 Hello, so we continue from where we left off last time we were trying to derive the Fourier expansion of a certain periodic function cooked up out of the Gaussian e to the power minus t squared we took the Gaussian e to the power minus t squared that is a rapidly decreasing function and we use this function to create a 2 pi periodic smooth function and we were about to compute the Fourier series for this function and use the point wise convergence theorem. So, let us look at what we are done. So, go to the slide we look at equation 1.23 look at this function f of t which is displayed in equation 1.23 last time I explained to you that this function is a smooth 2 pi periodic function and so the Fourier series will converge everywhere it will even converge uniformly but we do not need that right now. So, we computed the coefficients it is an even function. So, the sin terms will be 0. So, b n's will be 0 we computed a naught and this is the calculation for a naught and we found that a naught is 1 upon 2 root pi and we can calculate a n's in a similar fashion and it is an exercise for you to compute the a n's and check that the Fourier series is exactly what is displayed in the slide 1 upon 2 root pi plus 1 upon root pi summation n from 1 to infinity e to the power minus n squared by 4 cosine n t. So, we put back the definition of f of t and we look at this display 1.24 summation n from minus infinity to infinity x of minus of t plus 2 pi n squared equal to 1 upon 2 root pi plus 1 upon root pi summation n from 1 to infinity e to the power minus n squared by 4 cos n t equation 1.24 which is displayed in the slide this is a remarkable identity on the left hand side you got the one set of exponentials and the right hand side you got another set of exponentials with a cos n t thrown in you can put in t equal to 0 and we can see what you get. So, the more general version of this is what we are now looking for and the more general version of this will give you what is called as a theta function identity instead of the function f of t instead of 1.23 or rather in this equation 1.23 you throw in a parameter tau you throw in a parameter tau under the exponential. So, you look at f of t equal to summation n from minus infinity to infinity x of minus tau times t plus 2 pi n the whole squared. So, that is a modified the function f suitably of course the series must converge. So, the tau if the tau is real it must be real positive if it is negative then the series will diverge. So, you want the tau to be real positive or you can allow the tau to be complex, but then the real part should be positive. So, the tau must have positive real part. So, if tau has positive real part then following exactly the same ideas again you see this function of the left hand side summation n from minus infinity to infinity x of minus tau times t plus 2 pi n the whole squared that function is a 2 pi periodic function with respect to the variable t tau is fixed throughout the discussion. For now tau is going to be fixed you should think of tau as a parameter and you should think of t as a variable with respect to t it is a 2 pi periodic function. So, you can go through the same steps again and you can compute a naught a n's b n's again it is an even function. So, the b n's will be 0 you can calculate the a n's you will need the formula for the Fourier transform of the Gaussian that we computed last time that will be needed without that you cannot proceed. So, that is exactly why we did that Fourier transform computation the Gaussian. So, you can prove this equation 1.25 by appealing to the point wise convergent theorem you find the Fourier coefficients write down the Fourier series and you equate the original function with its Fourier series why can we do that because the function whose Fourier series you are trying to compute is actually very smooth and we know the point wise convergence theorem is true for hurdle continuous 2 pi periodic function here it is much, much, much better than the holder continuity it is actually infinitely differentiable. So, equation 1.25 is the equation that we are talking about in this equation 1.25 let us put t equal to 0 and the cosine term drops out and in place of 4t let us put u. So, the left hand side reads summation n from minus infinity to infinity e to the power minus pi squared n squared u on the right hand side the 2 has been taken under the radical 1 upon root u pi summation n from minus infinity to infinity x of minus n squared by u equation 1.26 this equation 1.26 is a very important equation ok. So, I already told you how to go about proving 1.25 and 1.26. Now, we define formally the Jacobi theta function the literature on theta function is very vast there are several different types of theta functions and so what we are displayed here is usually denoted by theta naught as this function theta naught of tau which is equal to summation n from minus infinity to infinity e to the power minus pi n squared tau where the real part of tau is positive 1.27 that is a displayed equation what we have derived what we have derived here instead of u I replace u by u upon pi if I replace u by u upon pi 1 pi will cancel out of here and 1 pi will appear here and 1 upon u pi will 1 upon root u pi will get replaced with simply 1 upon root u. So, the equation 1.26 with u replaced by u upon pi will exactly give you equation 1.28 this equation 1.26 has been recast as equation 1.28 this is called the Jacobi theta function identity the theta function I said the real part must be positive and now I can think of this theta function as a function of t but think of the theta function as a function of t it is holomorphic in the right half plane of course then you will ask me what is the meaning of root t in complex analysis when I write root tau I must explain which branch of the square root I am talking about and below I am explaining which branch of the square root I am talking about. So, the root tau appearing in equation 1.28 is defined as root tau equal to square root of mod tau mod tau is a positive real number and this one is a unique positive square root plus x of i by 2 argument of tau tau is in the right half plane. So, take the principal argument the principal argument means that the principal argument will vary between minus pi by 2 and pi by 2. So, that is the definition of root tau and with this prescription the theta function that you see in equation 1.27 is holomorphic in the right half plane this theta function is going to play a very important role in what is to come. In fact, it is very fundamental in number theory I will give you some nice references for theta function there is a reference of Bellman which is very readable and very beautiful account of theta function. Now, we need some information about how the theta function behaves as tau goes to infinity and how does it behave as it goes to 0. Now, look at infinity now look at equation 1.27 this you have got this exponential factor and as n becomes larger and larger this this exponential function will decay more and more and so for positive for real positive if tau is real positive as you go down the infinite series it is going to decay very very rapidly. The only term which does not decay at all is when n equal to 0 and n equal to 0 you just get 1 and so as tau goes to infinity. So, as tau goes to infinity the behavior of the theta function is like 1 it behaves like 1 and of course I just take in the first term 2 times exponent of minus pi tau the others are even smaller than this. So, I can take this 2 exponential of minus pi tau outside and what will be what will remain is going to be some bounded thing. So, big O of 1 means this stuff that I am ignoring here is bounded. So, this is the behavior of the theta function for real tau large real tau as tau goes to infinity we need it for real for real tau only we do not need it for complex values of tau. Then we also need the behavior of the theta function when t is real but close to 0 as tau approaches 0 plus along the real axis. For that you need the functional equation 1.28 by staring at 1.27 you will not be able to understand the behavior of theta function near tau equal to 0 because when tau is close to 0 what happens is that although the n becomes larger and larger when tau becomes indefinitely close to 0 there can be some balancing between this n square the large n square and the small tau there could be some kind of balancing and all the terms in infinite series may actually end up contributing. So, analyzing the behavior from 1.27 is going to be pretty difficult, but from 1.28 it is very easy because when tau is close to 0 1 upon tau is very very large and when 1 upon tau is large positive real then we know the then we can use the previous discussion and so consolidating we see that when tau is positive and close to the origin 1 upon tau is very very large and so by using the functional equation 1.29 and the previously discussed behavior we can consolidate these two discussions and we can say that theta naught of tau behaves like 1 upon root tau plus something which goes to 0 very fast faster than a constant as tau goes to 0 plus as tau goes to 0 plus this term blows up and the second term will go to 0. So, this is the behavior of the theta function near the origin we shall need this very soon because we are going to be manipulating integrals an infinite series involving involving these things the theta function identity 1.28 is equivalent to the celebrated functional equation of Riemann. So, let us recall what is the Riemann zeta function the Riemann zeta function zeta z is defined to be the infinite series 1.31 you see the infinite series 1.31 in the display zeta z equals summation n from 1 to infinity 1 upon n to the power z this series will converge when real part of z is bigger than 1 that you have studied in your elementary courses the p series test as you would known as it is known in elementary courses. So, applying the p series test you see that zeta zeta z the series converges absolutely when real part of z is bigger than 1 and it converges uniformly on compact sets and so, the sum function is a holomorphic function in the half plane real part of z bigger than 1. Now, we want to extend the zeta function analytically just as the gamma function can be extended analytically strip by strip by integrating by parts the zeta function can also be analytically continued by converting this summation into an integration we will do that later and by successive integration by parts I will indicate how to do that. And the theorem says that the function zeta z extends analytically to the complex plane minus 1 1 is a pole it is a simple pole and the residue is 1 actually and satisfies what is non-trivial is that this analytic continuation of the zeta function satisfies this beautiful functional equation 1.32 that you got gamma z by 2 zeta z pi to the power minus z by 2 you simply replace z by 1 minus z and you get the right hand side the left hand side goes to the right hand side the right hand side becomes the left hand side under the substitution z going to 1 minus z of course this is not going to make sense when z is 0 or when z is 1 because the zeta function has a pole at 1. So, this the left hand side has a pole at 1 the right hand side as a has a pole at 0. So, z equal to 0 and z equal to 1 has to be left out and because the zeta function is analytic on the complex plane minus the minus 1 the gamma function here has a pole at 0 minus 2 minus 4 etcetera will it when you put z equal to minus 2 this becomes gamma minus 1 when you put z equal to minus 4 this becomes gamma minus 2 gamma function has a pole at 0 minus 1 minus 2 etcetera. So, this so this pole must get cancelled out because right hand side when you put z equal to minus 2 observe that this gamma factor has a problem when you put z equal to minus 2 this particular factor does not seem to have a problem. So, it must be compulsorily follow that the zeta function must have a 0 at z equal to minus 2. So, that that 0 will cancel the the pole of the gamma function. So, the zeta function has trivial zeros at minus 2 minus 4 etcetera these zeros cancel the poles of the gamma function and of course, you know the famous conjecture of Riemann that other than these trivial zeros the non-trivial zeros the trivial zeros are minus 2 minus 4 etcetera and the non-trivial zeros they all lie in the strip they all lie in the strip 0 less than real part of z less than 1 Riemann conjecture that they actually lie on the line real part of z equal to half that is a famous Riemann hypothesis which even today remains an open question. So, we shall derive this celebrated result of Riemann from the theta function identity that we have just derived. We have derived a very important theta function identity namely 1.28 from using 1.28. In fact, this 1.28 is basically the Riemann the Riemann's functional equation in a very different avatar when you this functional equation and the Riemann's functional equation are related via the Mellon transform. Normally that is the way it is expounded, but we shall avoid the use of Mellon transform completely. We shall directly prove that how 1.28 implies 1.32. We shall be following this book of Richard Bellman a brief introduction to theta functions which was published in 1961. It is an old books published more than 60 years ago, but it is a very beautiful book very readable book and it would be a marvelous project to read Bellman's account. So, let us begin with a very very simple observation. Let us begin with a simple observation equation 1.33. One can directly one can directly perform this integration. One can directly perform this integration. How to do this integration? Put n squared pi t equal to u or rather think of Laplace transforms integral 0 to infinity e to the power minus lambda t t to the power s minus 1. That is a Laplace transform of t to the power s minus 1. You know that the Laplace transform of t to the power s minus 1 is gamma function divided by t to the power s, but evaluated where evaluated at this parameter value. So, this integral 1.33 is readily computed it is gamma s upon n to the power 2 s into pi to the power s or if you like put n squared pi t equal to u and you reduce this integral to a gamma integral and you will get it. So, 1.33 I will consider it as done very elementary. Now, put n equal to 1 2 3 etcetera and add. So, summation n from 1 to infinity integral 0 to infinity e to the power minus n squared pi t t to the power s minus 1 dt equal to gamma s is common pi to the power s is common summation 1 upon n to the power 2 s n from 1 to infinity. What is that summation exactly zeta to s? So, when I put n equal to 1 2 3 and sum I get on the right hand side gamma s zeta to s pi to the power minus s. Of course, the real part of s has to be bigger than half otherwise this infinite series will not converge. In order to make this infinite series converge it is not enough to take real part z bigger than 0 I am compulsorily have to take real part of s bigger than half. The other thing that you need to do is the summation which was outside the integral has now gone inside the integral in 1.34. When you exchange an integral and an infinite series you need to justify the exchange of limits. You could use the dominated convergence theorem from measure theory if you have studied the dominated convergence theorem or you can do some elementary estimates and whatever be your favorite way to justify the exchange of two limiting operations use that idea and you justify this exchange of summation and integration and you can take the summation inside. When you take the summation inside you begin to see the theta function staring at you. It is not the complete theta function. This should have become the theta function if the summation goes from minus infinity to plus infinity. If it were minus infinity to plus infinity you will get the complete theta function. You get part of the theta function. So, now let us try to see how this summation is related to the theta function. So, summation appearing within the integral is of course related to the theta and we must invoke the functional equation for the theta function that we have derived. We have taken a lot of trouble to derive the functional equation for theta and that has to be used. Now first before you do any of those things let us replace the 2s by s. Let us replace the 2s by s. So, this will become zeta s, gamma s by 2 and so on. So, let us do that simple change and let us write let us rewrite the previous equation as integral 0 to infinity t to the power s by 2 minus 1 dt, gamma s by 2 zeta s pi to the power minus s by 2. Because I made the change of variables now the real part of s is bigger than 1. So, what is the g of t? The g of t is basically this stuff that appeared under the integral sign. The summation that appeared under the integral sign summation n from 1 to infinity e to the power minus n square pi t I am abbreviating it as gt. But this gt is exactly one half of theta naught t minus 1 equation 1.35 you can easily verify the right hand side is one half of theta naught t minus 1. Okay, so let us put this gt in the integral. Now what we need to do now is to take this integral from 0 to infinity break it up as an integral from 0 to 1 and an integral from 1 to infinity. So, integral 0 to 1 gt t to the power s by 2 minus 1 dt plus integral 1 to infinity gt t to the power s by 2 minus 1 dt that is the right hand side gamma s by 2 zeta s pi to the power minus s by 2 real part of z is bigger than 1. Okay, so let us proceed further flag this 1.28 this is the most important identity beautiful identity 1.28 this equation has to be expressed in terms of our function gt that is a very simple algebra that is pure algebra and that algebra will give you gt equal to minus half plus 1 upon 2 root t plus 1 upon root t g of 1 upon t. So, the theta function identity in terms of the new function g simply 1.38. Now substitute this expression for g in the integral that was there from 0 to 1 in the previous in the previous page in this integral the first integral from 0 to 1 gt t to the power s by 2 minus 1 here we use the expression that we are just obtained this 1.38 we just put that over here then what will happen we will get 3 terms we will get t to the power we will get t to the power s by 2 minus 1 t to the power s by 2 minus 1 that factor is there that factor has to be multiplied with minus half after that you have to multiply the 1 upon 2 root t and you have to perform the integration from 0 to 1 I have performed the integration from 0 to 1 that is that comes out to be 1 upon s into s minus 1 then this particular part 1 upon root t already t to the power s by 2 minus 1 was there because 1 upon root t it became t to the power s by 2 minus 3 by 2 and the g of 1 upon t so that is the integral the other integral I have not touched so far and right hand side of course I have not done anything now exercises show that this integral 0 to 1 g of 1 upon t t to the power s by 2 minus 3 by 2 dt is integral 1 to infinity g u u to the power minus half minus s by 2 du well no prices for guessing I made a change of variables I put 1 upon t equal to u that is what I had done the integral from 0 to 1 became integral from 1 to infinity so far checking that the left hand side equal to the right hand side is just a simple substitution 1 upon t equal to u what you need to further verify is that the right hand side of 1.40 the right hand side of 1.40 is an entire function of s what do your entire function of s it is holomorphic throughout the complex plane how do you check that this is holomorphic throughout the complex plane you can differentiate under the integral sign as many times as you want every time you differentiate with respect to s you are going to pick up a log u factor no matter how many times you pick up the log u factor this integral will converge it is exactly for these things that we derived those 2 behavior of the theta function these 2 things 1.29 and 1.30 these expressions were derived exactly for such purposes of proving that this right hand side is an entire function the other thing that we have to do is that the following integral appearing in 1.39 is holomorphic in the plane real part of s is bigger than 0 so this is going to be holomorphic on this larger half plane so the left hand side of 1.39 so this part we know exactly has a pole at 0 and it has a pole at 1 this middle part is an entire function and this third piece is holomorphic in real part of s bigger than 0 but the right hand side was defined for real part of s larger than 1 so this equation 1.39 this important equation 1.39 has given us an analytic continuation of zeta of s on the larger domain real part of s bigger than 0 right because the left hand side makes perfect sense on real part of s bigger than 0 except at s equal to 1 except at s equal to 1 so that is the most important conclusion that the that we are analytically continued the zeta function on real part of s is bigger than 0 okay and the left hand side of this equation doesn't change doesn't change when I replace s by 1 minus s if I substitute if I make the substitution s is going to 1 minus x the left hand side doesn't change that you can check very easily that I just combined the two terms that appear I combined the two terms I combined the two integrals that I got this 1 to infinity and this integral from 1 to infinity I just combined them into one integral and I found that I get this so this left hand side of that equation doesn't change under the substitution s going to 1 minus s so the right hand side should also not change and when I replace s by 1 minus s on the right hand side of 1.41 what do I get I get the celebrated functional equation of Bernhard Riemann equation 1.42 in the vertical strip between 0 and 1 I have established the functional equation of Bernhard Riemann how to extend it for the whole of complex plane minus 0 1 I will explain in the next capsule I think this is a very good place to stop this capsule and thank you very much.