 This lecture is part of Berkeley Math 115, an introductory undergraduate course on number theory, and will be about a sketch of the proof of the prime number theorem. So we recall the prime number theorem says that the number of primes less than x is approximately x over log of x. In the previous lecture, I gave some background to this and gave a rough overview of the proof. And what we're going to do is to go through the various steps mentioned in the previous lecture in a bit more detail. So as I said, the first main step is to show that zeta of s has no zeros if the real part of s is greater than or equal to one. So when the real part of s is greater than one, this follows very easily from Euler's infinite product, that this is the product overall primes of one over one minus p to the minus s, as I discussed in the previous lecture. So the problem is to show that it's got no zeros on the boundary of this region where the product converges. And the proof of this is quite short but involves a rather clever idea which is not at all obvious. And the clever idea is to look at this function. You take zeta of s minus two i t times zeta of s minus i t to the four times zeta of s to the six times zeta of s plus i t the four times zeta of s plus two i t. And you may wonder where this is funny looking function come from. Well, the exponents come from binomial coefficients. So you recall that if we've got a number z then z plus c to the minus one to the four is equal to z to the four plus four z squared plus six plus four z to the minus two plus z to the minus four. So we have these exponents, these coefficients one four six four one which are appearing in the exponents one four six four one. So how do you use this? Well, we recall the infinite product for zeta of s. And if you take the logarithm of that we find that the logarithm of zeta of s is sum over all n and all primes of p to the minus ns divided by n. We're just using the usual formula that the logarithm of one minus x is x over x minus x over sorry plus x over squared over two plus x cubed over three and so on. So if we apply that to the product formula for zeta we get this. And this means if we take the logarithm of this funny product of zeta functions we get a sum over all p and n of p to the minus ns times p to the i n t over two plus p to the minus i n t over two all to the power of four divided by n. Here we're using the fact that this thing here is p to the two i t plus four p to the i t plus six plus four p to the minus i t plus p to the minus two i t. And so all we're doing is substituting in the sum for zeta of s for each term of this product. And this expression looks like a bit of a mess but it does one really nice property. You notice this is always greater than or equal to naught zero for t real because this expression is a number plus it's complex conjugate. So it's real and the fourth power of that is real and so on. So this expression here is always greater than or equal to zero. Well, if the logarithm of something is always greater than or equal to zero then that something is always greater than or equal to one. So we've got this key inequality that this expression here is always at least one. Now that's why we write it down. Well, now let's look at it in a bit more detail. So we've got zeta of s minus two i t times zeta of s minus i t times zeta of s four times zeta of s to the six times zeta of s plus i t four times zeta of s plus two i t. This is greater than or equal to one and here we need the real part of s to be greater than zero. So it's going to be greater than one so that everything converges. And now let's fix t and assume that zeta of one plus i t is equal to zero and this implies zeta of one minus i t is equal to zero because it's just the complex conjugate of zeta of one plus i t. And now let's take a look at what happens at the point s equals one. So all of these become meromorphic functions. So at the point s equals one, this thing has a pole of order six and zeta of s plus one plus i t has a zero order four and this also has a zero of order four. Well, if we've got all together, so this has a zero of order at least eight and a pole of order at most six. So altogether this function has a zero of order at least two at s equals one. Well, this contradicts the fact that it's greater than or equal to one or s greater than one. I mean, if it's at least one for s greater than one, it can't suddenly become zero at s equals one. So this is a contradiction. So zeta of one plus i t cannot be equal to zero. So that's the first step of the proof of the prime number theorem. The next step is to prove Newman's Tauberian theorem which I'm not actually going to do that because it involves a slightly tricky bit of complex analysis. So we're going on to step three, which is to show the integral from one to infinity of psi of x minus x over x squared dx converges. Here, we just recall that psi of x is the sum over n less than or equal to x of lambda of n where lambda has the property that lambda of a power of a prime is equal to the logarithm of the prime and is zero otherwise, as I discussed in the previous lecture. And what we do to prove this is to be observed that the derivative of zeta of s over zeta of s which we recall is equal to sum over n of lambda n over n to the s can also be written as sum over n of lambda n times the integral from n to infinity of s over x to the s dx by elementary calculus which turns out to be s times the integral from one to infinity of psi of x over x to the s plus one dx. And you're using that formula for psi. So we have zeta prime of s over zeta of s minus one over s minus one is equal to s times the integral from one to infinity of psi of x. Minus x over x to the s plus one dx, okay. And now we're going to apply Neumann's Tauberian theorem and Neumann's Tauberian theorem says that if the integral from one to infinity of f of x times x to the minus s dx converges for the real part of s greater than one and if you can extend this to a holomorphic function with no poles for real part of s equal to one then it converges for s equals one. Well, so if we take f of s to be psi of s minus x over x then we see from this that this function here has no poles for the real part of s equal to one because we showed that zeta of s is no zeros for real part of s equals to one. So Neumann's Tauberian theorem works for this and implies that this function converges for s equals one which shows that this interval converges which is what we needed to show. So now step four we want to show that psi of x is asymptotic to x and this follows easily because if psi is a increasing function so if psi of x is any increasing function and the improper integral integral from zero to infinity of psi of u minus u over u squared du is less than infinity then this automatically implies that psi is asymptotic to x and the proof of this suppose that psi of x is greater than lambda x for some x with lambda greater than one then the integral from x to lambda x of psi of t minus t over t squared dt is going to be greater than or equal to the integral from x to lambda x of x lambda minus t over t squared dt which is greater than or equal to some constant k so if the integral converges this can't happen for arbitrarily many values of x because then the integral will be bigger than k plus k plus k plus k plus k and so on so this condition here can't happen for arbitrarily large values of x so psi of x so the limb's soup of psi of x over x is less than or equal to one a similar argument shows that the limb inf of psi of x over x is greater than or equal to one which shows that the limit of psi of x over x as x tends to infinity is actually equal to one which is what we're trying to show that psi of x is asymptotic to x so this is a slightly tricky piece of analysis but really has nothing to do with number theory it's just a theorem about arbitrary increasing functions so now we've got psi of x is asymptotic to x and from this step five deduces the prime number theorem which says that x over log of x is asymptotic to the number of primes less than x and this follows easily from the following key step which is that log of x is almost constant well you probably don't think log of x is almost constant because if you've seen graphs of log of x it kind of looks a bit like that and that is clearly not constant no we're near being constant well that's because you haven't looked at the graph of log of x for really large values of x so suppose I look at it at a really large scale so here's nought and here's ten to the ten here's nought and here's ten to the ten so what does the graph of log of x look like well it looks like this it's indistinguishable from the x-axis then it's indistinguishable from the negative y-axis there's it's got a very sharp bend in it well that said that log of x is approximately zero so I want to focus in a little bit more on it so let's expand the scale of the y-axis a bit so what I'm going to do is I'm going to look at nought here's ten to the ten and here's nought and here's let's just go up to a hundred and then the graph of log of x looks something like this it's very nearly log of ten to the ten for all values until you get very very close to zero that's because whenever you go down by a factor of e log of x goes down by one well if we go down by two or three factors of e we're already down here but that means that log of this number has only gone down very slightly to two or three or something so we hardly notice it and the graph of log of x looks almost like something with the right angle in it if you look at something on a very big scale and you can see that what this is saying is that log of x so if x is less than ten to ten then log of x is approximately log of ten to ten unless x is very very small so the logarithm of x is very close to being constant when x is very large so now we've got psi x is asymptotic to x and what's psi of x well that's lambda of two plus lambda of three plus lambda of four and so on and what's lambda of n well lambda of p to the k is x equal to log of p so we've got a sum over prime so we've also got a sum over prime powers and now we notice that we can ignore prime powers and that's because prime powers are pretty rare so the number of what's the number of prime squares p you know two squared three squared five squared up to p squared less than x well it's going to be at most the square root of x because the number of squares less than x whether or not their prime squares is going to be at most the square root of x and similarly the number of prime cubes is going to be at most the cube root of x so the number of prime powers less than x is going to be at most the square root of x plus cube root of x plus fourth root of x and so on up to plus some kth root of x where two to the k is equal to x because there's no need to go further than that so k is going to be about log of x times some constant so the number of prime powers less than x that aren't primes is going to be at most you know about log of x times x and this will be very so log of x times root of x and this will be very much less than x divided by log of x so we can ignore prime powers and if we ignore prime powers we see that psi of x is going to be about log of 2 plus log of 3 plus log of 5 and so on and since log of x is approximately so log of p is going to be approximately log of x when x is large as long as p isn't too small so this is going to be approximately log of x plus log of x and so on where the number of terms here is just the number of primes less than x so we see that psi of x if we ignore prime powers and pretend log of x is constant is going to be about log of x times pi of x so since psi of x is asymptotic to x this gives us that pi of x is going to be asymptotic to x over log of x so that ends the sketch of the proof of the prime number theorem if you want to see the details I'm adding a link to a paper by Don Zagie which in particular gives a proof of Newman's Tauberian theorem and all together his proof of the prime number theorem is only about four pages long ok next lecture will be about the Dirichlet's theorem about prime numbers in arithmetic progression