Fourier type integrals using residues

welcome today we’re going to talk about Fourier type integrals using residues okay so let’s talk about remember that what we have here is integrals of the form so here if we have like a Fourier transform the Fourier transform I mean there’s many different alternate versions of it but they’re all of the form this so that’s represents the transform it’s a real integral of integral on the real line of f of x times our complex exponential e to the I Omega X DX okay so so of course that this exists provided F follows it with certain restrictions on its intergrity and so the we’re going to talk about a special class of f of X functions that a residue theory our residue theory allows explicit answers explicit computable results okay and so of course what we’re talking about is that f of X of a rational type of the form P of x over Q of X where both of these numerator and denominator are polynomials okay and these polynomials they have you know a real coefficients real coefficients and and and another thing is we want Q of X should have no real valued zeros so clearly if Q had a real valued zero then there would be a singularity that was actually on the real number line in which case that there would be questions of whether or not the interval would converge and so we don’t want to consider that kind of thing right now so what we’re talking about then again is is something of the form this way and what that means is that and also that so what we mean then is that of course we’re going to get complex conjugate roots for Q and that means I can if I draw out my complex plane here you’re gonna have roots like this that all over the plane like so okay and so I’m with these as we seen from before what we can do is of course take an R that’s a capital R that’s big enough so that we can create a contour and a goal the first branch of the contour integral is going to go like that and then it’s going to come around in a semicircle over the top like so and it will enclose all the roots in the positive complex plane or the the top half of the complex ation plane I should say and so of course we’re gonna call that contour C and C is going to be broken into two pieces I’m gonna call it CX this part that’s on the bottom there and also C are the radial part that goes over through the complex plane now of course what we know from this is that this integral C over the over the contour C that encloses all of those roots in the top half of the plane of f of z at times e to the I Omega X DX is going to be equal to two pi I times the sum of all the residues of fe2 the I Omega X or sorry Z that should be a Z up there Z Z equals Z K for all of those all of those singular points that are in the plane on the top half of the plane okay and so the goal will then of course be to take R going to infinity and then be able to establish that this these these residues can all these residue values that are accumulated can be attributed

just to the integral along the real line and not to this radial component here as well so that’s the goal all right so let’s see how we can do this so we need one more one more assumption we need to assume that Omega is positive okay and for for many of our analyses involving Fourier series or Fourier analysis we can always assume that Omega is is of one sign so we can just pick the positive side and make Omega positive okay so with that result we can get a solid solid solid answer here so with that assumption I should say okay so let’s just talk about an example so let’s let’s try this example here of f of X is equal to 1 over x squared plus 1 of course so again our Fourier transform of f is going to be e to the I Omega X over x squared plus 1 all right and now clearly we can rewrite this we can factor the denominator into X plus I and then X minus I okay all right and so when we draw the complex plane here we have a I and then a – I down there so there’s the complex conjugate roots and we can put our path going around like so so what we can do of course is now that encloses that path we’ll call that C and so what we do is we examine f of Z e to the I Omega Z the Z and say that’s equal to 2 pi I times the residue of of F of Z e to the I Omega Z ok so the residue is going to be at I right it’s because it’s enclosing that one right there ok so let’s actually compete what that is so we can rewrite this function here on the inside now in terms of this we can rewrite it as as e to the I Omega Z all over Z plus I that corresponds to the singularity in the bottom half of the plane and so this function of course is analytic in side C and then we have our singularity in the denominator like that and we know using the residue theorem that that what we do is it’s 2 pi times e to the I Omega Z Z plus I evaluated at Z equals Phi that is the value of the residue and that’s because we have a first power of a simple pole ok so let’s calculate what that is that becomes a 2 pi all over 2i e to the negative Omega ok because I times another I makes a negative 1 so those cancel their and we get a pie e to the negative Omega and that’s true provided our capital R is bigger than one and because we’re gonna take our to be you know big very big now we can ensure that for every value beyond that we’re always enclosing the singularity and this this contour interval is always going to be the same value pi times e to the negative Omega for Omega greater than zero we’re going to use this assumption a little bit later to do something alright so the question then in this course is that this contour integral is actually of f of z e to the i omega z DZ is actually we can break it up into two pieces one is the negative r – r eats the negative Omega I Omega x over x squared plus 1 DX plus integral C R that’s this top half and that’ll be e to the I sorry that should be a positive I there I Omega Z over Z squared plus 1 D

C ok and that that whole thing provided R capital R is bigger than 1 we have is equal to PI e to the negative Omega so the goal of course is to as R gets big we need to analyze analyze this when R goes to infinity and prove its it you know it goes to 0 ok so and that means then of course that as R goes to infinity it means that this this integral here then can be associated with that value and this contributes nothing because it goes to 0 ok and then we have successfully computed or very clearly successfully computed the value of this integral case because again this is the objective is to what this value is and if this thing here goes to zero this this contour this radial contour the semicircle in the top half of the plane if that goes to zero then the value that we’ve accumulated here from this contour interval can be associated with that improper interval okay so let’s see if we can do that and so um we’re gonna we’re going to use several tools from complex variables analysis to see if we can analyze this thing so I’m going to write down what those tools are again what we have is redraw it in the plane we have our contour see our up here then closes I goes from R to negative R and on that whole contour Z will be equal the modulus is Z will be equal to R ok all right and we’re going to use two tools so again we need to prove that this integral see our e to the I Omega Z Z squared plus 1 DZ it goes to 0 so the tools the tools we’re going to use is that of the integral bound so and let me talk about what that is again so in general what we have here is if I have some function G of Z and I have a contour C and I say suppose that M of R is so I’m going to redo that G of Z the modulus of G of Z for Z that are on our contour which I’ll call CR is less than or equal to M R where that is may be parametrized by our capital R then we’ve seen this theorem before that we can take the modulus is going to be equal to M R times the L which is the arc length of C R in this case of course C R is equal to PI times R because it is a half circle of radius R ok so there we go that’s our that’s our that’s our one of our tools that’s the first tool the second tool is that of the triangle inequality and this is a in order in order to find em are right there we need to be able to find this mr and so we’re going to use that and so recall if we have something of the form Z plus a over Z plus B where a and B are complex numbers and Z is a complex number as well we can always bound it using the two versions of the triangle inequality we use the upper bound for the numerator and we use that lower bound for the denominator where we take the difference okay all right so we’re going to use that a whole lot in order to find this bound okay because recall that of course that Z is equal to R okay in our particular case so let’s do that let’s let’s find this bound here e so let’s uh let’s uh that’s so what we have here is we look at the

the integrand inside of here we have e to the I Omega Z all over Z squared plus one and we know I’m just gonna leave that numerator alone for right now okay and I’m gonna look at the denominator and we can use that that the difference bound triangle triangle inequality yep there we go like that all right of course I know that that is equal to the modulus e to the I Omega Z all over R squared minus one and that’s because R is bigger than one we can do that we can take off the absolute values the next thing we need to do of course is because again what we’re doing is of course we’re finding that bound on the curve itself so we’re only looking on the curve in which case we know that Z the modulus of Z is equal to capital R the next thing we need to do is look at examine this e to the I Omega Z which is e to the I X plus I Y which is equal to e to the I Omega x times e to the negative Omega Y and that of course is we can break that up into two moduli like that this of course because this is a pure imaginary number is always going to be 1 and recall of course up here that y is always greater than or equal to zero and we also said that Omega is greater than zero in which case this thing right there is always going to be less than or equal to 1 all right so now we have a nice bound there we can continue on and state then that that we have this integrand here is is less than or equal to 1 over R squared minus 1 all right so let’s take this result and finally bound our our thing so again we wanted to look at this this this we wanted to take this part of this component of our contour in a goal and now we have a bound for it it’s 1 over R squared minus 1 and then we have to multiply by the arc length which was PI R so what we have here is pi times R to the first power over R squared minus 1 of course we know that as R goes to infinity then this thing goes to 0 in which case we finally conclude then that the limit R goes to infinity from negative R to R of of e to the I Omega Z sorry not Z x over x squared plus 1 D X is equal to it’s going to be equal to what we had found from the residue theorem e to the I Omega ok all right so that’s good but we have a little bit more of an issue here because of course this is equal to the principal value integral of e to the I Omega X x squared plus one so we we need to find out is it equal to the actual improv I’m going to put a question mark there is it equal to the actual improper integral of e to the I Omega X x squared plus 1 DX and the answer is yes because of course that e to the I Omega X is equal to cosine Omega X plus I sine Omega X and so this part right here is even this is odd and that’s the imaginary part but it turns out of course that we know that now if I break this integral into two pieces this principal value integral into two pieces I get one piece over the reals I can just write that as an abbreviation of cosine Omega x over x squared plus one DX that’s the real part and that has to be associated with the real part on this side and also on and then in addition then of course we have the imaginary part sine Omega X x squared plus 1 DX that’s going to be equal to you know the two halves sorry that has to be from there there sine Omega x over x squared plus 1 DX plus 0 to infinity of the positive part of the integral and what we have here is that this of course it

converges and this converges to an equal and opposite sign in which case they both met at to 0 which explains why that works so actually our integral this integral that we wanted to compute is actually equal to just the cosine part because of course that is the real valued part which we have there but anyway from this result we can actually say yes I no longer question this is an even function and so the principal value this the even part of this function which is equal to the principal value it’s gonna eat the equal to pi e to the negative Omega okay so that gives us a nice result and shows a good way of computing these integrals of Fourier type in later videos we’ll talk about how do other theorems that allow us to establish this these results a little bit more smoothly so we don’t have to do so much analysis all right so thank you very much