 that we had this quantity which we want estimate and we got this expression on the right hand side for this which is actually quite nice because you can clearly see that some of the terms we can estimate x is of course, x this is some constant whatever it is and this is a function of x which is actually shrinking x x increases this goes down. So, this will not substantially contribute to the quantity and this is bit of an unknown. So, this is something we need to understand, but even if we understand this and we get an expression on the right hand side what we have is that this integral equals this expression on the right hand side and if you recall what we want is that part of this integral from c minus i r to c plus i r that is what is equal to psi x plus an error term. So, we want to get rid of the remaining term here remaining terms here. So, let us look at this once again we have delta d this if you recall was 4 consisting of 4 integrals e minus i r to c plus i r c plus i r to minus u plus i r this is what we are interested in the first one of this and. So, we want to get rid of the remaining 3 and the strategy will be same. So, for choose a large in a value of u and r so it becomes absorbed in the error term. Now, let us pick let us say this one or any one of them we will last 3 of them in order to show that the total integral contribution is very small the standard strategy is take the absolute value which is bounded by absolute integral of the absolute values and inside the integral the integrand is x to the z by z which we can easily bound using taking absolute value, but zeta prime over zeta is something again of an unknown quantity. We want to get hold of the absolute value of zeta prime over zeta in this ranges as the integral varies and we want to put an upper bond on the value of zeta prime over zeta because only that will give us an appropriate error. So, we come back to the issue of understanding zeta prime over zeta, but now this time we do have bounding. So, to bound zeta prime over zeta we are going to make use of what we have learned about it the we have derived some expressions for this in particular if you recall the functional equation that was pi to the minus z by 2 gamma z by 2 zeta z equals. So, we know that this equality holds for all z. So, all that we need to do now is take the log and differentiate what do you get? We get pi to the minus z by 2 that becomes minus half log pi then we get gamma prime z by 2 gamma z by 2 plus prime z over zeta z equals half log pi this is the log characteristic differential of pi to the minus 1 minus z by 2 plus gamma prime. Now, let us pick up one of these integrals. Let us say this one this is the farthest in a sense is both imaginary and real parts are very far away. So, this should be easiest to bound. So, for that we just look at the real z being minus u which you assume to be very large negative number. Now, for such a z 1 minus z the real part of this is going to be very large positive number is u plus 1 actually. Now, when 1 minus z has a large real number part that means on the when you look at the zeta function at this point it is somewhere far to the right of the complex axis or the imaginary axis and there we do understand how this zeta function behave particularly zeta function there is bounded zeta at any of this point is a bounded quantity. In fact, if you look at the zeta function this is where we want to study the value of zeta function and this is 1. Now, beyond 1 for any quantity beyond 1 when the real part of z is more than that more than 1 then the zeta function in absolute value is bounded and as you go further to the right that absolute value actually keeps on decreasing because what you get is this and as you increase real z the exponent of n keeps increasing and so the sum keeps on decreasing. So, it is the entire on the entire part here beyond this line zeta is bounded and is bounded by some absolute constant. Similarly, if you look at zeta prime z what is zeta prime z again on this side we can use the this expression infinite sum some expression for zeta and zeta prime z would be what when you differentiate differential of 1 over n to the z which is same as e to the minus z log n you get minus log n times this right and this gives you an absolute value of bound this and again if you see as for any real z greater than 1 this is also bounded because the log n in the numerator hardly makes a difference because it is n to the 1 plus tilde and as real z keeps on increasing this quantity keeps stays bounded actually keeps on decreasing. So, both zeta prime zeta zeta zeta are bounded by some constant in this part therefore, in this equation this quantity whenever real z is minus u is constant this constant. So, what we are left with therefore, is equals. So, which tells us that if we can bound gamma prime over gamma at all the values then we can get estimated bound on gamma zeta prime over zeta for this part of the integral and gamma is a much better behaved function than zeta. So, that gives us a handle in getting this bound, but still we need to again go back to gamma. So, we started with gamma function to derive the functional equation now we again have to take into gamma function to derive an upper bound on zeta prime zeta. So, we go back to gamma function what would be a guess of this bound. So, what how does the gamma function behave at least on positive integers gamma n is n minus 1 factorial and we know n minus 1 factorial behaves like is bounded by let us say e to the n log n. So, that gives some clue of show you should sort of expect gamma z to be of the form e to the z log z well if everything and we will show that is indeed. In fact, there is something more interesting that we can show, but the roughly that is what we can show, but in order to show that it is not enough to say because on positive integers if we have like this therefore, on the entire complex variance going to behave like this because we need to actually see because the behavior of gamma function on positive integers does not quite reflect behavior of gamma function on the entire complex we already saw that on negative integers gamma function is actually unbounded which is sort of unexpected. So, we will now address in fact, that also show that it cannot be everywhere bounded by e to the z log z because certainly for z equal negative integer is infinity. So, certainly not upper bounded by this. So, we have to be little more careful in analyzing this. So, let us go back to gamma function and here I am going to prove a very interesting property of gamma function which is sort of a functional equation for gamma function gamma z times gamma 1 minus z is exactly pi over sin pi z this is quite remarkable one can do a sort of a quick verification that this equation makes sense by noticing that this gamma gamma z diverges at every negative integer what about gamma 1 minus z where does this diverge. In fact, gamma z diverges as z equals 0 also and if you look at z gamma 1 minus z when does it take value 0 minus 1 minus 2 it turns out it takes these values at exactly z being equal to 1 2 3 4 5 6. So, together the product diverges at all integers and if you look at the right hand side pi over sin pi z this also diverge precisely at all integers whenever there is an integer sin pi z is 0 and therefore diverge. But it actually shows something more it also shows that gamma does not have a pole at any other point in fact this is something I hand waved in the last class when I was trying to show that the 0 free region of zeta function I said these are 0s because of the poles of gamma function and I also said that gamma function has no other poles. Therefore, zeta has no other 0s on this region or in this region but I did not actually prove it I sort of hand waved it because again intuition often fails and we see whatever seems like should hold does not always hold. So, you have to be very careful, but this theorem indeed proves that the poles of a gamma function are precisely at negative integers including 0 because the right hand side has no other poles only at integers and this also has both at least at integers. Except of course, there is one more possibility that gamma z may have a pole and where is gamma 1 minus z as a 0 both have the same order and they cancel each other. But this we can show that gamma z does not have a 0. Let us do that if gamma z equals 0 then what happens zeta if you go to the functional equation if gamma has a 0 gamma z by 2 is a 0 then zeta has to have a 0 in fact if you look at this then zeta has to have a pole there, but that may be in fact we derive that zeta has no other pole using the fact that gamma has no 0s and this actually I did argue out last time. So, just use this the kind of recurrence for gamma function that gamma what is gamma z minus 1 is gamma z over gamma z over z no z minus 1. So, this also 0 now here I am assuming that z is not an integer because z is an integer I know that gamma z has no 0 that is well clearly understood. So, this is gamma z minus 1 0 therefore, gamma z minus 2 is 0 z minus 3. So, at all the way is 0 similarly what about gamma z plus 1 that is z times gamma z that is also 0. Similarly, so gamma z plus 1 z plus 2 z plus 3 z plus 4 all of them are 0. So, that entire if say z was here and this is 0 then all these points separated by integers are 0 now is that possible that is not possible why I will just leave this as a bit of an exercise. What one can show is that the can one show is that gamma z at a large enough point is going to be at least as large as gamma at one of the integer points below it that is if the real part of this is fairly large then this value gamma at this value in absolute terms is going to be at least as much as gamma at one of the smaller integer points and which we know is not 0. So, I will not do that is a simple manipulation here just use the definition of gamma function to derive it. So, this is not possible. So, hence gamma has no zeros but it does have certain number of poles. So, let us now prove this theorem and we will use the definition gamma z times gamma 1 minus z equals or is gamma z t to the is that the right definition no thought there is a z minus 1 here and now let us do a variable substitution t equals assuming here u to be a constant d t is u d v. So, that means this integral is happening inside this part. So, that d u is outside. So, u is a variable outside. So, as t varies from 0 to infinity v will also go from 0 to infinity and t by u is v. So, you get v to the z e to the minus u v plus 1 d t is u d v and t is u v. So, u gets cancelled out here and what we get is v to the z minus 1 e to the minus. Now, this playing around has the idea is that now we can do one integral easily. If you look at the variable u the only occurrence of this in the integrand is e to the minus u v plus 1 and we can integrate that. Of course, we have to exchange the order of integration but that we can do. There are standard theorem which guarantee that you can do this z minus 1 what is the result of this integral e to the minus u times v plus 1 that is going to be minus 1 over v plus 1 e to the minus u v plus 1. And when u is 0 this is when u is infinity this is 0. Of course, when this is u is 0 then this is minus 1 over v plus 1 and there is a negative sign. So, that is gets cancelled and we are end up. So, now we have to carry out this integral where e to the z minus 1 here z is a complex number v goes v is runs over in reals from 0 to infinity. Now, this looks like a bit of a messy integral because there is if only this was only not v plus 1 if it was just v then it would become trivial. But since it is v plus 1 we can either do it through trigonometric function. So, you substitute for v is something like tan square theta. So, 1 plus tan square theta gives you something and you get something and it simplify this. But that is like being clever we know a non clever way of doing this integral which is again through complex analysis. So, to evaluate this integral what we do is we consider this over an appropriately chosen domain. Now, what is the domain that we need to pick that of course, depends on the integrand as well we are interested in this going from 0 to infinity. But if you look at the integrand is it analytic over the entire complex plane z being a fixed complex number of course, if z is an integer then yes it is analytic over the entire complex plane except for a pole at w equals minus 1. But if z is not an integer of course, if z is a negative integer then another pole at w equals 0. But if z is not an integer for example, suppose z is 3 by 2 then in the numerator we have square root of w. Now, square root of w is not analytic over the entire complex plane we notice this long ago because the phase changes as you move around in a circle you end up with you sort of move up. So, there is a you can actually associate a Riemann surface which goes up and then for square root it one more round will take you back to the original Riemann surface. But here it can be any arbitrary complex numbers in general it is not going to be analytic. So, there is standard way of analyzing this which is again putting a cut on the complex plane. So, here the cut that we put is starting from 0 to the entire real positive line we cut out. Now, on this domain this function becomes analytic because now there is no problem of two different values at this point I mean you can easily define the angles or arguments around this and without any problem of getting of course, as you increase from here and go around here the argument will keep on increasing. But it will never you will not get two argument on the same point because this line has been cut out. So, with this domain we know that the integrand is analytic on this, but we need to choose a domain specifically for this over which we want to integrate and that we do the following once we what this cut. Now, this cut works for any such w to the z whenever you have this duplication of values if you take out one cut starting from origin all the way along any direction actually then basically you are what you cut out is all the rotations around the origin which is the what causes you this duplicate values. And besides these points you can now associate a unique value same thing with log z that if you cut out one line then you can uniquely define log z I will not uniquely define they still exist many definition, but at least there is it is with every definition it is analytic on the reduced domain. Domain we choose is of course, this is minus 1 and we pick a large number r here and the corresponding minus r here and take a circle around this of course, this circle will not complete because this line is cut out. So, just before this line the circle start this and makes a tiny circle around the origin because origin is not also not available and then comes back top. So, that is the domain what is the value of this integral inside this domain this given by the cautious formula there is only exactly one pole assuming r is large enough. So, the encloses minus 1 there is exactly one pole of the integrand in this domain and. So, the this equals residue at that pole and the residue is well multiplied with w plus 1 and take the limit w going to minus 1 minus 1 to the which is e to the pi i z minus. So, we know the integral and now we split this using our standard method into this integral into 4 parts epsilon to r that is this part this line plus this integral around this circle w equals r. Of course, this is taken slightly different way that you do not complete the circle just stop little before then you go from r to epsilon again these integrals are taken to mean I am sort of omitting the slight perturbations here because you do not really have the real line available to you. So, the numbers cannot be exactly epsilon and r it has to be epsilon plus a tiny imaginary amount to r plus a tiny imaginary amount and similarly this circle will start from that and then come back, but stop just a little before r. So, r minus a tiny imaginary amount epsilon minus a tiny imaginary amount and finally around this, but this is now counter clockwise sorry this is a clockwise these are counter clockwise. So, now let us bound away some of these integrals. So, that this integral around the circle look at the absolute value this would be absolute value w is r is that. So, just take absolute value everything inside here first doing the substitution that w is r e to the i theta. So, the integral goes from 0 to 2 pi again little more than 0 little less than 2 pi and this is w to the z minus 1 the absolute value of this is bounded by this quantity r to the real z minus 1 and the occur this absolute value here and d w is r d theta and then take out the absolute value. So, you get this and this is less than equal to 0 to 2 pi r to the real z minus 1. Now, look at this denominator and since we are looking at upper bound what is its smallest value smallest value has to be at least r minus 1 cannot it has to be at least this much. So, this is of course, this is all when r is large r r minus 1 take care of themselves r to the real z minus 1 and this goes to 0 provided for real z less than 1. So, we will choose our z when integrating this such that real z is less than 1 and this is something we need because we are not interested in this particular integral. Similarly, if you look at the fourth one and absolute value of this this is less than equal to again 0 to 2 theta epsilon to the real z. So, we have epsilon e to the i theta again absolute value of w plus 1 now what is the smallest value of w plus 1 now again 1 minus epsilon in the same fashion and then epsilon d theta again following the same structure and since now we take epsilon to be very small. So, this is actually order epsilon to the real z because 1 minus epsilon is close to 1. So, you just multiply epsilon to this epsilon to the real z this also I want to send to 0 and this will go to 0 for real z greater than 0. So, which means that here z we pick in this strip any z in this strip is fine. Now, let us look at the remaining two integrals the first one was epsilon to r second was epsilon to r to epsilon plus r to epsilon now here I had going to be a little careful analysis here because what happens what is happening here is here you see that w is starting from roughly 0 epsilon and going to r as you would expect normally. But when you come down here and w is going from r to epsilon it is actually not r to epsilon it is r e to the 2 pi i little less than that actually 2 epsilon times e to the 2 pi i little less. So, that is the distinction I am going to make. So, when I stick that in basically I will get of course, again this is taken with a little bit of perturbation here. But that perturbation we can make very very small essentially make it vanish. So, we do not worry too much about that. So, that is the sum of the two integrals that we are left with and if you just change the order and add the two things up the only thing that survives as it should as you would expect is that this because when if z was an integer then this will not survive this will also go away and then the everything is as normal. But if z is not an integer then this will not go away this will give you a phase shift. And now take the limit r going to infinity and epsilon going to 0 where yeah there e to the 2 pi i is 1 w you replace w by e to the w times e to the 2 pi i and d w is d w times e to the 2 pi i e to the 2 pi i is 1 that is goes away. In fact, one part we can simplify this a little bit of course, because that minus 1 is goes away. And now taking limit epsilon going to 0 and r going to infinity we get the integral that we want and this is equal to what we wanted was 1 minus e to the 2 pi i z 0 to infinity now we do not have to set up this is equal to because the left hand side there is precisely equal to this integral when w goes from r epsilon to r then it is real actually. So, we can just replace w by e this equals we just derived this formula that this is equal to this integral and this part was equal to that integral. So, 2 pi i times this is what we are looking for. So, what we get is tau z gamma z sorry now 1 minus z equals 2 pi i and this is equal to e to the pi i z this we know this is just take the corresponding cos in sin terms. So, you get sin pi z minus 2 pi sin pi z, but notice that this whole thing the entire analysis required z to be in this strip. So, what we have so far shown is that whenever z is in that strip between 0 and 1 then this equality holds. Now on your left side is analytic over the entire complex plane except for pole set at negative in tears. The right side is analytic at all entire complex plane except for pole set in tears and they agree on this strip. Now you invoke your uniqueness of analytic continuation they must agree everywhere. So, we have now this question and we want to now coming back to our original problem which was of trying to get a estimate of gamma prime over gamma. We want to use this to get this estimate only it is not quite sufficient to get that estimate yet, because we have only a product of gamma 2 gamma values giving us this. And we want in fact to get an estimate of a single gamma, because we can take a log and differentiate, but that will give you gamma prime over gamma of z plus gamma prime over gamma 1 minus z. And there combined sum is bounded by some quantity this is we know how this is bounded, but that is not quite sufficient because it is still a sum. So, there may still be divergences which might give us some strange things. However, what we can conclude from this is this and this gives if you look at the absolute value. What is the sine pi z in absolute value? Sine pi z is a difference of two exponential functions e to the i pi z and e to the minus i pi z. If you take their absolute value they are pro like e to the order z times gamma 1 minus z. How does gamma 1 minus z grow? Is there an upper bound to this? Well, you can do that in the following way. One has to look at the real part of this. So, if 1 minus z has real part which is positive then you bring it below using the recurrence equation. Suppose, it is positive. So, then 1 minus z is positive you want to keep on subtracting once from this. So, you want to go to minus z minus 1 and so on like this. Suppose, k is such that the real part becomes very small. Then we can again invoke that property of gamma function which I just asked you to prove that gamma z for a when the real z is positive is sandwiched between the two gamma values which are integer not too far away from those. So, when z the real part of this is very small then it is some small number in absolute value. So, you can just write this as e to the order z and ignore the z part. How much is this? z plus 1 up to z plus k minus 1 this is a product of how many of z k of them and k is what? k is bounded by absolute value of z. So, this is something like z to the order z. Now, this is interesting because this tells us that 1 over gamma z grows its growth is always upper bounded by this which is something we can completely believe in because gamma has no zeros. It has poles, but 1 over gamma therefore has zeros as those points. So, there is it is always upper bounded by some quantity and of course, it has no poles. So, this means 1 over gamma z is an entire function of order 1. I did define entire functions in the long term. Do you remember what are entire functions? Entire functions are analytic over the entire complex plane no poles those are entire functions and order 1 is something I have not defined which I will define now. So, now of course, we must realize that our main job which was to say something about the prime counting that keeps getting interrupted. I do something then interrupt go into a diversion then come back do something more interrupt another diversion. So, now we are in another diversion which was analyzing gamma function and now this is now I am going into the second order diversion that is interrupt the analysis of gamma function and created diversion into the theory of entire functions, but I limit myself to entire functions of order 1. One can easily generalize it to entire functions of any order any of you seen inception. So, it is a bit like that not quite as complicated, but still like you pause a level and go down a level and then do things there. So, we will come back I promise that I will come back all the levels and then close it this thing by the end of the course. So, what are entire functions? We know entire functions. So, f is entire everything analytic on the entire complex plane. We say f is entire of order k if absolute value of f on z is bounded always by e to the order mod z to the power k or we allow a little more than k plus 0.1. So, this obviously means its entire because it is at every point on the complex plane its value is bounded. So, now of course which means entire functions of order 1 are when your k is 1. Now these functions are the most well behaved analytic functions and therefore, we can say a lot about them. The simplest one of these are entire functions of order 0. Those one can show and the when I do the analysis for order 1 you can use those tools to show for order 0 that these functions are polynomials. The polynomials grow as a polynomial of degree d will grow as about z to the d mod z to the d which is in the exponential when you take the exponential it is e to the d log z and log z is bounded by any z to the epsilon. But we are not interested in that we are interested in entire functions of order 1. Now what we are target is to derive an expression for entire functions of order 1 and we will see that entire functions of order 1 have a very fixed expression you can practically write down what the function looks like. But before I prove that let me further specialize. So, if I say an entire function of order and has no 0s every anywhere then f can have or f will have precisely this form e to the a z plus b for some a z a and b being complex numbers. Clearly this is an entire function of order 1 e to the a z plus b and it clearly does not have any 0s and this theorem says that that is all if an entire function is of order 1 and has and has no 0s this is the only possibility and the proof is quite interesting is not straight forward is defined the function g as log of f minus log of log of f z minus log of f 0. Now whenever we take log with some alarm bells will start ringing because log is not such a simple function here. So, but it turns out because of the fact that f has no 0s anywhere the log is defined very nicely see what why was log causing a problem for example, log z log z was causing a problem that we travel around circle around a circle around 0 you do not end up at the same point you go a level higher if you think of it in terms of Riemann surfaces and you keep high by high. But that was if you think about it carefully with I will not explain it too much, but if you think about it you realize that that is happening because we were travelling around 0 which was a 0 where z is 0 that is we are we are looking at log of f z and we are travelling around a 0 of f z and that is when all this funny things happen. If you took log z the function log z itself and took it made a circle around z equals 1 a small circle around z equals 1 everything will be perfectly defined analytic there is because again if you think about it in terms of Riemann surface if you just travelling around z equals 1 you will not change the level you will come back to the same level you only change level when you travel around z equals 0 and that happens because it is a there is 0 at that point now because f has no zeros the log is very cleanly defined you there are no funny things happening here. So, we can actually we do not have to actually assume some funny Riemann surface for defining this function we can define it over the complex when itself and conclude that it is still analytic everywhere because there is no 0 g that is defined everywhere perfectly and because I am taking difference of two logs I can take any definition of log the difference will be again the uniquely defined. So, function g is also entire and what g 0 is 0 again force by the definition now the fact that f has order 1 this means that absolute value of z is bounded by e to the z to the 1 plus epsilon what is real of part of g z and this the z is bounded by e to the z to the something. So, we can say that this is bounded by this is a bit of a dampener because what I would ideally like is an absolute bound on absolute value of g z in terms of absolute value of z and meaning will the reason will be clear very soon, but what we get directly out of the definitions is that real part of g has this bound, but now we use the fact that so what we want is to put a get a bound derive a bound on absolute value g and we will use the fact that g is analytic everywhere and this is general property of analytic function actually if it function analytic everywhere which is an entire function then if you can bound it is real part roughly the same bound hold on its absolute value also. So, let me just collect this fact in a lemma not for f z the f z make send z to the complex part or the that is the complex part the argument could be large of course, you can reduce it sure, but then you lose the analyticity, so continuity is broken if you shrink it the f z is growing in a continuous way. So, it is argument part is also growing if you use the fact that subtract 2 pi i if it is goes above 2 pi i then you lose the continuity and then g when g does not remain analytic and then nothing will work. So, that is a very good point, but one has to keep these issues in mind. So, if you can put a bound on the real value of analytic function g. So, this here the function g has to be analytic only in this disk z f less than equal to 2 r in that case when a side is smaller disk the absolute value of g is not too much bigger and this is a neat trick this lemma basically I mean the proof of this is a neat trick. So, what you do is let me see if I can figure this out define this one g hat z. Now, is g hat analytic in fact g hat we can conclude that g is inside this disk why you only have to worry about the poles what are the poles of g hat is z equals 0 a pole no g z is also 0 g 0 is 0. So, z equals 0 is not a pole. So, that pole is taken care of, but what about this other one when g z is 2 m then it is certainly a pole right, but we see assumption of the lemma says the real part of g z is at most m inside this disk. So, g z is never 2 m therefore, that is also inside the disk that pole does not exist. So, inside the disk g hat is analytic. So, there is a lemma of Cauchy integral formula which says that for an analytic function defined on a disk the value of the function the absolute value of function inside the disk is at most the value maximum of the function value around the disk around the edges of the disk. So, that is the function attains is maximum only at the circumference of the disk and that is easily proven if it attains is maximum somewhere inside then you look at that point and take a tiny circle around this within the disk. Now, that circle now you apply Cauchy integral formula and that will you can express this value which is inside the disk which is a maximum as a integral of the values around at the circumference which are all smaller and that will tell you that there is not possible. So, if you just think of Cauchy integral formula is taking the average of the values around the circumference then it will you can immediately realize that the value inside is only the is bounded by the values around the circumference. So, I will just invoke that g hat w is less than equal to g absolute value g hat z when z is 2 and what is this absolute value of z is 2 r. So, this cancels what is absolute value of g z all that is actually equal to square root of alpha square plus beta square where you assume that g z is alpha plus i beta divide by 2 m minus alpha whole square plus beta square square root. Now, notice that use the fact that real z is at most m real g z is at most m. So, alpha is at most m which means 2 m minus alpha is at least m and of course, beta is the same. So, that means, numerator is at most is less than equal to the denominator because this quantity is at most m and this quantity is at least m. So, this is less than equal to 1. So, this says that g hat everywhere inside the disk is at most 1. So, now let us relook at this expression and write g in terms of g hat we get g z equals on the side 2 r g z equals g hat z time z. Now, for absolute value of z equals r what is absolute value of g z absolute value of z is r. So, this 2 cancel out absolute value of g z is let me just write it for once this is at most 1 this 2 cancel out. So, you get m divide by 1 plus 2 r g hat z. So, now in the denominator there is 1 plus something. So, how small this quantity can be? So, which is certainly if you can say that if you look at the maximum value of this second part and if you can bound that to be less than 1. So, this quantity in the denominator is 1 minus the maximum value. So, let us look at this for absolute value z equals r this is r. So, this is half here and absolute value of g z is at most 1. So, absolute value of this second quantity is at most half. So, absolute value of this is at most half. So, this means that going back to our function because a real g real part of g was bounded by this using the next lemma we conclude that equals an absolute value of g is also order. So, we get a bound 1 absolute value of g and now the life is very simple. Just using this we can easily derive that g must be of the form a z plus b. In fact, not even a z plus b g must be precisely a times z why? g is analytic function over the entire complex plane. Therefore, it has a power series representation which is valid. So, this power series expression for g is valid over the entire complex plane taking it around 0 and now let us just look at r to the i theta I will write it in terms of polar coordinates. Just multiply this thing quantity by e to the minus i k theta integrate from 0 to pi you get this. What do you say about right hand side? This integral when m is not equal to k this is 0. So, this is non 0 even only if m equals k. So, just writing it the reverse way when m equals k we get this. So, you get c k r to the k and then the value of this integral is I think 2 pi i or 2 pi that is equal to just take the absolute value and what is this? Absolute value of this is on this fold for any value of r. So, what do you get out of this? Absolute value of c k times r to the k is order r to the 1 plus epsilon k is at most 1. If k is 2 then you get c k c 2 absolute value of c 2 times r square is order r to the 1 plus epsilon for any value of r that is why not true or actually I should say c k equals 0 for k greater than 1. So, this means g z is c 0 plus c 1 z but g 0 is 0. So, g z has to be exactly c 1 times z now recall the definition of g that is log of f minus log of f 0. So, this gives f z to be e to the f 0 now well f 0 that is simply f 0 times e to the g z f 0 and that is precisely the form we were looking for. So, all entire functions without zeros in fact this can be extended as I said for higher order entire functions also order k entire function will have a degree k polynomial in z in the exponent that is all it has to be e to the a degree k polynomial. Now, but what about the rest of the entire function entire functions with zeros this is the entire function without zeros which have a very nice formula and as it turns out entire function with zeros also a very nice formula. So, that is an x 0 does not have a 0 or we use firstly to define g you cannot even define g if f is a 0 and you cannot define g everywhere. So, that means g will not be analytic over the entire complex plane and then therefore, the power series representation for g will not hold for the entire complex plane here if you recall we require it to hold over the entire complex plane because this equation will fail to hold only when r is large enough. So, that is important, but another point to note here is I do not need it to hold this bound on this bound on absolute value of g and that is the fact that will come use of use to us later on that this bound on g does not need to hold for all values of r as long as it hold for infinitely many values of r ever increasing then it is good enough to derive this conclusion because all we need is a large enough value of r for which this equality holds which will force c k to be 0. By the way if f is an entire function of order 1 with a finitely many zeros then its expression is very clear let me just write that as a theorem and with an f z is simply counting multi series. So, this is straight forward because if there are these are the zeros of f you divide f by this product then it becomes a function with no zeros it still remains an entire function. So, therefore, it has that form and what is more interesting is an entire function with infinitely many zeros because it has no poles but f is 0 at z i. So, now if we have infinitely many zeros then what should f be that this one was easy to guess that removes all the zeros and it separates out all the zeros and the rest is entire can we write the same thing for here can we write for example like this we cannot for very obvious reason can you see that yes exactly does not for it most of the z it is unbounded to this product infinite product and just to take a particular example suppose z i is 1 2 3 4. So, infinity and you take z to be 0 then this product would be 1 times 2 times 3 times 9 4 or this unbounded product. So, it is does not even capture the entirety of function f. So, this is there is you although you would like to write it something like this we cannot no it is not 0 everywhere there are entire functions with infinitely many zeros 1 over gamma we just showed it is an entire function and it has infinitely many zeros at all negative integers. So, this was a beautiful insight by Hadamard in late 19th century when he studied this entire functions and gave a complete classification of that and the key insight was to guess what should be this form the idea is remains the same that you want to just take the product of z minus z i is in some way, but still to make it converge and what he came up with is the following. So, this is the part which takes care of all the zeros product over i greater than equal to 1 minus z over z i. So, these are these are this is where so, this part captures the zeros, but to ensure the convergence we multiply it with e to the z over z i and then take the product over all i. So, this part and we will see when we see the proof of this theorem is an entire function with precisely zeros at z 1 z 2 z 3 and so on. And therefore, when you divide f by this you get an entire function with no zeros which has the value of the form e to the a z plus b that is a very good point yes. So, I should stick that in 20 minutes zeros if I have zero is zero let us say of order k then you just multiply this by z to the k in the beginning why does not make it unbounded we will see next class we will see.