Computing Improper Fourier Integrals Using Complex Analysis Techniques

greeting students and welcome back to another video on complex variables in this lesson we're gonna learn how to

compute improper integrals that come from Fourier analysis and by improper integrals from Fourier analysis what I

mean is integrals of the form the integral from negative infinity to infinity of f of X times sine ax DX and

the integral from negative infinity to infinity of f of X times cosine ax DX where a is some positive real number now

if a were negative you could still take out the negative sign from the sine and the cosine expressions such that the

number multiplying X and the sine and cosine terms would always be positive thanks to these convenient identities

so although the integration technique and justification for that integration technique rely on a positive a you can

always still manage a negative a by taking out the negative sign using either of these identities now let's

start by recalling the manner by which we integrated improper integrals in a previous video the links in the

description in that video the functions we were integrating were just rational expressions which didn't involve sine

and cosine and also satisfied a number of other conditions what we did there was replace the X by the complex

variable Z and then perform a contour integral over the closed contour C where C was a curve enclosing this semi

circular area in the complex plane now this closed contour integral equals the contour integral over the

semicircular arc C sub R plus the integral over the line segment on the real axis now the way we calculated our

desired improper integral was to first compute these two contour integrals in terms of the radius capital R and then

take the limit of both sides as capital R approached infinity and then finally isolate this integral over the line

segment of the real axis to get our desired answer now the technique we're gonna use to integrate Fourier integrals

is quite similar to the technique we use to perform improper integrals of just rational functions however we can't use

the exact same method where we replace the X in our improper integral by Z the reason for that is if you take either

the sine or the cosine of X and replace that by the sine or cosine if a complex number Z then the moduli of

the sine and cosine will depend directly on the exponential of a Y the imaginary part of Z in fact you can actually

derive these formulas for the moduli of cosine a Z and sine a Z I'm not going to do that here but this is what they look

like now since the hyperbolic sine is just the exponential of a Y minus the

negative exponential over 2 this confirms that the sine and cosine of a complex number Z depend directly on the

exponential of a Y this is actually a problem when you're looking at this integral over the semicircular arc

because as the radius capital R approaches infinity in this limit Y also approaches infinity and when Y

approaches infinity the moduli of the sine and cosine of the complex number Z now start getting prohibitively large

because of their dependence on the exponential of a Y so instead of straight-up just replacing the X by the

Z in our Fourier integrals we actually replace the f of X by F of Z but then for the sine or cosine terms we like to

replace them by the exponential of I this is because if you split the Z into X plus iy where x and y are real numbers

then if you take the magnitude of the exponential of I a Z you'll end up with the magnitude of the exponential of I a

x times the magnitude of the exponential of negative a Y now in this case even as our approaches infinity this modulus is

going to remain finite on the integral over the semicircular arc the reason is that this first factor the exponential

of AI ax just involves the sine and cosine of a real number so it's magnitude is going to be just 1 in

addition the second factor the exponential of negative a Y will also not approach infinity as capital R

approaches infinity when you're looking at the integral over the semicircular arc because the semicircular arc is on

the upper half plane where Y is always going to be positive so this capital R approaches infinity Y will become a

progressively large positive number and the exponential of a negative of a positive number will always approach

zero as that positive number approaches infinity it will remain finite so it's almost as if when you take the

exponential of I a Z which is just the combination of cosine AZ and sine AZ the e to the a Y in the modulus of the sine

term cancels with the e to the a y in the modulus of the cosine term so the modulus of the whole combination no

longer involves the exponential of a Y therefore by combining the sine and cosine to form the exponential we've

effectively nullified the exponential of a Y which causes problems let's go over the technique of dealing with these

improper Fourier integrals the first step is to replace the f of X in the integral by f of z once you do that

you'll need to ensure that f of z is a rational function or the ratio of two polynomials P and Q we would also need

to ensure that P and Q have real coefficients and no common factors that the degree of Q the highest power of Q

is greater than the degree of P and finally that Q of Z has no real zeroes but at least one zero above the real

axis the reason that it can't have real zeros is that if Q did have real zeros and we were integrating it over the

entire real line then we might run into some issues with keeping our integral finite so to avoid those issues we

impose this fourth condition step two is to replace the sine or cosine terms by the exponential of I a Z as I just

talked about and convert the real integral you're working with to a contour integral over a semicircular

section C of radius capital R over the upper half of the complex plane now this semicircular section consists of a

straight line from negative R to R as well as a semicircular arc that all again label C sub R it should be easy to

see that the integral over this entire closed curve C equals the sum of the integrals over the line segment from

negative R to R and over the semicircular arc C sub R and this would be our step 3

breaking down the integral over this closed curve C into an integral over the line segment plus the integral over the

semicircular arc I'll call this equation 1 now step 4 would be to let capital R approach infinity in that case equation

1 now involves an improper integral instead of an integral over a simple line segment if you use the residue

theorem you should be able to evaluate the left-hand side because it's a closed curve you're integrating over and if you

use jordan's lemma you should be able to cancel the semicircular arc integral on the right and this would allow you to

determine the improper integral of f of Z times the exponential of IAC now the final step once you have this improper

integral is to take its real part if your original goal was to find the cosine integral and to take the

imaginary part if the original goal was to find the sine integral this makes sense if you look at the Euler formula

if I take this exponential integral and use the Euler formula on the exponential then it breaks down into a cosine

integral and I times a sine integral so if the integral on the left is a complex number the real part of that complex

number would be equal to the cosine integral while the imaginary part of that complex number would equal the sine

integral and that's why step 5 is the way it is for a cosine integral you take the real part for a sine integral you

take the imaginary part hopefully all those steps make sense but it's one thing to write down and explain the

steps and another thing to implement and apply those steps so let's do that let's solve an example problem the problem is

relatively simple we want to integrate cosine x over x squared plus alpha squared from negative infinity to

infinity where alpha is some real number essentially the problem over here is to integrate the rational function f of X

which is 1 over x squared plus alpha squared times the trigonometric function cosine X let's go through each of our

steps the first step is to make sure that our f of X satisfies these four conditions which it does it's a rational

function with real coefficients and no common factors between the numerator and denominator the denominator has a degree

of 2 while the numerator has a degree of 0 and the denominator has no real zero so all the conditions are satisfied step

two if we go back up again is to replace the cosine by the exponential and put everything in terms of Z with the

integral being made into a contour integral over a semicircular region C with radius capital R step three would

be to break up this contour integral into an integral over a line segment and an integral over a semicircular arc

CR using this diagram above as your reference for what C and C are mean step four is to take the limit as capital R

approaches infinity the integral over the line segment from negative R to R now becomes an improper integral now the

contour integral on the Left which I'll label a can be evaluated using the residue theorem which I'm going to do on

the side here now the residue theorem states that the integral of this function over a closed curve C equals

the sum of the residues of the poles of that function that are contained in the curve C so if we want to apply the

residue theorem our first task is to find the poles of this function and that's pretty easy just set the

denominator equal to 0 and solve for Z you'll get two solutions I times alpha and negative I times alpha now the only

pole that's contained in the curve C which occupies the upper half plane is the solution I times alpha so that's the

only residue we want to find the next thing we do is find the residue of the function we're integrating at this pole

I times alpha that's relatively easy because this is a simple pole we just multiply the function by Z minus I alpha

and then evaluate the result at Z equals I times alpha when you do this you end up with e to the negative alpha over 2i

alpha so now if you plug in the residue and apply the residue theorem you'll find that the integral of this function

over the closed contour c is pi times e to the negative alpha over alpha let's go back to our original integral

equation and write the value of this integral a that we just evaluated all that we've got left now is this integral

over the semicircular arc I'm going to label this integral as B and also go to the side and evaluate it the idea here

is to use jordan's lemma to say that as capital R approaches infinity this integral approaches 0 but if you

remember the previous video then in order to apply jour Don's lemma we need to make sure that three conditions hold

the first is that the function we're integrating is analytic everywhere beyond a certain distance from the

origin which it is the function only has two poles Z equals plus or minus I times alpha and

if we go beyond those poles the function is otherwise analytic it's continuous and differentiable the second condition

is that we're integrating over a semicircular arc which we are in this case C sub R is a semicircular arc the

third condition is that the function we're integrating has an upper limit on its magnitude over the semicircular arc

C sub R and that this upper limit approaches zero as capital R approaches infinity if all these conditions hold

then we can safely say that this whole integral B approaches zero as capital R approaches infinity

according to jordan's lemma now the only thing stopping us from straight-up applying sure Dan's lemma is the upper

limit condition so we're gonna have to find an upper limit on the magnitude of the function we're integrating so we'll

start by looking at the magnitude of the function first we'll split up the numerator and denominator magnitudes

because we're allowed to do that the magnitude of the product of complex numbers is the product of their

magnitudes and since we're integrating over a semicircular arc we can write Z using the polar representation of

complex numbers as Z equals capital R times the exponential of I theta where capital R is obviously fixed it's the

radius of the semicircular arc which means that our complex number Z on the semicircular arc pretty much only

depends on theta now the magnitude of the exponential if you evaluate it is the exponential of negative R times sine

theta and this is from my previous video on jordan's lemma so I suggest you go back to it in case you want to find out

how I made this jump now in addition the magnitude of the denominator is this square root quantity if you actually

substitute Z in there and then carry out the algebra now to get the maximum possible value of the magnitude of our

function we need to find the highest possible value of the numerator which is just 1 over the semicircular arc where

theta varies from 0 to PI and we also need to find the lowest possible value of the denominator in the only way you

can get that lowest possible value is if you let Co sine of two theta equal its lower bound

of negative one so after simplifying we can therefore say that the upper limit the maximum possible value of the

magnitude of our function is one over R squared minus alpha squared it's easy to see that this maximum possible value

approaches zero as capital R approaches infinity because of the capital R squared in the denominator so therefore

we can use jordan's lemma to say that our integral over the semicircular arc approaches zero as capital R approaches

infinity we can now go back and cross off this integral B in which case we can see that the integral from negative

infinity to infinity of e to the I Z over Z squared plus alpha squared is just pi times e to the negative alpha

over alpha and this should cover our step four the final step is to take the real part of this expression because our

initial integral involved a cosine and since the integral of this exponential is already a real number we can conclude

that the integral from negative infinity to infinity of cosine x over x squared plus alpha squared is pi times e to the

negative alpha over alpha anyway that should do it for this video I'd like to thank the following patrons for

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the Faculty of Khan signing out