 So, this is the lemma we are going to prove today amongst other thing of course, and I might have to come back and modify the statement of this lemma this square root u could be 1 over square root u, but we will see good. So, the Fourier what is Fourier transform that we start with definition. So, it is defined function f n as f s. So, all I have done is extend the definition of function n from integers to all numbers. Now, Fourier transform can be defined for just about any function as long as it satisfies some reasonable convergence property, but there are two distinctions depending on what the function is like. If the function is periodic with a certain period, then you only consider the that period the interval of the period you know to define the Fourier transform, because after that it is all same. On the other hand if the function is a periodic you just take that the period is infinity and the constant the whole function. So, in this particular case this is not periodic there is no clearly no period here. So, we just define the Fourier transform to be t integral from minus infinity to plus infinity just take the entire span for the function f of s e to the minus to multiply the function by e to the minus to pi i s t and integrate from minus infinity to plus infinity. And of course, one has to see whether this there is this is well defined and for that one can or one should show that the absolute value of this integral converges. In case of the function we are looking at certainly it does convey this is really very very rapidly decaying function. So, that is no problem now question is what is this will we know is minus infinity to plus infinity e to the minus pi s square u e to the minus 2 pi s t. And what I want to show is that this is actually same as the original function that is the target or not quite the same there is a up to a multiplication factor by square root of u. Now, for that we will use a little trick just observe here there is a pi here a pi here. So, pi can be taken out in common. So, let us actually just write this in a slightly more simple form then you have s square u plus 2 i s t s square u plus 2 i s t this almost suggest that we should try to complete the square and see what we get out of this what you need to complete the square minus t square by u. And of course, plus t square by u now this last k has nothing to do with s. So, we can just pull it out it is basically this expression is s square root u plus i t by square root u. Now, all then we need to do is now we just let z be s square root u plus i t by square root u. So, so far s u t u s u t were all real now we move into the complex number let me let z be this and express this integral as a in terms of z. So, where does z go from when s goes from minus infinity to plus infinity what happens to z. So, it just goes from minus infinity plus i t by square root u plus infinity e to the minus pi z square that is what you get here and what happens to d s. So, u t are fixed they do not vary. So, just take d z is d s root u. So, d z by so we have integral this form which is somewhat simpler except that now it is a complex integral. Now, we know some ways of handling complex integrals. So, let us try to see what happens with this if you want to integrate it from minus infinity plus this to plus infinity plus the same quantity that is like a where does this go from t by square root u which it is maybe somewhere here and the integral is along this line minus infinity plus infinity. So, we can use the standard tricks that is just integrated from minus r to plus r and then we will send r to infinity. So, we consider minus r plus r and this integral again will write as from here to here. So, let us define a contour whichever and let us take a rectangle and let this be the domain d. So, what is the integral of this e to the minus pi z square d z along the domain boundary of this domain d this you should know there is no pole the integral has no pole anywhere. So, this is 0. So, this implies that 0 equals now if you see that you are traversing it counter clock wise minus r plus t by square root u plus t by square root u. Now, if you consider out of these the second and fourth integral these two are more or less same similar to each other except that you have one is minus r one is plus r going from here to here another is going from here and you have you are integrating e to the minus pi z square. If you look at the absolute value again using the similar ideas that we have been doing what is the absolute value of this e to the minus pi z square as we move along these two vertical red line e to the minus pi is r square. So, at least e to the minus pi r square and since this is not in the denominator we can say that it is bounded this integral is bounded by at least the integral and is bounded by 1 over let me just write this. Let us just pick one of them say r plus this is less than equal to and then this is of course, bounded by order assuming t is and v is t u v to be constants. And as r goes to infinity this goes to 0 and the same happens with the fourth integral and this gives that the first and third integrals are equal when you take the limit. This has nothing to do with integral this comes out and then you are integrating d z is not even r this is actually v minus t by square root u. So, that is just one assuming that v t u these are all fixed numbers and it is r that I am sending to infinity. So, it is so that that basically says that the second integral and the fourth integral vanish as r goes to infinity which means that sum of first and third integral is 0. So, third integral is in the opposite direction minus 2 plus if you swap this you basically get integrating along this line is same as integrating along this line which therefore, tells us that if you look at this you are integrating along this line instead of that if you integrated along any other line horizontal line from minus infinity to plus infinity the result will be the same. So, that is the first step of our evaluation of the integral. So, what we can say is therefore, in particular this equals let us forget about the complex part just integrate along the x axis the real x. So, this is a further simplification we are now at the simple integral over the which is a complex integral we are now got it as a real integral. Although it is sometimes easier to evaluate real integral through complex integrals, but in this case it is easier to just evaluate the real integral directly and this is integrated by a very neat trick actually the integral of this is 1 and to show that let us just give it in a minus infinity to plus infinity e to the minus pi by square d y this is a normal distribution basically the and you are just saying that this is the probability mass is 1 which is all. Can you prove this that this integral will be 1? It is very simple actually just look at i square this is against nice trick as I said right. So, multiply i with itself you get double integrals and this is now an area integral over a plane where you take the dx dy is a rectangle and integrate along value along that and is add up all. So, it is integral over the entire plane two dimensional plane not a complex plane, but normal r 2 right. Now, this integral I can rewrite using polar coordinates. So, polar coordinates will be r and theta how r will be going from 0 to infinity and theta will be going from 0 to 2 pi what happens to the integrant e to the minus pi x square plus y square x square plus y square is r square that is minus pi r square. What happens to dx dy? dx dy is a tiny rectangle that I am going to change with here is a dx dy. This I will replace with in the polar coordinates. So, this is x in the polar coordinates I will replace with r and this is d theta with this piece. So, it is r and this is d r and this is d theta this area is r d theta d r is pretty standard, but I thought I will just mention it anyway. Now, you see that this is because there is no d theta to you get 2 pi here 0 to infinity e to the minus pi r square r dr and this you integrate you get 1 the integrating this is pretty straight. Just do a k equals r square and so now going back to where we started from what did we get we showed now we have shown that this integral is 1 and therefore, my Fourier transform f hat of t is equal to 1 over square root u e to the minus pi. So, I do need to change the statement this is minus. So, that is the complete proof of the lemma. So, of course, what I have shown is much more general thing this Fourier transform holds at every point not only at integral point, but all I need to use it for is integral point. So, that now if you recall we started with this infinite sum and each term of the infinite sum was a e to the minus pi n square u and we took one term and then calculated its Fourier transform, but it is really this infinite sum we are interested in. So, let us now get back to the infinite sum w what was it w u. Let us look at this particular function little carefully. Let me define another function associated with let us say capital F of let us define f of v as again a generalized version of w where again u is something I fix and then we have this infinite sum minus infinity plus infinity to the minus pi n plus v whole square. Now, for this we make this observation f v is f v plus 1. Now, is this obvious this is very obvious because replacing v by v plus 1 you just get n plus 1 plus v whole square n going from minus infinity to plus infinity. So, n plus 1 will also go from minus infinity. So, this is a periodic function with period 1 and now again I will work Fourier analysis on this function which is a different one than the previous this is a periodic function. So, for periodic function the Fourier analysis shows which I will again show you in 2 minutes that f v can be written as this infinite sum. So, it is a infinite sum of e to the 2 pi m v these are all each one of this is a periodic function with period 1 and this you get this each one of them multiplied by C m which we call the Fourier coefficient corresponding to this particular this is really standard transformation of the function as a representation of function as a sum of sine waves. How do we get this well that is pretty simple let us just define C m out of this C m would be for this let us see how do we identify or extract out C m. So, this gives So, this is equal to of course, 0 to 1 you substitute this make this n is to separate this out now again using uniform convergence of things we can swap the sums with integral and this integral is easy to evaluate this is going to be 1 if and only if m equals n otherwise is going to be 0 and so this is equal to C what is that C n. So, this is this integral is non-zero exactly when m equals n. So, all the terms in this infinite sum will vanish except when m equals n and then it in that case this integral takes a value 1 and so you get exactly C n. So, that is why you can express this function which is periodic in terms of this infinite power series. Now, what do I want to show you if we was this infinite sum let us look at the C n in a slightly different way see this also can be shown easily that C n actually equals this integral I will not show I will leave it for you to work this out. So, this is equal to 0 to 1 if you substitute for this for every m going from minus infinity to plus infinity and every was e to the now this integral with this infinite sum you can merge is an integral going from minus infinity to plus infinity and what happens to these guys see v was going from 0 to 1. Now, I will make v go from minus infinity to plus infinity what happens to this making sense here what was the Fourier transform difference see what I want to show is that this is equal to and actually the exact form of feta n does not matter this is a general result that Fourier coefficients of this periodic function which is an infinite sum the capital F was defined as this infinite sum and this is a periodic function with period 1 for such a function the nth Fourier coefficient is actually equal to the Fourier transform of its nth term can you see how to prove this. So, this is C n that is how you start with and if you plug this in okay let us leave it let me give you we are spending too much of time on this let me give you this as an assignment anything that I cannot prove I am going to give you as an assignment it requires some manipulation of this. Now, let us get back to now we are now close to what we want to prove. So, if you recall f v is this infinite sum C m e to the minus 2 by i m v right f v by definition gives me I am going from minus infinity to plus infinity of e to the minus pi m square u m plus m plus v square u and this is equal to m going from minus infinity to plus infinity C m is this feta n. So, feta n is we already know 1 over square root u e to the minus pi m square by right times of course, e to the minus 2 pi i m v. So, we get this relationship now in this relationship again we had proved a much stronger relationship then we need just plug v equals 0 what do we get in from minus infinity to plus infinity plug v equals 0. So, we get e to the minus pi m square u equals this multiplier is all 1 when v is 0 this is exactly what we needed because this is w and this is 1 over u square root u w. So, we can conclude that w u is u to the minus half and the whole point of this exercise was to derive this this relationship between w u and w 1 by u and let us go all the way back where we started from that same let w with w 1 by w u w 1 by u and why did w u arise out where did this come from. We had this capital W u which is related to small w u by this formula and this capital W u actually occurred here in this relationship with of zeta function and gamma function in this inside this integral. So, let me just lift all of this and write a fresh there, but let us first establish capital W u was half of small w u minus half this is what it was capital W u is half of small w u minus half. So, we can rewrite this relationship in terms of capital W what do we get w u is therefore, 1 plus 2 capital W u is equal to u to the minus half 1 plus 2 capital W 1 by u with this case here the w 1 by u and this is what will be eventually interested in is half of and this is the relationship we should remember this is the take away from this. Now, go back to that Riemann equation which are not Riemann equation, but the zeta function equation and let us create a new page what did we have zeta z pi to the minus z by 2 that gamma z by 2 equals this integral which says 0 to infinity u to the z by 2 minus 1 w u and let me refresh your memory that we had a problem in the convergence here when real z is less than 2 because then this part sort of diverges as u comes close to infinity 0. So, the problem really is when u is close to 0 then this may not converge and that is what we want to get along get away from. So, let me split it in 2 integrals this is a bad integral 0 to 1 this where bad things happen plus 1 to infinity there is no problem in 1 to infinity in 1 to infinity this integrant converges absolutely. So, this integral converges to something a sensible amount no matter what the value z is because in this range 1 to infinity w u which is a e to the minus u or or worse that is or even faster decaying one that will dominate u to the z. Let us consider the bad part 0 to 1 and let us do a change in variable replace u by 1 over v what happens to this when u goes from 0 to 1 what happens to v goes from infinite comes from infinity to 1 u to the z over 2 is becomes u to the sorry v to the 1 minus z by 2 w u becomes w 1 over v what happens to d u it becomes d v over v square with a negative sign are you with me so far. So, since there is a negative sign outside and the limits are also reversed we can flip that both you got from 1 to infinity v to the minus 1 minus z by 2 because there is v square dividing this w 1 over v. So, everything is sensible here except w 1 over v, but we know how to handle w 1 over v that is 1 to infinity v to the minus 1 minus z by 2 w 1 over v we just go back and stick this is therefore half u to the v to the half 1 plus 2 w v. So, I will leave it as a workout at home.