 Today we are going to prove prime number theorem and this essentially follows from everything we have done so far plus the Midsim problem here thought. So, what did the problem if you remember one of the Midsim problem was that if you consider a rectangle here and zeta prime z is order log square while absolute value. This is on the boundary or the question did I ask was the question about the boundary or was it in overall and the step 2 was. So, yeah once you calculate over the boundary it is pretty straight forward. So, these are the two results that we will make use of and. So, now come with the theorem statement that zeta z 7. So, what does this say in terms of the picture that we have. So, we know that zeta is not 0 on the right of the line z equals 1 rather sigma equals 1 and what this theorem is saying is that for a little less sigma just a little less and that less becomes lesser and lesser as we increase t right as t goes to infinity this is converging towards 1. So, it is something like curve like this it goes towards infinity it reaches 1. So, that is this is the curve and then in this region entire region. So, no 0 of the zeta function lies in this region which automatically translates to the prime number theorem. What is the corollary of this that psi of x is x plus the 0s were at only at sigma equals half then it would be error would be square root of x times log square. But now they are not they are away from out therefore, much closer to 1. So, what will come out is something like x to be 1 minus small o of 1 something very small there inversely proportional going very decaying towards 0 but going up. But I will not do that part because this is sufficient to essentially imply all of this. So, let us do the proof of this. So, we will just look at this case that at the height t you have this we have this curve or a rectangle on the boundary of which we know how the zeta behaves fine. So, let us first do something very nice which is let us start with the definition of zeta z and here we will. So, always assume z is sigma plus i t. So, we fix the height t and vary the real part between 2 and coming towards becoming little less than 1. To begin with we will just vary it start from 2 and go towards this side towards 1 but not quite reaching 1. So, in that region we know what zeta that the using the original definition this is the product over all prime p this is 1 over 1 minus for 1 less than sigma less than theta. Take log on both sides and now expand this log also around 1 and then you get another series here which is m greater than equal to 1 I think and this is log of 1 minus x is x x square by 2 x cube by 3 that is how it goes this m p to the m z and let us combine this and write them as which I can. So, this is running over all prime speed these are running over all powers m essentially. So, this will run through all numbers n except for number 0 or number 1 1 will not occur here because m equal to 0 is not all other powers are there. So, this is but this is running over all prime powers to get rid of prime other numbers we stick a lambda n here which is 0 whenever n is not a prime power. So, that takes care of that when n is prime power this is log of. So, if n is p to the m this is log p and. So, this becomes n to the z and what do we do about this see when n is p to the m then this is p to the m z what is log n m log p and what is lambda n log p log p log p cancels and you get 1 over n. So, that is an alternative expression for log of zeta when real part is written. Now, we will do a little bit more work here real part of log zeta n to the sigma. So, I can write z c z is sigma plus i t that we have to keep in mind for the entire proof. So, I can break this n to the z is n to the sigma plus n to the i t and. So, this is the cos of n to the i t is e to the i t log n and when you goes from numerator denominator numerator become negative. So, cos of minus but that is also cos of plus that is no problem. So, this is cos of t log n. Now, by the way this is also equal to log of absolute value of zeta log of absolute value of n number is really the real part of log of. Now, we use a simple trigonometric identity which is involving cos 2 theta cos theta cos 2 theta is 2 cos theta minus 1. So, let us I just want to write it in some in a way that because positive. So, just involve cos 2 theta cos theta and 1 and create a perfect square how do we do that. So, we get cos square theta by this I need to get the rest of it make this a perfect square 2 cos square theta plus what else do we need plus theta plus 2 minus 4 cos this make sense. So, this means 3 plus 4 cos theta plus cos 2 theta is 2 times cos theta plus 1 which is greater than equal to 0 this is the key point. So, let me explain what I am trying to do if you look at this expression this is greater than equal to 0 log of n is greater than equal to 0 because n is greater than 1 n to the sigma is greater than equal to 0 sigma is between 1 and 2 this is the only quantity that can be negative. So, I want this expression make to make this expression always greater than equal to 0. So, what do I do to ensure this well the simplest way is to add some things to this. So, that this cos becomes bigger than greater than equal to 0 always and that is this is the trick which we use. So, using this what we can say is 3 log zeta of sigma plus 4 log zeta of sigma plus i t plus log zeta of sigma plus 2 i t is greater than equal to 0 that is all and this gives us. So, now these are all logs and sums of logs. So, obviously I can write this log of absolute value of zeta sigma cube or zeta sigma whole cube let me write it this way zeta of sigma plus i t to the 4 equal to 0 and this of course means that the product of absolute values of this quantity is greater than equal to 1. And this is a very important for us because now we can use this to argue some interest essentially argue the prime number. So, what this is saying is that sigma nor notice that sigma is approach we can is between 1 and 2. So, I can take it towards 1 right, but looking at the right hand side of this product always stays greater than equal to 1. So, you can take it as close to 1 as possible it will stay product will stay greater than equal to 1. In fact, now then we I can use the continuity to actually take it to 1 and the product will stay greater than equal to 1 even at sigma equals 1. Because sigma zeta function has an analytic continuation at sigma equals 1. So, once I take the analytic continuation take those values this product has to remain greater than equal to 1. Now, suppose now one more thing is that how would that products stay greater than equal to 1 that is one thing as sigma goes towards 1 zeta of sigma goes to infinity. So, that will really make it bigger and bigger. So, that will I try to satisfy this and as long as these two characters remain bounded is going to shoot up and problem is solved. The problem may occur when zeta of sigma plus i t is a 0 or rather zeta of 1 plus i t is a 0. In that case that 0 will try to cancel out this in this diverging. Now, who will win that is that is the next question one is going to shoot up one is going to send it to 0 that is also easy to decide who will win. You can look at the expression and tell me who will win the order see order of 0 and order of pole what is the order of pole at zeta 1 1. So, zeta cube will have order 3. So, it will be zeta of 1 or zeta of sigma around 1 will behave like 1 over sigma minus 1 and the cube of it will behave like 1 over sigma minus 1 cube. Now, if this is a pole sorry if this is a 0 at sigma equals 1. So, sigma minus 1 will be a factor of this as sigma tends towards 1 and since this is a fourth power it will be a 0 of order 4. So, that 0 of order 4 will cancel out this pole of order 3 and they will still be left 1 0 of order 1. So, these two will together still go towards 0. So, now comes a third character now this third guy may still want send it to infinity or at least keep it above 1 provided this also has a pole of order at least 1. If it is not a pole of order 1 then this whole thing goes to 0 right, but does it have a pole of order 1 no zeta function has only one pole which is a sigma equals 1 nothing over else. So, this will be bounded this does go this is together is a pole of order 3 this is a 0 of order at least 4 if not more. So, this product will go towards 0 if zeta of 1 plus i t which is not possible we just this inequality tells us this is not possible. Therefore, and t was up to here. So, it says that on that line sigma equals 1 the zeta function there is no 0 of the zeta function lying on that line. So, this at least already shows part of the theorem we it does not quite show the full theorem because full theorem says that actually there is a small region to the left of that line also where there is no 0, but at least it shows that on that line there is no 0. Now, to show that we can go slightly beyond this line well we use this the two things that we learnt starting from this expression. So, we have already argued from this expression, but let us continue for our argument a little more. We have zeta cube zeta of sigma cube and this fourth power of this and this product is greater than equal to 1. So, let us rewrite this I am interested in understanding the absolute value of this this is greater than equal to 1 over. So, now assume sigma to be just a little more than 1 let us fix it may be some constant c 1 divide by log of t to be this is just a little more, but no problem this inequality is holds there. At this value of sigma what is zeta of sigma like well zeta of sigma at least the absolute value behaves like 1 over with residue is 1 with this plus a another the Taylor series if you higher power which sort of go vanish away when this is. So, for this sigma is really very close to 1. So, we can assume that this is plugging this value of sigma because I have already fix sigma. So, zeta of sigma will be so this minus 1 1 over this. So, this is about order log to the next. So, this gives an estimate of this how does this group how does this group how about this this is where I am going to use the result on that line entire line actually the value was zeta of z was order log t z is sigma plus i t and sigma being anywhere between 1 little less than 1 to 2. So, at height 2 t it will still be order log t the absolute value. So, zeta of sigma plus 2 i t and zeta this is bounded by order log t. Therefore, this is greater than equal to just plug in everything here this is bounded this is I mean this is an upper bound on this the log t is upper bound on this. So, inverse size that is these are the lower bounds and then you multiply this out log to the 90 cube 27 plus 1 more is order log to the 20 h t this is fourth power and this gives minus 28 why minus 28. So, at that point at that sigma this is at least 1 over this. Now, let us go to the next and let us look at very closely and this line a little here and a little there. So, I am going to look at these two points this is sigma little more than 1. So, at this point I know that zeta is at least greater than equal to 1 over log to the 70. I also know that throughout this region the derivative zeta prime is order log square t for equal to alpha with an equal to 0. So, 2 minus sigma sigma is a little more less more than 1. So, 2 minus sigma will be little less just the same quantity less than 1. So, that is and zeta prime sigma plus i t is greater than equal to order. So, a function which changes from has certain value here and with a certain derivative along this way as it travels around this question is can you attain a 0 in this region. The answer is no what is the distance between this distance how much is this sigma was 1 plus 1 over log t to the 9. So, this is about twice that this is order 1 by log to the 90 and this is bigger than log to the 1 over log to the 70. So, in that tiny window as you move around this and we are the in the worst case you are always shrinking that is a derivative is always taking you down with that speed. And you start with a certain number here and it is go down with that derivative how low can you get what derivative into the distance essentially that is the lowest we can get. So, that is basically log square t into log order log to the 90 which is 1 over log to the 70. Now, it is only a matter of fixing the adjusting this constant that we pick initially. So, this constant is up to us to pick this constant that we picked here C 1 is up to us. So, we pick it really small constant. So, that this derivative I mean this distance the constant here is so small that when you multiply this with this it is still order 1 over some constant divide by log to the 70 that is maximum it will shrink. And that is smaller than this number in the reduction in quantity will be at most this. So, C prime is a fixed constant C is up to us to choose and that is it that shows that in this we can I can start from here and go all the way up to this without finding a 0. That is the prime number of primes asymptotically approach x by log x.