 So, let us work out your idea. So, let us say in psi x prime p contribute p x flow. Now, log p x flow times log p equals of course, log x of p flow times log p. Now, this is log x over log p plus order 1. There can be a it can differ from log x over log p by at most 1. And the in the worst case it will differ by something close to 1. Then psi x equals sum over prime p less than equal to x log x. This is equal to pi x log x plus order log of product of m p less than equal to x. Now, what do we do with this quantity? The error term how do we estimate that product of all primes less than equal to x. So, if I want to x log x less than that is not very nice. See the error term we want to be square root x log square x order psi x by log x, which means order x by log x plus something plus error which is. So, you do not want any constant. No, no, no way you want x by log x. If you look if you wanted constant it is very easy to show that pi x for example, or psi x is between 1 by 5 x and 5 x. So, you did not have to do any of this you know Riemann hypothesis and complex analysis forget all they just do simple counting. One can show with a little bit of cleverness this is what Shevishov showed long even before Riemann that pi x is greater than equal to x by 5 log x and less than equal to 5 x by log x. Clearly that is a good point let that be an assignment that is show that pi x is x by 5 log x is between x by 5 log x and 5 x by log x and just do simple counting nothing else. Of course, here we have been very pessimistic in approximating the floor here. We are all saying that x log x by log p floor is always log x by log p plus order 1 can bound this by 1 sure, but that is does not help. There are would be times for different p's this will be very close to 0 the error and there would be times when it is closer to 1, but then it is going to be very hard to say how for how many p's is closer to 1 how many p's is closer to 0. So, unless we do an estimate of that and some clever way of doing that do not think we can much about this. So, this is likely not to where in this approximating this you mean, but this already looks pretty high. In fact, if forget order log p here if this was just order 1 then what do you get pi x log x plus order pi x and that is already too much because what we are looking for is a much tighter relationship. So, if psi x is x plus order square root x log square x then I want pi x to be also the error term to be also close to square root x. In fact, yeah now it is I think this has to fail because see this floor unless it is exactly unnecessary it has to fail that is not right, but it is much more likely to fail because even if the log x has a slight differ slightly from a multiple of log p then error would be at least 1 over log p and if error is at least 1 over log p then multiplication by log p will only it will give you a error order 1 for 1 prime and then when you sum up you get order pi x here and then that is too much. So, we have to do something else and I tried something last time which clearly was wrong because it we cannot differentiate the error and I promised that I will find out about it today and do it, but I had no time in the morning to find out. So, I am in the same position I was this way, but we can still try let us see the issue last time was that the error we had approximated certain error and we said psi x if you plug this in and differentiate then we get this, but we cannot differentiate the error. So, let us go step back and before arriving at that error we had some expression for psi x which and that error part we then we do an approximation and got an error. So, instead of doing an approximation for the error let us go back to the original expression for psi x differentiate it there because there was an exact equality and then do an approximation maybe that will help. So, let me rejig your memory what was exactly psi x no that is that is one thing, but psi x right in the beginning we derived expression for psi x which was something like is c minus i r to c plus i r and of course 1 over 2 pi i here then we have minus zeta prime z over zeta z x to the z by z d z plus an error term what was that error term and now I am not saying that get an approximation of that of something like I think that was like x by r. No, so I do not want this because this I cannot differentiate I want that expression which we approximated to get this log square x x log square x by r there was some infinite sum if I remember correctly summation over here and the summation goes from n greater than 0 this was the exact term. So, this is what we work with y order was also what what approximation got us the order go back where y order because delta is also approximate. So, let us go back further let us what was the delta's business. So, there was an approximation there to yes no no no it is delta which said there is like from c minus i infinity to c plus i infinity that integral of x to the z by z is precisely the delta. But when you truncate the integral at from we just look at it from c minus i r to c plus i r then you get an approximation of delta and that approximate truncation error is what we were trying to we did estimate if I remember correctly. In fact, we did as you see there how do we calculate this this definition of delta that was by looking at this rectangle two rectangles right one c minus i r to c plus i r then going to positive side and once going to the negative side and then we said that if you evaluate this integrals on the three sides three edges it tends towards zero. But now one side of rectangle we can send towards zero that this these arms which go from c minus i r to u minus i r that u can go to infinity but i r stays there. So, integral along these two branches one the vertical one and two horizontal ones vertical ones is what I want the horizontal one is the error and the horizontal one stays the error stays. So, what is that integral value actually we can just write it as may be just as integral you see let us may be this may be a good idea to revisit the whole thing and just write it as integral. So, first of all this answer with the question is this exact apart from the problem in the approximation in delta is this part that you get is this exact there is no approximation involved here except for the delta function one which is like e plus i r 2. So, these are the two integrals which are the error terms for delta approximation actually this integral is c plus i r to infinity plus i r. So, it is going on the right hand side depending on x whether is less than one or greater than one actually there will be a integral going to the left hand side also. But is it not that already here this is what I am confused about this sum is this not already incorporated we approximate these integral to get this order let us just start from the basics zeta prime or zeta is summation n greater than equal to one lambda n by n to the z right. And moreover we also know that psi x equals summation n less than equal to x lambda n which we said is summation n greater than equal to one lambda n delta x by n right if I remember and delta x by n is 1 over 2 pi i integral c minus i r to c plus i r plus infinity plus i r to c plus i r to infinity plus i r plus infinity plus i r to c minus i r x to the z by z. So, that is delta right. So, we get therefore, 1 over 2 pi i this is x by z divide by n by n to the z and this should be minus. So, that is a good part and then yes these integrals are what giving me that approximation. So, this is exact. So, this is the error part which are these two integrals. Now, after that we focus on this part and we derived an expression for this as well what was that that was in terms of the zeros the residues in that big rectangle and that came out to be I think x minus zeta prime 0 or zeta prime 0 minus or plus I do not remember whatever it is. Then there was another minus of log of 1 minus x square these are all trivial zeros and then there was a plus over rho x to the rho by rho rho's are the trivial non trivial zeros and this is it that is the expression for that integral. I am not sure about the signs here and may be not not sure about this one also is it this probably it, but these are this is not important anyway these are these are very tiny small number in any case plus this error fine. Now, this is exact no approximations anyway now let us differentiate psi. So, psi x or psi t whatever it is whose then what is d psi x by d x 1 this vanishes what happens to this this is a like 2 x upon 1 minus x square what happens to this it is the rho minus 1 and what happens to this x to the z by z now what is the approximation what is the error. So, error is this whole thing it is 1 plus the error. So, what is the error 2 x by 1 minus x square yes yes this one yeah or is 1 over x square. So, then this will differential differential of this would be 2 by x cube which is same as saying that is multiply whole thing by x cube then you get 2 over x cube minus x now 2 over x cube minus x it is bounded I mean for any sensible value of x this is order 1. So, this is gone what about this x to the rho minus 1 now when you can look at the error you just look at the absolute value. So, which is we just look at x to the real part of rho minus 1 assuming the Riemann hypothesis to be true this would become x to the x minus r and sum over all rho such that imaginary part of rho is less than equal to R this we have already estimated. So, this is then we do some calculations for this we must have because when we approximated the error here how we did 1 over rho take it from me it is like R log R will prove it is the number of 0 actually there is very nice expression we can very precisely define or you know write down the formula for number of 0's of zeta function at height up to height R. So, it cannot out to be order R log R that takes here of this now what remains is this how do we estimate this. So, z is like whatever that c it goes c to infinity and c to infinity is. So, the imaginary part is really playing the no role is c to infinity both sides right and when we look at the anyway the abstract value you get t. So, basically what you get is order c to infinity sum over n greater than equal to 1 lambda n over n to the z. Now, n to the z is also of course bounded right n to the z is in absolute value it is we get n to the n x to the t minus 1 dt t is the one that is going from c to infinity t is the variable parameter that is being integrated on. See this is going the right z is integral for only the real part is varying the imaginary part is always the fix and when you take the absolute value the imaginary part anyway goes away. So, it just becomes an integral of this kind integral c to infinity you have x to the t divide by n to the t. So, this is simply integrated as 1 lambda n can be taken out and then integrate c to infinity and this would be of course I have strictly speaking I should have split this sum into two parts n less than equal to x and n greater than x because depending on whether n is less than equal to x and n greater than x the definition of delta function itself will change in the sense that this integral I have just said it is going from c to infinity the other thing occurs that is n is less than x then this would go from c to minus infinity if n is bigger than x then it goes from c to infinity. Remember that delta function for n less than x delta n over is 1 because you are going from the negative side and then there is a pole that you are pulling at n inside the rectangle if n is bigger than x then you go to the right hand side where there is no pole and then delta value is 0. So, this sum actually this integral should come I have should I come inside the sum and with this two spreads but in effective the effect of that is going not going to be anything see you look at this integral. So, here think of n always being bigger than x because only then this integral will converge otherwise if this goes to infinity and n is less than x it diverges because and that is fine because whenever n is less than x this integral the limit changes to minus infinity which has the same effect. So, integrating this gives you what same thing x by n to the c actually 1 over log this is more or less the same thing more or less I call exactly the same expression that we got earlier except that there is something there has to be something missing here there is an r that is missing what happened to the poor of of course, this is x to the t minus 1 there is an. So, instead of an r there is an x that is the only difference that is happen. So, we can use exactly the same analysis to derive that this is order log square x x log square x divided by capital R but now is divided by x. So, you just order log square x because the earlier error which was this is order x log square x by r and this is in this expression the only change now is instead of this r we have an x and that is it. So, this is it becomes the x log square x by x which is order log square x and hence is 1 plus now also remember that we are finally, plugging in r to be square root x when you plug in r to be square root x we simply get the error to be just order log square x you agree with me is there any question you got got to ask now if there is any doubt perfect. Now, let us come back to your current analysis and what we had was d psi t by log t integral going from 1 to x pi x let me just stick order pi now plug that in 1 plus order for psi prime x t now we know 1 plus order log square t. Now we are back to the good situation this is trivial this error is 1 plus order log square x or I should say trivial something is wrong you know when you say 1 plus order log square x what does it mean this means not sense this is much smaller than this error is completely bizarre yes we miss many things r is of course, we miss but this is also not good we did this estimation and we said this is equal to order r log r by square root x completely unacceptable this does seem to what what does it mean this will be a 1 over square root x when you take the absolute value and then sum overall let us just account of all the zeros they have to be r log r and then we missed the r again the when you differentiate this with respect to x you lose that cancel out that z in the. denominator that is how you lose that r r log is also when r is square root x you order log log x that is too bad too big yes that is worse but even this is you can get rid of that even this is 1 here in fact even here I should not be writing order 1 I should simply be writing something like order 1 over x cube because x is a parameter I will take the absolute value. So, absolute value of x to the rho minus 1 is square root x for any rho when the Riemann I have this holds that comes out square root x is common then you just sum over all rows the number of such rows is as I said is less than equal to imaginary part less than equal to r is r log r. So, this is of course this is also wrong whatever I have done is completely wrong. So, what is going to work this seems to be I do not see how we can improve on this because here just look at this you have to take the absolute value when you take a square root x and then r log r is does come out you cannot avoid it the moment it does come out and r is we are always fixing it to be square root x or can we any play around there remember other if you play around the value with the value of r then psi x itself is going to change. But how does it matter how does it matter that is a very good point what is stopping us from choosing a better value of r because you see this equality holds for all values of r. So, let us choose a value of r which optimizes the error. But here we can say fine we can choose the value of r which optimizes the error but what about here there is no r here at all because r is gone when you take the absolute value where r simply goes out and then you end up with log square x good another assignment problem there is clearly a fix just that we have not been able to find it. And there is a simple fix is not anything complicated maybe you can well you cannot disprove Riemann hypothesis but maybe you can disprove this connection between Riemann hypothesis and prime density.