 And calculating step-by-step is a good sign today. Hey, we've got the formula of calculating correlation functions of the operators we will need to scale up in the sector for the C plus. So let's start. Let's write a few 3-point functions of that. Now, later when we actually start calculating 4-point functions and so on, we will use the fact that when we insert operators at different points, it doesn't matter where we insert them. But for the first few calculations, we'll actually check the signals working. OK, so one of the active vertex operators is e to the pi k1 at x1 at sigma e to the pi k2 at x2 at sigma, sigma 1 sigma 2, and e to the pi, let's call it k3 at x2 at x and x to the x and x to the x and x to the x. Remember that these are, so this is the correlator we want, together with the c of sigma 1, c of sigma 2, c of sigma 1. Because we have exactly 3 vertex operators, we have to do no integration. So all three operators are fixed. OK, so we're going to be calculating the correlation function, which that's starting at. So let's first calculate the correlation function and then figure out what else we need to add to get the schedule. So first, the correlation function. Now, let's first do systematically and then see what, how we can do that more simply. First, remember that all these operators are in master. So they have a equation alpha prime k1 squared by 4 is equal to 4. They're all dimension 1, 1 operators, and alpha prime k1 squared by 4 was the left and right dimensional part of x. OK, so now, let's use our general formula. Our general formula for the correlation functions of many products of int of i, a, dot, x is, was product over i, n, and j, product of the pairs i, j, and z, i, j, mod squared to the power alpha prime k i, dot, k, j by 2. OK, so this is the x one. Let's just write it down in this case and set 1, 2, mod squared to the power alpha prime k1, dot, k2, z, 1, 3, mod squared to the power of k1, dot, k3 by 2, alpha prime to the power k2. I think this is pretty complicated. It depends on how this k1, k2, k3, and how it helps to cancel. You remember that the answer should be independent of z1, z2, and z3. However, if you think about it a little more, this answer does not depend on k1, k2, k3. Let's see why. It does not depend on k1, k2, k3, because k1, dot, k2 is equal to k1 plus k2, something squared, minus k1 squared, minus k2 squared. I took conservation. k1 plus k2 is just k creation. And each of these k's, so there's the master of addition. Because these brackets, it's simply alpha prime by 4. It's just 4 by alpha prime. So minus 2 prime. The answer again actually makes it work. It's actually z1, 2, mod squared, z1, 3, mod squared, z2. So this c is the c-controver. The c-controver is simply z1, 2, mod squared, z2, 3, mod squared, z3, 2, mod squared, x-part, c-part. And so here it can. That's the answer. The other function, more precisely, is the answer to the partition function. So let's call that c as 2, time. That is, there's an undetermined sort of determinant factor outside everything. It's called cx2. And every time we compute the correlation functions on the sphere, we get a similar factor. So since we're not being very careful about how to value this determinant at the final cell, there will be some overall normalization constant, which will depend only on the manifold in which we're doing the calculation. So in some constant that depends on the fact that we're working on the sphere, and then that rules the calculated state of the string that we have. So something that we would provisionally do, we say, we added, we used that. We said that not only do we use the scattering attitudes of given states, we need to put vertex operators of these states on the sphere. But when you do this in final diagram, you know, vertex insertion in a particular normalization, you also put a factor of a coupling constant. So what we're going to do is to leave open the fact that every time we add a vertex operator of a given sort, we should be multiplying this vertex operator with some as yet undetermined normalization. Open the fact that the actual scattering amplitude in string theory is not the correlated function of these vertex operators normalizes either. After all, who told me to normalize the vertex operators like this? But the actual vertex operator, let me call it g-tac non. And if this normalization would be different, then it is equal to i-tac non. We will let it use consistency to determine what g-tac non is in terms of something else. So with all this in place, the real vertex operator. The thing that we need to put into the sphere diagram is in order to compute. So being complete in general, not committed to any particular normalization, the final scattering amplitude that we compute is cs2 times g-tac non will be q times 1. It is a string. Now of course, we're going to do a more complicated calculation. Let's look at a couple of features of this scattering machine. You see, the z1, z2, z3 dependent states in the final answer. In any of the many, why was it obvious that this had to happen without doing any calculation? It should become complicated. But suppose we were just conforming few cases. And I gave you, let me take most of them. Suppose I gave you the 3-point function of these operators involving only x. And then, clearly enough, knowing very clearly about the operators, you can immediately predict the z1, z2, z3 dependence of this 3-point function. Can you tell me how? Exactly. Just by scaling. You say, suppose you were an operator, three operators that transform h1, h1, h, actually include just an analytic idea. So suppose they're h1, h2, and h3. And I want to compute the correlation function of these three operators. Then, scaling variance, determine that they transform like 1 over z1, 2 to the power of h1 plus h2, 2, and h3 by 2. I'll write it down. This is just a general result of conforming impactions, depending just on the weights of these operators. Since we're going to do a view of string scaling, all of the x operators have weight 1. So h1 plus h2 plus h3 by 2 is, this is once, where is it? Yeah, we're going to take half. Just 100. Let's add up all the weights of this thing, that h1 plus h2 plus h3 by 2. Yeah, so you can write it this way. It would be, because if we add up all the weights, so we just add up the net weight of each of these. Yeah, h1 plus h2 plus h3. And we get that. Sorry, yes. So these dimensional values are the same, this way. The basic point is that, suppose we were to complete this, we would have a, well, I think this is going to be one set squared. I mean, that's something. Oh, yeah. This is not that simple. So another set we do in string theory, and h1, h2, and h3 are h equals 1. This is just 1 over z1, 2 to the power 1. 1 over z2, 3 to the power 1. 1 over z, 4 to the power 1. And that cancels space. And this is the answer of the six. Perhaps the even easier way to say it, is that c times v is a primary operator weight 0. So the 3.1 to the 3 primary operator weight 0, that's going to be something just, z1 to the power 0, z2 to the power 0, z1 to the power 1. So the z1, this was necessarily automatic. We didn't need to do all this complicated stuff in order to see that it didn't benefit of z1, z2, z1. All of this work was necessary just to get the number. So now that we've understood what the three tachyons are doing with this, in a calculation list, let's try now how about two tachyons with a boundary, or a massless closed string state. So what's the vertex operation of it? It's good. So we've got some g, so we've got gt then squared, times gm, g to the massless states, OK? Times e to the power k1.x times del mu, del bar, del bar, x mu. And let's put a polarization, e mu. Times e to the power k1.x. Times e to the power k2.x times e to the power k3. And in the first list, everything is at sigma 1. Everything here is at sigma 1. Here is at sigma 2. So this is the thing that we really reached in time. Let's remember the Marshall conditions here. We have k1 squared equal to 0. But k2 squared is equal to k3 squared is equal to 4. Because those guys, that one, is getting the massless states. And then we also have the C stuff that we insert. So let's say, so how does this work? So now we're going to compute these correlation functions along with an insert, insert of an x. If you remember, we have this nice rule, which I missed. Calculate whatever you could have got with e to pi, pi k, a dot x's, and multiply by the factor for each insertion. You multiply by the factor minus r by 2 alpha prime times, and in this case, we've got only one insertion. So I'll just write it out for that one insertion. So sum over k2 mu plus k3 mu by z1, 2 plus k3 mu by insert 1. Again, we've got this thing, the insertion at point z1. And then we're supposed to sum over all other k insertions to do it. We should multiply by such a factor plus the factor of alpha 1 mu and some oscillator that makes a contract, which you can think of as del x mu tilde, which you have to contract according to the usually contract. One of these oscillators, so the usually contract in the rules just give you zero. There's nothing but the factor of x. So that guy is just throwing away. This guy is throwing away. And we take this and multiply it with what we would have got. Then we've got this, and then we've got x0, x0, and we get from the z part. This is with the mu, and then we have to multiply the whole thing. Just the expectation values of z to the power of x. And that we didn't know what that is. That's z1, 2 to the power of z1, 2 to the power of x. So let me write just the amplitude part. There's also the analytic. So z1, 2 to the power k1 dot k2 by 2 z2, 3 to the power k2 dot k3 by 2. And z1, 3 to the power k1 dot k2. We use the same formula as k1 plus k2, the whole thing squared minus k1 squared. Minus k1 squared minus k2 squared divided by 2. So that's k3 squared minus k1 squared minus k2 squared divided by 2. But k1 squared is 0. And k3 squared part can't say k2 squared. So this still is just a claim to k1 dot k2. Because it's k3 squared minus k2 squared, each of which is equal. It's just kind of a fix. You know, it's any field which has k1 squared equals 0. This thing. Yes. So, right. Similarly, k1 dot k3 is equal to 0. However, why won't k2 dot k3? k2 dot k3 is k1 squared, which is 0. Minus k2 squared minus k3 squared, each of which are 4 by alpha. Divide by 2. k2 dot k3 is equal to minus 4. That's what we are talking about. Yeah. From these factors, from the e to pi k dot x of fractions, it is a factor of 1 over z to 3 to the power 2. That's this one. Is this clear? Now, it looks pretty complicated. This one looks pretty complicated. Well, I'm going to simplify the little bit. We've got minus i by alpha prime to 2. And then we've got k2 z1 3 plus k3 z1 2 over z1 2 z1 3. So, the power of k2 and k3 is there explicitly. Because because these powers k2 and k3 are explicitly, it doesn't seem to be a simplified numerator. It's hard to imagine half root. But I think it would be better if that. Numberless it is. So, what is transfer? So, what is the equation? So, e mu nu k mu should be which k? Of k1. Very good. e mu nu k1 mu is equal to z. And the conservation of this. k3 is equal to minus of k2 plus k1. But the k1 part has to come to be there. Because it comes to be there. Complex to 0 over here. So, there is. So, basically here, the factor of k2 is the same as a factor of minus k2. So, we can write this in a more more symmetrical fashion by replacing. Every time we have k2, we replace it by k2 minus k3 by 2. As far as the contraction with the ease concerns. Same thing. And then we have k3 by replacing the opposite of that. So, let's assume that minus i over k2 is equal to minus k3 by 2. So, we can write this as something like that. So, let's assume that minus i over k2 by 2 let's say by 4 and then we get k2 by 3. k2 by 3 is k minus k3 into z1 by 3 minus z1 by 2 over z1 by 2 z1 by 3. So, now this is equal to z2 by 3 over z1 by 2 z1 by 3. And now we combine with this. So, the next factor is 1 over z2 by 3 z1 by 2 that's what we get from the C's. We have something similar from the complex function. And so, in the end what we have in the end we have factor. We have alpha prime by position alpha prime by 16 k2 by 3 0 destruction. Since you take on this we come back to this. The number behind this is a little bit, right? Oh, because we have a square here. But it just seems to lack a square unless I mess up with something here. Okay, okay, we come back to that. Alpha prime. The alpha prime gets from this rule which we have this rule. We have one of these factors which is the reach of the left and the right. I don't see how you can not get an alpha prime square. Let's get back to this when we have a check. Anyway, this is up to our determinant. So, there's gt gt squared g masterless. That is the exercise. What we'll do is calculate the three masterless projects that we have. Whereas the same factor we have whenever we do a calculation and some determinant which we are not we are not doing that. Okay, now we're going to do the calculation of three master states. So, it's a little more messy but let's just at least understand the structure. Okay, first we're going to have e to the pi k1.x e to the pi k2.x e to the pi k3.x But now each of these k1, k2, k3 squares to zero. So, the path that you get just from the contraction of the exponential is one. k1.k2 is in k1.k2 square minus k2 square minus k3 square which is 0 minus 0 minus 0. So, you get nothing from the contraction of the exponential. Okay, so that part will be clear. So, all that remains is the path that we get from oscillator contraction. So, now from the oscillator contraction what we have here is i by 2 alpha prime and then k2 over z1 2 plus k2 mu and 2 plus k3 mu over z1 3. This is going to be contracted with the polarization vector. Okay, so as we saw before in the process of this contraction we replace this by k2 minus k3 by 2 and so we get the 4 and replace this by 3 mu divided by z1 2, z1 3 and then z2 3. Okay, plus we have L alpha tilde at signal. So, we have a product of three such factors and they're not expected. So, the other factor is i by 2 alpha prime and now second. So, this will be k31 mu z31 over z2 3, z2 1 plus L x tilde mu at sigma 2. k12 mu z1 1 2 z31 z32 plus L alpha prime. This is done. Mu alpha. This was to take all the weak contractions with this. But what can we do? We have to take all the numbers of oscillators. So, what we can do is we can contract these pair ones. So, you see when we do this when we just take the product of these. The second term, the next three terms when we take product of one of these and we contract it with any tool. This is awesome. Let's just check the z1 2, the z dependencies in each of these. If we take the product of each of these you see clearly what we're getting is 1 over z1 2, 1 over z2 3, 1 over z3 1 in the denominator. The product of that, the factor of that things where we normally have a factor of it in the numerator. On the other hand, when we take the product of this sky with this with the contractional least. This will give us a factor of z2 3 squared. 1 over z2 3 squared. Long is twice the parenthesis. Long which is the weak contraction of the x this is twice the parenthesis. Yet, that multiplied by the z dependencies here once again give us 1 over z1 2, 1 over z2 3, 1 over z, z. There was a z2 3 in the numerator here. It becomes z2 3 in the denominator because you have a 1 over z2 3 squared. So, all of the terms of the same z dependencies exactly the z dependencies will cancel with the z dependencies from the c side. So, the z dependencies are used. And then you just have to keep trying what we get when we contract in index space. In index space, contracting this guy and this guy so this was mu1, this was mu2, this was mu3. It's just a delta value. The index structure of what we're going to get is k1 mu1 times delta of mu2 with mu3 plus k2 mu2 times delta of mu1 with mu3 and so on. So, now that you see the structure just make right down the answer. Is this okay? We're going to get a sum of four terms both from the left and from the right and so let's try to do the way which we'll see next too. We're going to sit down as e1 mu mu e2 alpha beta gamma delta in fact with t mu alpha mu beta delta these factors is the factor you get from the left and from the right. These are the factors from sum of four terms and the sum of the four terms is for k2 3 eta alpha gamma plus k3 1 alpha eta... k2 1 gamma eta mu alpha plus alpha prime by 8 k2 3 mu 3 to 1 alpha prime k1 to 2 See many of these terms not shown. This term is described in this contract. This term is described with these two contract. This term is described with these two contract and this term is the product of these three. Some index factor on the left, index factor on the right compared to the first index and the second index of the fourth index. And all of this now is going to come out of life by the CS2 that G master started to kill dying to save. Now you see that already you start to get really interesting things. The fact that that three-tachyon amplitude was independent of k more or less was the statement that the three-tachyon vertex would be represented in a Feynman diagram with justify Q. The fact that the single graviton with two tachyons has some k dependence will, as we will see, come out of the fact that this term is an interactional with an OMG effect reaction that involves the coupling of graviton to the tachyon field. And this interaction will come out of you see this interaction is the stress tensor from the tachyon interacting with the graviton. That's the t-square term with the stress tensor with the tachyon which interacts with the graviton and that's what's been reproduced this interaction. Whereas this free graviton interaction is simply the free graviton interaction from the Einstein interaction. You know if you just take the Einstein action if you just take the Einstein action and you know expand it as eta plus h and compute the cubic terms in that process using physical state conditions plus what's happening to the gravitons you just do that born calculation classifying calculation of three graviton waves coming in interacting with each other. This is the structure that we get. We probably see this in body data once we start to look at the effect when we start to look at the mass of the graviton. Okay? We've got a little more information here than just from the graviton because we've done all the mass of the states of string theory together. We've done the graviton, the BVU feature of the graviton. So this computation here has already a lot of information. It's telling you how the mass of the states of string theory interact with each other and I'm telling you that at least partly we'll have a lecture seeing this in body data how this thing works. A non-effect of action. There's a lot of information in the calculations and this information will soon summarize in terms of an effect. Point functions are relatively boring. All those three-point functions are relatively boring. Already these three-point functions just in that polarisation information and the dependence on domain terms already these three-point functions have been that interesting information about what these mass of the states of string theory are and how they interact. Exactly how we interpret this in terms of the domain that we've got here as a Lagrangian and what we've got here. These are three-point function calculations I want to do with you guys in class. Any questions or comments about it before we move on to format. Why not two gravitons and a calculator? Now, good. Now, I think that probably should be right. You see the tracheon potential in string theory starts at this point. I was there, honestly, I didn't try to calculate this but I I found suspicion that this should be right. Basically, you see, because what the effect of action of the tracheon is quadratic coupled to gravity. So, let me say this again. Had two gravitons times tracheon, not gravity that's gravity back now would have sourced the tracheon. Inconsistent in looking at pure gravity distribution string theory, which is not the case. So, I think maybe I haven't done it like I haven't done it in this round I'm sure I've done it at some point but it's a good exercise. Calculated, I think you should try it. Now, other questions or comments. Okay, very good. It is to turn states to the sphere but turn from three-point function calculations to four-point function. Anyway, these people encounter one of the classic formulas of string theory. A formula that came worth the string theory and it's a close thing version of Venetian. The Venetian. It's called the Venatian. It's a formula that was discovered before string theory. And it was studied, in some sense, led to the birds. All right. This thing was capturing an amplitude of four, just say, and it stopped. It's just four dashes. Okay. So, what do we have? We've got e to the pi k1.x, e to the pi k2.x, e to the pi k3.x, e to the pi k4.x. And now, to keep life simple, what I will do is assume this formula result that we've seen that things don't matter. They don't matter where we keep the fixed vertex operators. So, it's the fixed three vertex operators and specialist, specific points. And to try to layer it, we soon see that the three specific points that make life simplest. Actually, in this layer, we can just keep them all up. Actually, it's easy as it gets. But there is a standard to choose one of them to be zero. The other to be one and the third to be infinity. This infinity, we will choose to be some m which we will have to write into that. And this formula, we will just write into that. The three fixed vertex operators. Those are the three fixed vertex operators. And this is the the chat that's going to be. What do we have? So, what do we have? So, let's compute Firstly, I'm going to be taking m to infinity. So, the first thing I want to see is that all the dependence on n, that this doesn't give me i to zero infinity. Okay, this is a little bit part of the test that things don't depend on where you get some things. So, let me compute all the dependence on the location of the third wave. So, from the c's, what do I get? From the c's, I get z23 to the power a2.a3 which, since we can say three things that would be like m to the power by 2 k2.3 by 2 and I get z31 the dependence on m from the coordination function of the c's. Okay, from c's. m from the contractions. From the contractions, what do we get? This is let's just Yeah, you saw that. No, you see, now the momentum conservation No, your community. You're completely right. So, this is just m to the power there's alpha prime here, there's alpha prime here. So, each of these is, as you said minus 2 by 2, right? As we've seen before. So, we get m to the power minus No, I'm sorry, I'm sorry. That's wrong. Let me say it wrong. You see, no, it's not right. You see, because that when we have three tagions the sum of two tagions sum of momentum of two of the powers which was a momentum of the third one that's not true anyway. You see, what we have is momentum conservation tells us that k1 plus k2 plus k3 plus k4 is equal to 0. So, if you get k2 plus k3 that is not basically wrong. k2 plus k3 is equal to k2 plus k3 minus k2 squared minus k2 squared minus k3 squared by 2, that would not be true. But k2 plus k3 squared is nothing simple. This is scary. So, what we get from k3 interacting with all these other things What is the alpha prime? There's no C contraction between k3 and k4 because k4 is not fixed by k4 See that? Yeah, this is not working out well. Okay, I messed something up. Let me just go through it completely and I'm going to check. Okay. So, okay. So, now that's what it depends on the name that comes from this guy turning into the part, you know, the attraction of the axis. So, what do we get? We get into the power alpha prime by 2 sorry, sorry, sorry, sorry. I'm mixing up. Let me stop this again. Here, sorry, sorry, sorry. Sorry. Okay. So, from this, let's do the C's. So, the C's are just a factor of m from this guy and m from this guy. Because from the C's we get z3 2 and z3. So, from the C's we get because z3 2 and z3 1, I would take and put it up. This is what we got from the C's. Then what we get from the axis is what it is. But that equals all operators. Yes. So, there are C's, right? There are 6, but we're only keeping track of the number of contractions, m and something. Okay. So, that gives us m to the power alpha prime by 2 into k3 for k1 plus k2 plus k2. So, this is equal to and now we can use the Marshall condition. Okay. Now we can use the Marshall condition. So, k3 squared is equal to 4m, right? So, this is equal to m to the power minus 2. Okay. So, all the independence is finished. So, what remains? What remains? Let's see. From the C contractions we get a z2 minus z1, the whole thing. A z2 minus z1. But z2 is 1, z1 is 0, z2 minus z1 is 1. Okay. So, here's another here from the C contractions. So, from the x contractions where one contraction between this and this and that's also 1 to some power. So, you don't care about that. The contraction between this and this and the contraction between this and this. Okay. So, that's what goes on. So, the contraction between this and this is e to the power is now, this is z minus 0. So, it's just... So, let's say that the location of this is z. So, this is z to the power k1 dot k2 alpha prime by k4 alpha prime by z minus 1 only the whole number. Now, multiply and mod squared. And z to the power alpha prime by 2 k3 dot k4 This one? Yeah. So, mod squared and stuff. So, I'll find the answer is d2z d2z mod z Let's try it with mod z alpha prime k1 dot k4 So, let me do the theory. As you're writing, a mod invariant is exactly as your input. Is it just... Okay. So, now let's pause from the formula to just examine how you write it in this situation. Okay. So, what are the invariants in the four particles? So, how do you write them? Well, basically, there are two invariants but in order to write things electrically, we will define them all. Let's define k1 plus k2 all things squared is equal to s k1 plus k3 all things squared is equal to t and k1 plus k4 all things squared is equal to d. So, we've taken two combinations of two moment down. We couldn't take the combinations of two other things. But any other combination like k2 plus k3 would by momentum conservation directly be the same as k1 plus k4. So, we don't need to take any other combination. This is three ways of dividing four objects into groups of two. I think they're okay. These that will be independent. No, they're not because of the mass check condition. Okay. So, let's check. Let's add these things together. Okay. So, we add the three of these together what we get? We get k1 squared three times plus k2 squared plus k3 squared plus k4 squared. And then plus two k1 into k2 plus k3 plus k4 but that's minus k1. One squared plus k2 squared plus k3 squared plus k4 squared which is equal to 16 by alpha. Okay. Now, I think we're going to do invariance in the scattering. And these two invariance are easy to understand. You can put it in two particles scattering. So, the invariance is essentially the center of mass energy and the angle at which the scattering vanishes is the energy at which it scatters and the angle at which things come out. But in terms of these invariance. So, let's quickly do that. We can do that nicely. That's very simple because k1 dot k4 k1 dot k4 is equal to k1 plus k4 dot k2 minus k1 squared plus k4 squared by 2. And we're going to just alpha prime like this. This is alpha prime. This. But that is equal to alpha prime times u by 2. And now this is 4 by alpha prime. So, minus 4. Similarly, we can write this guy k2 dot k4 as t. Right. So, this thing is g2z. Okay. z to the power alpha prime u by 2 minus 4. 1 and z to the power alpha prime t by 2. Now this formula looks like it picks up t and u as special. You know, and discriminates against z, for instance. But actually, that's not true. You can change variables in this formula from z to 1 by z. Okay. And in the new variable, you get the same formula. But when u and t are replaced by the guy that was special for the Japanese infinity, which was k3. So, what was special? Here. The guy that was special here was guy at 1 and we got 1, 3 and 1, 4. So, we get there 3. Then we get 3, 4. So, the guy that was fixed to infinity, the guy that was fixed to infinity was k3. And so, we will get this is also because you're getting the state of system. Yeah. Yeah. S and D variables. Yeah. Sorry. You get, I'm not 100% sure. You get, you get two others. This is you because you will have k3. k3 was the one within. Here you look at these two formulas. You have, you don't have 3 in S and D. You can write them as, you know, k3 plus, depends on the moment of validation. Yeah. That's probably right. Actually, I'll, you get the answer. So, basically, this formula is completely symmetric. You know, it's completely symmetric in S, T and U as it should be because these things are interchangeable under the real of the moment. Okay. Now, the key point that we want to make, that I want to take care of. Okay. So, we're going to have to stop there. But, let me make two, three points. Okay. The first point I want to make is, is the problem. What I will make is that you might, when you look at this formula, wonder about whether this defines a convergent. This power was smaller than 2. And this power was smaller than 2. So, suppose this power was smaller than 2, then this is the integral of the divergent. See, if this power was smaller than 2, smaller than 2, then the integral of the divergent at 1, some of these two powers was larger than minus 2. Then the integral of the divergent is 3. Because the infinity would go like z to the power sum of the powers and that, I'd better go a diode faster than 1 by 6, right? No. You know, I'd better get a convergent. Okay. So, the first, now, certainly for some values of u and t, and for x, do you think? Of course. This integral does that. But, the first question that you might have in your mind is, are there any values of x, t and u for each of the convergent? And the answer is yes. You see, suppose this was equal to 2, sorry, this was equal to minus 2, so that this was, the integral here was at the edge of that energy, and this was equal to minus 2. So the integral here was at the edge of that energy. Then, at infinity, we've got minus 4, which is a safe ocean of income minus 2. Okay? So we can make this minus 2.5, minus 2.5, sorry, we can make this minus 1 and minus 1, which are both safe. But, minus 1.5 and minus 1.5 which are both safe. We convergent at these two points, and we still be minus 3 at infinity. There's a range in the space of s of u and t, in which this integral is a convergent. If we go outside, as we will see, though we will have to see this next time, as we will see, we will be interested in this function as an analytic function on the whole x, u. Or on the whole x, u. You know, an analytic function of complex u and t, for instance. The way we're going to think, make sense of this formula is by evaluating it, where it makes sense, where it's given by the convergent integral. That way, it was an analytic function of u and t. We will find that where it starts to stop making sense, that's because this analytic function has some singularities. But the singularities will all turn out to be points. Because they're points, we can analytically continue on. So basically, we're going to by analytic continuation, by using the factors, well defined in some range, we're going to evaluate them in that region, and then by analytic continuation, find the other side. One method of this procedure, we will get an answer beyond the range of convergence of this integral. But the answer will have singularities, which will turn out to be poles. Now these poles will be extremely physical. The poles will represent the singularities due to exchange of intermediate particles. That's the problem. The next job is to take this, well, first thing that we evaluate is exactly what I mean. It's possible to evaluate this integral exactly in terms of gamma functions. But most presently, it is starting the poles structure by the answer. The singularity structure by the answer.