 Hello. So, we continue with this chapter on Fourier transforms and this will be the last capsule in this chapter. Here we shall discuss a remarkable formula discovered by the great Indian mathematician Srinivas Ramanujan. A few words about Ramanujan. You know that the movie has been made a biopic on Ramanujan and I hope you have seen this and if you haven't seen it already, I urge you to see it. It is a very well made movie and the movie itself is based on a book written in 1991 by Robert Conigal, the man who knew infinity life and genius of Ramanujan. Well, the gamma function featured prominently in the works of Ramanujan. If you look at the notebooks of Ramanujan which has been published by Springer and we just flip through the notebooks, you will see large number of formulas the gamma function features. One of these formulas concerns the Fourier transform and that is a formula that is displayed in this slide. It is a Fourier transform or the absolute value square or the gamma function restricted to vertical lines in the right half plane. And this formula appeared in 1915 in the messenger of mathematics and this was cited by G. H. Hardy in his collected papers, volume 7 and you can look at page 98 in following. The formula of Ramanujan as Ramanujan proves it is quite tricky and we shall offer a transparent proof of this based on the principles that we have studied so far. Now, in the next few slides we will get to the proof of the Ramanujan formula. So, what is the Ramanujan formula? Look at the slide integral minus infinity to infinity mod gamma a plus i t the whole squared exponential of minus i t chi d t that is a Fourier transform. What is this Fourier transform can be obtained in closed form namely square root of pi gamma a gamma a plus half hyperbolic cos to the power minus 2 a chi by 2. A is real and a is positive. Well, it is a significant that a is positive because you look at the thing on the right hand side. The hyperbolic cosine grows exponentially fast. So, cos to the power minus 2 a decays exponentially fast and if a were negative that is no longer true. So, it is essential here that a should be positive. So, let us look at the proof of the Ramanujan formula. But before we plunge into the proof of the formula, we must make a few observations. Let us look at this function t maps to mod gamma a plus i t the whole squared as a function of the real variable t. This function decays very rapidly. So, rapidly that this function is in l 1 in fact, it is much better it decays exponentially fast and because it is in l 1, we can compute the Fourier transform using the formula for the Fourier transform. So, before you take this up, let us recall the sterling's approximation formula n factorial is approximately n to the power n into e to the power minus n into square root of 2n pi for large values of n. Now, this formula was given by James sterling way back in 1730 in his Methodist differential is almost 300 years ago. This formula is unarguably the most remarkable formulas in classical analysis. We are talking about a time when analysis has not developed the way it has developed today. Today, you know a lot of analysis, but analysis was in the very nascent stages of development. It is difficult to imagine how sterling would have hit upon this right hand side n to the power n into e to the power minus n into square root of 2 pi n. What will be the analog for the sterling's approximation formula when n is not a natural number, when n is a positive real variable? Well remember that n factorial is gamma n plus 1. So, the sterling's approximation formula reads gamma n plus 1 is n to the power n into e to the power minus n into square root of 2n pi. Simply replace n by x. What do we get? Gamma x plus 1 approximately equal to x to the power x into e to the power minus x into square root of 2 pi x where as x goes to infinity. This is an asymptotic formula and this also holds. Numerous proofs are known for this, but we will need a analog of this sterling's approximation formula not for real x, but for complex values of x. What value of x am I going to take? a plus i t minus 1, a plus i t minus 1 that is the value of x that we are going to work with. So, now let us look at this formula in the complex domain. So, in the complex domain we are going to assume that z lies in a certain sector r z less than pi minus delta. Remember capital A when I use capital A for the argument it is a principal argument the value of the argument between minus pi and pi. So, what you do is that you take the complex plane you take the negative real axis and cut out a small sector with opening angle 2 delta. So, the with the origin at the origin draw two rays going to infinity enclosing the negative real axis with opening angle 2 delta. So, the argument of z the principal argument of z varies between minus pi plus delta to pi minus delta. So, in that sector we are going to work and this delta is going to be fixed for the rest of the discussion and this sector is going to be denoted by r subscript delta r delta is this sector and the value of z is going to be very large modulus of z is going to be much bigger than 1. Now, what is z to the power z plus half? Remember there is x to the power x and root x. So, x to the power x plus half. So, we should replace it by z to the power z plus half. How do you define z to the power z plus half? It is exponential of z plus half log z where log z is ln mod z than real logarithm of a positive real variable plus i times argument of z where I take the principal argument. So, this is the way you define this piece z to the power z plus half rest is easy the exponential function is defined everywhere in the root 2 pi is an innocent constant. And so, now what we need to show is we need to show that limit as z tends to infinity while it remains in the sector r delta this limit is taken as z goes to infinity within this sector. In short as z goes to infinity z is not supposed to approach the negative real axis arbitrarily closely. The fact that z remains in this r delta forbids z from coming close to the negative real axis. So, this is the limit that we need to show this limit formula is what we need to show or in other words gamma z plus 1 behaves like z to the power z plus half e to the power minus z root 2 pi as z becomes large in this sector. The proof of this we cannot do it here I will give you a reference B. C. Carlson's special functions of applied analysis academic press 1977 pages 45 to 47 it is a very nicely written argument and you can read it. Since mod gamma a plus it is the same as mod gamma a plus it plus 1 times a squared plus t squared to the power minus half. Remember the functional equation for the gamma function gamma x plus 1 is x gamma x that is what we are used. It suffices to show that mod gamma a plus it plus 1 is in the Schwartz class as a function of t and we are ready to use the Stirling approximation formula. So, now let us look at the vertical line the vertical line a plus it. So, this vertical line will certainly lie in this region for any choice of delta and let us look at the approximation of root 2 pi gamma a plus it given by Stirling's formula. We can look at exponential of a plus it plus one half log a plus it minus a plus it. So, we need to estimate this the absolute value of this not to estimate the absolute value of this we need to look at the exponential of the real part of this piece a plus it plus half log a plus it minus a plus it this is what we need to look at ok. Remember that mod of e to the power w is e to the power real part of w all right. Now, I am going to take the real part and I am going to ignore this multiplicative constant exponential of minus a and we have to estimate this piece x of a plus half ln mod a plus it minus little t times capital R the principal argument of a plus it. So, we have got to multiply several things we have to multiply these two complex numbers and then take the real part and this it goes away when I take the real part and e to the power minus a is an innocent constant. So, all these can be ignored and when I write this out I get exactly exponential of a plus half ln of a plus it how does that behave that behaves like t to the power a times another factor which remains bounded that is very easy to see right ln of a plus it what is that ln of root a squared plus t squared and as t becomes very large that will behave like basically ln t for all practical purposes this piece behaves like exponential of a log t or t to the power a and whatever else is remains is a bounded factor. So, this piece has been taken care of let us look at the other piece exponential of minus t principal argument a plus it where does this go this converges to exponential of minus t pi by 2 times something that is bounded the second factor is bounded well why does this happen as t becomes large remember that you are in the right half plane you are in the right half plane and t is very very large t is very very large and so what is the principal argument a plus it it will converge to pi by 2 as t becomes very large. So, this this is a behavior of the exponential factor this proves that gamma a plus it decays exponentially fast and we can compute the Fourier transform of gamma a plus it squared directly using the definition. Observe that since t and a are both real gamma of a plus it squared modulus equal to gamma a plus it into gamma a minus it and now you need to use the beta gamma relation what are the beta gamma relation gamma p into gamma q equal to gamma p plus q into beta p comma q my p is a plus it and my q is a minus it and so gamma of p plus q simply gamma to a and a is constant and I am left with the beta factor beta a plus it a minus it what is the beta function by the way the beta function is given by beta p q equal to integral 0 to 1 t to the power p minus 1 1 minus t to the power q minus 1 dt for real part of p q bigger than 0. This integral let us make a change of variables let us put t equal to 1 upon 1 plus e to the power u and make this change of variables this integral transforms to this next display beta p q equal to integral minus infinity to infinity exponential of q u by 2 minus p u by 2 d u upon e to the power u by 2 plus e to the power minus u by 2 the whole thing to the power p plus q. Now remember one very important property of the beta function beta p q equal to beta q p that is very easy to see because of this integral simply replace t by 1 minus t simple change of variables will convince you that the beta function is a symmetric function of p and q. So, in this equation I can exchange the roles of p and q and I can add the two results and what am I going to get I am going to get beta p q equal to exponential of q u by 2 minus p u by 2 plus exponential of p u by 2 minus q u by 2 divided by e to the power u by 2 plus e to the power minus u by 2 the whole thing to the power p plus q the two which I am going to pick up here when you add the two has been brought over here and everywhere you see a u by 2 and so the u by 2 has been replaced by u and you get the next result exponential of q u minus p u plus exponential of p u minus q u upon e to the power u plus e to the power minus u the whole thing to the power p plus q du my p is a plus it and my q is a minus it we finally get gamma a plus it mod squared equal to gamma 2 a into integral minus infinity to infinity what happens with denominator p plus q is 2 a here you see e to the power u plus e to the power minus u the whole thing to the power 2 a and the numerator is exponential of 2 i ut plus exponential of minus 2 i ut this is a perfectly well behaved integral because a is positive and you see a denominator which is growing exponentially fast and the numerator is pretty innocent sum of 2 unit complex numbers now this integral as far as it goes is a perfectly well behaved docile object now we want to compute the Fourier transform you multiply by e to the power minus it chi and must integrate with respect to t remember that this gamma function decays exponentially fast so there is no problem in taking its Fourier transform directly using the integral so multiply by e to the power minus it chi both sides take the e to the power minus it under the integral and just combine the exponentials into one exponential and then let us write the integral so what do we get the Fourier transform is what gamma 2 a integral minus infinity to infinity dt what is the inner integral integral minus infinity to infinity the denominator you have got the exponential factor e to the power u plus e to the power minus u the whole thing to the power 2 a this grows very fast and the numerator e to the power it times 2 u minus chi plus e to the power minus it times 2 u plus chi d u it is very tempting to switch the order of integrals so let us do that and see what happens the t integral is done first and the u integral is done later and the u integral is d u by e to the power u plus e to the power minus u the whole to the power 2 a let us look at what are the t integral from minus infinity to infinity e to the power it into 2 u minus chi plus e to the power minus it into 2 u plus chi d t here we run into the same difficulty that we have encountered before namely we have to deal with an oscillatory integral this object here is definitely not in l 1 it does not even decay it remains this absolute value is 1 and so this object is not going to be Lebesgue integrable so how do you make sense out of this integral with respect to t if you go back to the proof of the Fourier inversion theorem you will see that the same situation happened there you had an oscillatory integral exactly of the same kind and how did we circumvent the problem we circumvented the problem by resorting to what we are described as e to the power minus epsilon t squared trick so that is the e to the power minus epsilon t squared tricks and you simply put in the e to the power minus epsilon t squared right here before switching the integrals and then once you put this e to the power minus epsilon t square then you are allowed to switch the integrals so that is exactly what we need to do we need to introduce this e to the power minus epsilon t squared inside do not switch the order of integration before doing so introduce this e to the power minus epsilon t squared and then switch you safely switch the order of integration the limit is outside as epsilon goes to 0 gamma 2 a and I done the switch my integral minus infinity to infinity du upon e to the power u plus e to the power minus u the whole thing to the power 2 a integral minus infinity to infinity e to the power i t into 2 u minus chi minus epsilon t squared plus e to the power minus i t into 2 u plus chi minus epsilon t squared dt here you see a sum of two integrals each of which is a Fourier transform of a Gaussian we are looking at the Fourier transform of e to the power minus epsilon t squared the Fourier transform is evaluated at 2 u minus chi what is the Fourier transform of the Gaussian you need to go back to the early parts of this chapter and you know the formula there is a root pi by root epsilon will come out as a constant factor it is another Gaussian x of minus 2 u minus chi the whole squared upon 4 epsilon in the other over obviously it is going to be x of minus 2 u plus chi squared upon 4 epsilon so that is what you are going to get and how convenient I am going to now put 2 u minus chi equal to root 4 epsilon zeta so what will happen you will just get e to the power minus zeta squared the du will become 2 root epsilon d zeta the root epsilon will cancel out okay so you perform this simple change of variables for here it is 2 u minus chi equal to root 4 epsilon zeta here it is going to be 2 u plus chi equal to root 4 epsilon zeta the root epsilon will cancel out and then you take the limit inside the integral the dominated convergence theorem will allow you to do that and you are going to get what you want namely you are going to get this e to the power chi by 2 plus e to the power minus chi by 2 raise to the power minus 2 a when you compute the remaining integral e to the power minus zeta squared d zeta is going to be root pi the root pi and the root pi combined to give you a pi but what is this object here we see the hyperbolic cosine appearing here so I have written this in terms of the hyperbolic cosine and there is a 2 to the power 2 a that will have to be adjusted and that is how we get this 2 to the power of 1 minus 2 a pi gamma 2 a hyperbolic cosine 2 to the power minus 2 a chi by 2 but you will argue that this is not the Ramanism formula that we advertised how do we get the one that we have advertised you need to write this gamma 2 a in terms of gamma a and gamma a plus half to do that we need to recall the famous duplication formula of Legendre which goes back to 1809 what is the duplication formula it is here standing at you at the bottom of the slide root pi gamma 2 a equal to 2 to the power 2 a minus 1 gamma a gamma a plus half so once you use this to get rid of the gamma 2 a and replace it by gamma a into gamma a plus half and you will get the Ramanism formula as advertised in the theorem that completes the proof of the formula of Ramanism for the Fourier transform of the absolute value of the gamma function restricted to a vertical line of the right half plane so a reference for this is the paper by G. K. Shanivasan and Debra Chakravarti on a remarkable formula of Ramanism archived their mathematic volume 99 page 125 to 135 which appeared in 2012. You can use the same ideas for a multi-dimensional setup and you can prove a variety of integrals of this form and these integrals are known as the integrals of Melon Barnes type integrals of Melon Barnes type were used by Hecke in certain problems in automorphic forms in number theory and they were used by Voronoi in analytic number theory and so I call them Melon Barnes Hecke type integrals and for a more detailed account of how to compute the integrals of Melon Barnes Hecke type you may consult my paper in which appeared in Expositions Mathematicae in 2013. Okay so now before closing let us look at the duplication formula I would like to share with you some thoughts on the duplication formula. First of all how do you prove the duplication formula? Let us start with the beta function beta pp what is it put it in the definition 0 to 1 t to the power p minus 1 1 minus t to the power p minus 1 dt well both the factors have exponent p minus 1 combine the two put t equal to sin squared theta put t equal to sin squared theta and then you are going to get a sin theta cos theta to the power 2p minus 2 immediately we must appeal to the double angle formula of course dt will give you a 2 sin theta cos theta that is how we get a 2 sin to the power 2 pi minus 1 theta cos to the power 2p minus 1 theta d theta 0 to pi by 2 use the double angle formula I need to introduce a 2 to the power 2p minus 1 so that is how I get beta pp is 2 to the power 1 minus 2p integral 0 to pi by 2 2 sin theta cos theta to the power 2p minus 1 into 2 d theta ok now we put 2 theta equal to phi so 2 d theta will become d phi and this becomes sin to the power 2p minus 1 phi d phi but now this integral is from 0 to pi remember the sin function is symmetrical about pi by 2 so the sin function takes a it starts from 0 it ascends to pi by 2 and from pi by 2 it descends to 0 and so the integral from 0 to pi by 2 and the integral from pi by 2 to pi will be exactly the same so it is twice the integral from 0 to pi by 2 sin to the power 2p minus 1 phi so now there is no cos term there is cos to the power 0 but cos to the power 0 is between 2 times half minus 1 but this is nothing but the beta function again it is 2 to the power 1 minus 2p beta p comma half so I have evaluated beta pp and I got beta pp equal to 2 to the power 1 minus 2p into beta p half now use the beta gamma relation use the beta gamma relation and write everything in terms of the gamma function and this is what you get and let us assume that p is real positive remember that when p is real positive the gamma function will be positive you are integrating a non-negative function you are going to get something which is positive and one gamma p factor can be cancelled out and we will get the duplication formula root pi gamma 2p equal to 2 to the power 2p minus 1 gamma p gamma p plus half now you appeal to the identity theorem in complex analysis we approved this identity for positive real p both sides are holomorphic in the right half plane and so you got two holomorphic functions which agree on the positive real line so they must agree on the right half plane so the identity theorem the result has been proved for complex values of p in the right half plane and some more thoughts on the duplication formula this was proved by Legendre in 1809 and generalized by Gauss in 1812 and this may appear mysterious but this will not be mysterious at all if you remember the reflection formula of Euler gamma z into gamma 1 minus z is pi by sin pi z this says that the gamma function is basically one half of the sin function in a multiplicative sense so any factorization formula for the sin is likely to have a gamma analog the function sin pi x has this factorization and the Legendre duplication formula is basically the gamma analog of this there is a corresponding thing with sub multiplicative factorization of sin there is a Gauss generalization for these matters and more you can refer to my paper with Ritesh Goenka gamma function and its functional equation which appeared in resonance volume 26 in 2021 the references there on the slide and I think this will be a very good place to stop this particular lecture thank you very much and with this this chapter gets over and we start with the next part of the course on Sturm level problems we will do that in the next capsule thank you very much