeta we do not know the order of. So, instead of this we look at the other definition of psi which is psi z is 1 plus z z minus 1 integral from 1 to infinity t to the z by 2 plus t to the 1 minus z by 2 and here w t is this sum and greater than equal to n e to the minus pi n square t. So, knowing this as a definition of psi can we estimate the order of psi what does this give. So, order of psi means absolute value of psi z we want to estimate this is less than equal to mod z square for this part and then integral going on to infinity t to the z by 2 what is that bounded width that is bounded by clearly t to the mod z by 2 and t to the 1 plus mod z then w t over t fine. Now, of course it is a sum. So, we can equal to therefore, write this as order the order we can always use equal to z square 1 to infinity t to the mod z by 2 because the first integral there is corresponding there is a sum the first integral is going to be less than equal to second integral because it is a it has a big higher power of t which then cancels out with this and then we get. So, we want to estimate this quantity and again it is not too difficult to estimate it actually if you look at this 1 to infinity t to the mod z by 2 w t t just plug in the definition of w t sum n greater than equal to 1 e to the minus pi n square t. Now, here we can exchange the sum with integral although both are infinite but again using the fact that this quantity is uniformly convergent we can exchange the 2. So, we will write it as n greater than equal to 1 integral 1 to infinity. Now, what do we do with this that there is a simple way to handle this actually this is mod z is some real number. So, let us this is at most to infinity t to the whatever is the ceiling of mod z by 2 in integral the integral the integer just higher than mod z by 2 e to the minus pi n square t d t right and now this this look familiar t to the integer e to the minus pi n square t d this is almost like the gamma function except that pi n square sitting in the up there. So, that we can get rid of very quickly when do a variable substitution. So, set u equals pi n square t. So, we get this is equal to n greater than equal to 1 when t goes from 1 to infinity this will go from of course, you can say 0 to infinity for u also gamma function is integral from 0 to infinity right gamma function integrates from 0 to infinity. So, 1 to infinity I can always replace by 0 to infinity and put a less than equal to here and here you get u to the mod z by 2 divided by pi. Now, just take everything out that you cannot integrate and this is now gamma function which becomes 1 over pi n square to be plus 1 and what is inside is simply the gamma of mod z by 2 plus 1 right which is of course, has nothing to do with n. So, this is a common factor outside and we get this sum now what happens to this sum this converges right it is like yours you are summing up 1 over n with a certain power that power is more than 1 that is clear it is at least 2 more than 2. So, that sum is a bounded quantity right. So, this is whole thing is order gamma z by 2 plus 1 which of course, is order 1 of course, I have not looked at this multiplication factor of mod z square which does not change the order at all good. So, the moment we establish it we can bring in the whole machinery we developed for analyzing the xi what can we write xi s is xi 0 0 or it is not right xi 0 is not 0 we already argue saw that. So, if there is no multiplier of z. So, this is like e to the a z plus b for some constants a and b then product which runs over all the zeros of xi which are precisely all the zeros of non-trivial zeros of zeta function. So, we will use the symbol rho to run over all non-trivial zeros of zeta function. So, whenever I write product over rho implicitly it means it is running over all non-trivial 0 zeta 1. So, that is a good short end otherwise I need to write that every time and we will not use the symbol rho anywhere else this is the only place we are going to then 1 minus z over rho e to the right. Now, that we have this product expression what is xi prime or xi that is easy that is a plus sum over rho and minus 1 over rho here divide by 1 minus z over rho here plus 1 over rho to this what we get. So, again this is a familiar expression we saw already over gamma prime over gamma, but of course there sum is over non-trivial zeros instead of integers. So, this is a little more difficult to analyze, but we already get some very interesting things. For example, if you recall we said that the entire function of order 1 in a radius of r is going to have at most how many zeros r to the 1 plus epsilon we can throw the same result on this also. So, now of course we know that in this case the zeros only line that strip between 0 and 1. So, at a height of r and that is a height we are interested in we are we are interested in integrating we are interested in this integral this is at r we are just interested in integral between from here to here to at height r and what we know as such in this region what is the maximum number of zeros is order r to the 1 plus epsilon that is directly from this result. Of course it is not useful, but still interesting to know that there is some immediate quantity we can put to the number of zeros. So, it is not something very arbitrary of course we still do not know whether there exist infinitely many zeros here. Although I have shown this product now this product is valid whether there are infinitely many zeros or finitely many zeros. So, that is okay, but we do not know if there are infinitely many zeros in fact it turns out that number of zeros at up to height r is I think r log r plus some error term. So, you can actually very precisely bound the number of zeros up to height r, but I will not prove that not yet because it is not useful for us. So, coming back to this that is a sum we want to estimate and we want to estimate this sum when z has a specific range of values. So, let us set some things up let us write z as for the range of values we are interested in as alpha plus i r because it is at the height r we are interested in and we know that alpha is between minus 1 minus 1 and 2 it is actually in between minus 1 and c and c is less than 2. So, instead of writing c every time let us just write this and we also let rho which is a 0 as sigma plus i t and sigma of course, we know is between 0 and 1. So, with this notation fixed let us analyze this sum a is constant we can just hide it away in order 1. So, what we really want is to understand this sum. So, let us just consider this sum or sum over rho. So, with the notation we just introduce we I can write this as and this again it is generally is easier if it get rid of the complex numbers from the denominators less otherwise summing this inverse of complex numbers there is very little intuition you can stick to this. So, let us just take this up using standard methods good. Now, let us look at this expression for a moment the first part can you bound it not clear the t here can become in ever bigger as rho increases the mod absolute value rho increases value of t also increases. So, it is not clear whether this is going to get bounded, but if you just look at the real part of this the first expression is that bounded that is because sigma is always between 0 and 1 and sum over rho of 1 over absolute value of rho squared that is bounded why that is likely we proved this using this fact number of 0's is bounded if you recall we proved that for entire function of order 1 if z 1 z 2 z i's are it 0's then sum over i 1 over mod z i to the 1 plus delta is bounded. So, certainly 1 sum over 1 over sum of 1 over mod z i squared is bounded and this is what exactly what is happening is summing over all 0's of this entire function of order 1 and 1 over this. So, if you just look at real part of this that is equal to this is order 1 plus here what can we say again we can forget about this complex part alpha minus sigma this is at most 2 because alpha is between minus 1 and 2 when the I have already specified the range we are interested in and sigma is between 0 and 1. So, this is at most 2 to alpha minus sigma for the moment plus r minus t squared actually I will be wanting an upper bound here. So, I want to replace this yes that is fine I want to replace this with a quantity which is now greater than this. So, let us keep it that way. So, what we get out of this is therefore, that if we put everything together all of this all the way then what we get is sum over rho alpha minus sigma and let us take the absolute value by now this course this part is small this part is small. So, what eventually we are getting here is sum over all 0's and then inverse of r minus t whole squared that sum right and where t is the imaginary part of all the 0's is gone. I am only looking at the real part real of this real part of this is equal to this right. So, this is at most the absolute value of this real part of this is real of a plus real of this right. So, I can say that real of this is real of xi prime over xi plus order 1 right a is fixed and that is also equal to this. So, I can say this sum is equal to real of xi prime over xi plus order 1. Now, real part of xi prime over xi is less than equal to absolute value of xi prime over xi. So, I am getting an interesting sum here on the left hand side which tells me about an upper bound on the behavior of inverse square of the complex parts only the complex parts of the only the complex part of the 0's. Of course, I know that this converges that we know any inverse of imaginary parts inverse of squares of imaginary part will converge because the real part is anyway bounded. So, it is going to converge, but with this analysis I can actually derive a relationship about with this sum and the quantity that it converges to and the quantity will clearly depend on r because that is only parameter here everything is being summed over right. So, that is what the target is. So, for in order to achieve that we need to get a bound estimate on this, but that was original problem anyway. So, what have we solved well the nice thing is that this relationship is available or is true for all values of alpha right alpha remember varies from c to minus 1 and for all values of alpha this available. In fact, forget about c when from 2 to minus 1 it is phi vary it is available and when I set alpha equals 2 then I know everything because at that z that is which is 2 plus i r I can bound zeta prime over zeta I can bound gamma prime over gamma and therefore, I can bound psi prime over psi. After all what is psi prime over psi is here that psi prime over psi once I stick in z equals 2 plus i r this is order 1. So, let me just put that in black and white for z equals 2 plus i r zeta prime over zeta is order 1 and hence psi prime over psi is order log of z what is the order log of z is log r right. So, all together we get 2 minus sigma in absolute value order log r now I just want to simplify this expression a little bit. So, all I will do is I will replace this expression with something which is smaller than this. So, that smaller expression will continue to satisfy this equation and to make the quantity smaller what I will do is I will make the numerator smaller and denominator bigger. What is the smallest value that this can take 2 minus sigma 1 because sigma varies from 0 to 1. So, I will replace this by 1 what is the largest value that 2 minus sigma can take 2. So, I will replace this by 4 and to make it even simpler I mean these are trivial stuff I will get rid of this 4 also how do I get rid of this 4 I will multiply this by with 4. So, that becomes bigger. So, this becomes smaller and then take this 4 this side and this is going to be a very important relationship for us because this is giving us something very interesting. Let us pull out again that diagram let us look at this region many zeros will lie in this region possibly between t and t plus 1 t is an integer some t and t plus 1. This expression is going to give me a value to how many those zeros can maximum number of zeros can be how just substitute for r capital t then every single zero in here. So, what I know this sum is order log t. So, every single zero in this region of course, will be counted by this will contribute something to this sum will contribute everything is positive here. So, everything adds up will contribute something to this sum what is that quantity is going to contribute a zero in this is simple at least how much what r is capital t. So, I know that this sum 1 plus inverse of 1 plus t minus small t whole squared is order log of capital t and this sum is over all the zeros all the way up. So, in this region when the rows are in this region pick any one of these rows it will have some value small t what is that value small t it is going to be between capital t and capital t plus 1. So, whatever that value is if you plug that in here this is capital t this is something between capital t and capital t plus 1 what is the minimum number that you get the maximum this difference can take is 1. So, this number the contribution of any of these zeros here will be at least half to this and all the numbers here are positive. So, how many zeros can be here at most order log t there cannot be any more than that because otherwise the sum will go beyond this and this is true for any t. So, this tells us how to choose my r of course, I want to choose my r to be a certain around a certain value, but within that see remember we wanted to an r. So, that to avoid any of these poles here because we cannot afford to integrate over a pole now that we know that there are at most log t zeros between t and t plus 1 we can conclude therefore, that there is there are at least two zeros. So, let us say t is the value around which I want r to be. So, what I do is look at this region I know that there must be a strip in this of horizontal strip of width about 1 over log t and there are no zeros in that strip. So, I will send my line r choose r to be the middle of this strip just cut through like this which will guarantee that this line is away from any of the poles by distance of at least order 1 over log t. So, that is what I am going to do. So, this instead of writing the distance to be order 1 over log t I am just writing log 1 over log r because t is say t is r plus minus 1. So, that is the first thing we learned from this. Now, this was obtained only by looking at the real part of this and we ran away from the complex part because that was diverging and this was simpler to manage. But we still are done because we have still not been able to bound zeta prime over zeta when we have done everything except bounding zeta prime over zeta in that region. So, that has to be done and for that we have to get an upper bound on this quantity and upper bound on this quantity for z varying from between in this region. So, we sort of in this analysis we just say because we cannot vary z in this region let us fix z as 2 plus i r and we know that zeta prime or zeta is bounded there. So, I do not have to work hard, but now I have to work hard. So, let us get back to this fortunately the hard work is not too much. So, we can manage it hopefully within this class let us see if we can. So, now after this let us come back to if I am not allowed to look at the real part of this then this quantity 1 over rho itself becomes troublesome because it is not clear whether some over rho of 1 over rho is bounded actually is not going to be bound which may not mean much because this quantity which has some negatives they may it may cancel out things and then eventually they get to some sensible value and we know that it does get to some sensible value it is bounded we just need to know what upper bound is, but for the sake of analysis we cannot easily act this. So, we employ another simple trick which is that we consider this quantity. So, we had that zeta prime over zeta that is equal to order log z plus this sum. So, what we do is this is a simple really cheap trick we already know we have used this actually that this is order 1 and actually in absolute value this is order 1 and this of course satisfy this equation as well. So, we put all this together our target is to bound this zeta prime over zeta get derive an upper bound of zeta prime over zeta we are going to use this, but this is hard to bound. So, all we do is we subtract zeta prime over zeta from zeta prime over zeta at 2 plus i r what is this equal to this order log r actually order log z here is also order log r because z is varying from minus 1 to c. So, it is always the real part is always bounded. So, this is always going to be order log r this order log r this is order 1 and so plus the nice thing is that sum over 1 over rho and sum over 1 over rho cancels each other out here. So, what we are left with is and since we know that this is bounded this constant in absolute value. So, this is equal to order log r plus we can just bound this absolute value. So, the problem now is to bound this quantity. So, let us focus on this of course, we are looking at the absolute value here this is equal to I should say less than equal to and this is equal to z is alpha plus i r where alpha is a varying quantity. So, if you just take that in get alpha minus 2 here alpha plus i r minus 2 minus i r. So, that cancels each other out the higher and now unfortunately I will have to expand this also what is this z minus p we know that this is alpha minus sigma squared plus r minus t squared square root as let us take square root outside what is this part that is simply the only change is 2 minus rho square plus r minus. Now, I need to get an upper bound. So, I will replace this with the smallest possible value of alpha minus sigma what is that 0 get throw it off what about this 2 minus sigma it is not 0, but assume it 0 throw it away and this 2 minus alpha I need to replace with the largest value 2 minus alpha largest value is 3 that is much simpler expression. So, now we are very close actually. So, if you see this is this familiar we just derived something similar to this where was it yeah it is here that sum over all rows 1 over 1 plus r minus t whole square is order log that is giving us a good upper bound what we have here is 3 forget 3 1 over r minus t whole square the only thing missing is 1 plus here, but for this we would have had our upper bound very neat order log r done, but we are not too far off because this I can write as again chip tricks 6 pi 2 r minus t whole square which is less than equal to which I will break as sum over rows such that r minus t is less than equal to 1 plus sum over rows and this we can bound see because absolute value of r minus t in this sum is bigger than 1 to twice r minus t I can surely replace by 1 plus r minus t whole square and lesser value. So, this is number bound and this is a sum over all rows actually except for a few zeros, but we know that even if you sum up over all zeros this is bounded by now that leaves out this but this is a sum only over finitely many zeros like r is there just r plus 1 and r minus 1 in that strip whatever the zeros are we are summing over those and this is the quantity we want to sum and this is where the choice of r plays a role I just fix the r right r was to be chosen. So, that it is away from every zero by about order 1 over log r what that translates to in terms of this t is represent the zero t is represent the height of zeros and r is the height where we cut that region and that height is away from all those t's of zeros by order 1 over log r. So, r minus t therefore, for all t's in this is at least 1 by log r some constant by log r and since this is in the denominator I can replace this quantity by r minus t by 1 by log r right. So, what we get here is order log square r and of course, this quantity has nothing to do with the sum. So, we are just summing over all the zeros in that strip how many zeros are there in that strip order log r. So, this is order that is it we have bound zeta prime over zeta in that region by cutting across it. So, now I am done therefore, the integral as we went along from c to minus 1 c plus i r minus 1 plus i r of zeta prime over zeta t z by z this is bounded by zeta prime over zeta I can replace by order log cube of r x to the z is replaced by absolute value of x to the z which is x to the real part of z. So, I will just I do not want to mess with this. So, I will just replace with x to the alpha mod z is at least r. So, I can divide this by r safely and we have d alpha d z is same as d alpha. So, this is order log cube r times the divide by r of course, here times the integral going from c to minus 1 x to the alpha t l and this is of course, familiar we know that this is equal to order log cube r by r x to the alpha would integrate to x to the alpha by log x. And then we have going from c to minus 1 so we get x to the c by log x plus 1 by x log x and now use the value of c c was 1 plus 1 over log x. So, this is actually x x to the c is essentially x constant gets hit away and this is clearly bigger than this. So, I can throw this away also. So, this becomes order x log cube r divide by r and this quantity is going to be bigger. Let us now I have to really go back well in the past now some one of you will have to dig that up. What was the original equation of psi x? Psi x is yeah that is right, but this we estimated to be is integral c minus i r to c plus i r and I think there was 1 over 2 pi i also somewhere zeta prime z over zeta z x to the z by z d z plus order the question is what is the error no we simplified that x log square x by r that is what we had and now we know that this integral this is equal to x now pull out this x minus this is the residues at all zeros all the residues all the poles. So, there was a residue at pole x z equals 1 which was x 1 pole at z equals 0 that was zeta prime over zeta 0 that is order 1 forget that then there were residues at all negative and we are integers what were the residues and there was a sign there also I think it was like x to the minus 2 m by 2 m. Now, I do not remember the sign whether it was plus or minus it was plus plus and then there is a negative right sum over all rows x to the rho by rho and the error terms now what are the error terms. So, we had this 3 error terms the integral from minus u plus i r to minus u minus i r that vanished went to 0 because we have sent u to infinity. So, that just leaves 2 integrals minus 1 to minus u what was that integral we estimated this last time it was something like r to the 1 minus epsilon was there somewhere at least let me pull this out. So, as long as r log x is common so as long x as x is bigger than r to the epsilon over x which is x square is bigger than x is bigger than r to the epsilon by 2 this is going to be bigger or anyway let us keep this in. So, now we put this here what do we get psi x equals I think this is 1 over 2 pi i also just take this out and push show that in here we get x plus by the way this is familiar what is this quantity sum over m greater than equal to 1 x to the minus 2 m divide by 2 m just differentiate this quantity this sum what do you get you get sum over m greater than equal to 1 x to the minus 2 m minus 1 and what is that sum let us take this out let us maybe we do not need to differentiate let us just take this let us just write this as half of this familiar to you what is log of 1 minus z or log of 1 minus y this is not talk about log of 1 minus y it is sum of m greater than equal to 1 y to the m by n and that is precisely what this is. So, this is equal to half of log of 1 minus 1 over x square and there is a minus also here there is plus here so this should be minus log of 1 minus y is minus of sum m greater than equal to 1 y to the m by n. So, this actually is a well known quantity. So, I can replace this with minus half minus so the rho by rho plus all the errors what are the errors add up to this plus this thus the original one x log square x and I need to minimize this error. So, what is the quantity at which this error is so my only parameter of control is r I need can choose my r to be just about anything of course, whatever value of r derived from here eventually I have to perturb that slightly in that band of t and t plus 1 to reach the rights avoid the all the 0, but that does not affect the error because it is very small perturbation. So, what is the minimum value of this expression for what r does it achieve r equals infinity of course, at r equals infinity this is this is 0. So, but what happens let us see what happens. So, this certainly implies that psi x equals x minus half I have been making a mistake here at least here I made a mistake this sum over rho is only for t less than equal to r actually minus r less than equal to t because that is a region we are integrating and of course, we can send r to infinity and derive this an exact formula for psi x is precisely this. Now, the first 2 expressions are well understood the last one is not well, but this also shows that this very starkly that the distribution of primes is dependent on the distribution of 0s of 0 zeta function non trivial 0 zeta how they are distributed is what determines, but this is unfortunately not very good in terms of estimating psi x. This is nice formula, but what about this quantity what what can we say about this quantity it is an infinite sum first of all and secondly, there is a rho sitting here which is in 1 by rho. So, I cannot even say that this is bounded if there was a power of rho which and worse than that there is x to the rho and so that all of these put together do not make this expression useful this is a very nice and interesting expression was not useful expression. In order to get a useful expression we need to limit r to certain limited or bounded value then we can get an error. So, this is of course, a trade off between what value of r we choose see suppose we just add up all the all of these quantities in the worst case they will all actually they will not get added let us say assume in worst case they all get added. What is the maximum contribution from 1 x to the rho over rho? Well this would be x to the rho will contribute square root x what am I saying x to the rho is going to contribute see the real part of rho is between 0 and 1. So, if the real part is 1 then in the real bad case this can contribute something odd like odd omega x and if it does contribute omega x then the first two expressions are sort of meaningless actually second one is anyway meaning second one is order 1 x is always going to be bigger than 1 and log of 1 minus 1 over x square is very close to 0 as x increases this actually goes to 0. So, it is actually only the first one is important if the real part of rho is 1 it might actually some of the such rows can get together and cancel out this x here and then we have no clue what psi x is going to be on the other end if these are going to be smaller than 1 then we can say that maximum contribution say if my real part of rho is sigma which is what we had assumed then the maximum contribution this can make is x to the sigma divide way of course, some absolute value of rho, but that is small than x to the sigma maximum contribution, but then the smaller the sigma is the smaller the contribution from this quantity is what is the smallest value sigma can take for 0 if this sigma takes value 0 then contribution here is 0 that is the best case, but that is also the worst case because if there is a sigma equals 0 by symmetry there is 0 at sigma equals 1. So, this sigma is going to be 0, but that sigma is going to be x so you gain nothing. So, taking sigma below half does not gain you anything because symmetry gives you bigger sigma on that side. So, at sigma equals half is when this contribution will get minimized and if all sigma were at sigma equals half all rows were at sigma equals half then this contribution will globally will be minimum to this quantity and that will be around square root x fine, but if that is so that is a very rough analysis to say that this error introduced by this in the best case will be about square root x, but then we have to look at this error how much is this going to be. So, let us try to estimate all of this what I am going to do is before estimating all of this just now we can already fix the value of r because anyway the error going introduced by this quantity is going to be about square root x and also notice that the bigger r is the more there is error because the more of these terms are there. So, you want to give r as small as possible to reduce the contribution from this, but if I make if you make r too small then this center error blows up, but now since we know that r is the error is going to be square root x. So, looking at this x y r x y r it makes sense to fix r to be square root x there about right if you fix r to be square root x this contribution is also about square root x and that is what we are going to do we get psi x equals x minus and we are now going to throw this away the second term because that is pointless and what happens to this term is become square root x log square x this is can be safely ignored and this becomes square root x log square x both of these actually terms match. So, that is also show that you can this is the best you can achieve in terms of choice of r when you want to choose it around square root x now we come back to the Riemann. So, this all this analysis was done by Riemann I mean everything that I have shown you today more or less was done by Riemann and it has taken me what 20 odd lectures to do that which he did in of course, 11 pages very densely packed of course he skip lots of things which were obvious to him which are not. So, obvious to the rest of the world, but now at this point is where Riemann made this hypothesis for all rows real rows that is the alternate formulation and the reason was to minimize the error here because at if all the zeros are at that line contribution of this sum to the error is. So, what is the contribution let us just add of everything up what is x to the rho x to the rho is absolute value of x to the rho is square root x assuming Riemann hypothesis of course. So, all this calculation is now assuming Riemann hypothesis then absolute value rho absolute value and now we take this sum of this from minus R to plus R t going from minus R to plus R 1 over this we can do in many ways one of the way we can do is to we just derived we derived that in one band of height 1 there are going to be only log t many zeros. So, if you just take that in here. So, I just replace this by summation. So, by the way there is symmetry around the real axis also of the zeros that follows almost why is this symmetry let me write this down we had this psi z which whose zeros are precisely non trivial zeros. So, if z is a zero what about z bar z bar is a symmetric point down there z bar would be z bars here t to the z bar z bar here right and what do I why is that a zero well because that is equal to psi z whole bar psi of z bar when you replace z by z bar everywhere is same as taking psi z and taking the complex conjugate of that. So, t to the z you take this conjugate what could be t to the alpha plus i beta what is the conjugate of that split that t to the alpha conjugate is t to the alpha t to the i beta conjugate is t to the minus i beta. So, t to the z conjugate is t to the z bar. So, whether you replace z by z bar or take the bar of psi z the answer is the same and that is why zero in fact that happens for any analytic function forget about this because any analytic function you can write when as a power series or something and again the same argument applied. So, now coming back to this. So, I can this is I can replace this with the order and some t from now zero or let us say one to r because between zero and one there are only finitely many zeroes. So, their contribution is going to be only finitely finitely many in the sense some constant times square root x right. So, we can that we can ignore and then we can just look at one less than equal to t less than equal to r of course, square root x comes out one over absolute value of rho. Now, I can replace this by some over t between one and r because how many rows are there in for between t and t plus one log t by absolute value of rho when rho is about t is how much rho is half plus i t what is the absolute value of this is also about square root of something which is around t and how much is this sum now log t over t t going from one to capital r. It is surely log square r log t is upper bounded by log r and sum over one over t where t goes from one to r is most log r order x square root x log square. Therefore, Riemann hypothesis is true then psi x equals x plus this is where we close this now we have completely derived the analysis of Riemann this is what is psi x if Riemann hypothesis is true if Riemann hypothesis is not true then psi x estimate gets worse because this error term then starts becoming bigger and bigger. Now, after doing all this you might believe that you know we started from estimating psi x and gone all done so much and then eventually we showed this implication if Riemann hypothesis is true then psi x and you might think that in doing such and this implication has such a long proof with so many steps that we will lose something that is lose something meaning that this if psi x equals this then Riemann hypothesis may or may not be true because we have this is a very long implication. So, probably there is a good chance that we lose somewhere the equivalence and only have single direction, but surprisingly that is not true if psi x equals this then Riemann hypothesis is true and that is a really simple proof I will show it next time.