How to Integrate Complex Logarithm Functions Using Branch Cuts

grieving students and welcome back to another lesson on complex variables in this video I'm going to talk about

integrating complex logarithm functions watching the previous video on branch point and branch cuts is essential links

in the description because that's what I'm going to be applying here I also recommend solving this problem with me

to better consolidate things our goal in this lesson is to integrate the natural log of x over 1 plus x squared from 0 to

infinity there's really three basic steps behind performing an integration like this the first is to write the

function being integrated in terms of Z the second is to set up the contour of integration using a branch cut to make

the natural log a single valued function as opposed to a multiple valued function and the third step is to set up the

integration and apply the residue theorem let's do the first step where we'll let f of Z equal lawn of Z over Z

squared plus 1 the function being integrated now the denominator here can be factored as Z plus I times Z minus I

which is going to come in handy later now this factorization means that if we draw the complex plane the denominator

of F of Z gives rise to two poles one at Z equals I up here and the other at Z equals negative I down here in addition

the natural log itself is undefined at 0 so the origin is also a pole now comes the second step putting in a branch cut

that'll make my function single valued and setting up an appropriate contour of integration in the last video I showed

you that you could make a branch cut in any direction but for this problem there are a few special considerations one

consideration is that we probably don't want a branch cut along the real axis because ultimately this real portion of

our contour integral will get us what we want for our answer in addition we don't want our branch cut such that our

contour goes through any of our poles if it does we can't use the residue theorem the residue theorem just doesn't work if

the contour has poles on it only when the poles are inside the contour does the residue theorem apply so keeping all

this in mind here's what I'll do I'll make a branch cut along the negative imagine

axis and keep the argument of my complex numbers Z between negative PI by 2 and 3 PI by 2 I also need to avoid the origin

so here's the contour that I'm going to make I'll make a large semicircular curve of radius capital R with center oh

that's going around then I'll make a small semicircular contour of radius Rho with center capital o that's going

around and I'll connect them like this and end up with this composite contour which I'll call see the contour integral

over this closed curve C is then the sum of the integrals over these semi circular arcs which I'll call C sub R

and C sub Rho the C sub Rho integral is clockwise plus the integrals over the real axis from negative R to negative

row and from row to capital R if we want to find our integral in question we should be focusing on this particular

integral and we should also take the limits as Rho approaches 0 and capital R approaches infinity when we do that this

is the integral equation we end up with for the integral that we care about I'll call this equation 1 now according to

the residue theorem we can write this closed contour integral over C as 2 pi I times the sum of the residues of the

poles of F of Z contained within the contour C but the only pole that's actually contained and C is the pole at

Z equals I so we just have to find the residue of F of Z at Z equals I let me do that on the side here Z equals AI is

a simple pole since Z minus AI only occurs once in the denominator of f of Z so all we have to do is multiply f of Z

by Z minus I and substitute Z equals I into the remaining expression when we multiply F by Z minus I we're only left

with Z plus I in the denominator and when we substitute Z equals I will get lon of I over 2i but we don't know what

law and I is or do we recall from the previous video that the natural log of a complex number is the natural log with

its modulus plus I times the argument of that complex number plus 2 pi n where n is some integer so in our case the

natural log of I is just the natural log of the modulus of AI the modulus of AI is 1 so it's

natural log of 1 is just 0 plus I times the argument of AI plus 2 pi 9 the argument of AI is PI by 2 and I can also

ignore the 2 pi n do you know why I've done that if we go way back up you'll see that the branch I created to make my

natural log single valued takes in Z's with an argument between negative PI by 2 and 3 PI by 2 so for Z equals I the

only option for the argument that I have which is also contained within this interval is PI by 2 that's why I picked

it and that's why I crossed off the 2 pi n I have a single valued branch I'm not dealing with a multiple valued function

anymore anyway once I substitute this law and I into my expression for the residue I end up with a residue of PI by

4 at Z equals I for the function f of Z plugging this into equation 1 gets me the following expression for my integral

now I can make further simplifications to all of this if I move this last integral from negative R to negative Rho

to the left hand side and if I write out the function f of Z in all of my integrals now in this first integral on

the left I'm just integrating over the positive real numbers so I can take out the Z and

replace it by a simple real number X for the second integral I can use the definition of the complex logarithm to

write out the numerator as follows as the lon of the modulus of Z plus I times the argument of Z again I'm not

including the 2 pi n here because this logarithm has been modified to be single valued we're only using a single branch

for this integral we're also integrating over the reals so again I can replace Z by X however the argument of Z is now PI

because we're working over the negative real axis again it's PI because we've restricted our arguments to lie between

negative PI by 2 and 3 PI by 2 to create our branch the only argument of Z which falls within this interval that

corresponds to the negative real axis is then PI now what we do is split up the terms in the numerator and split up the

result integrals to obtain the following the denominator the second integral is an

even function so it's the same no matter whether we're integrating from zero to infinity or negative infinity to zero

it's symmetric on a reflection over the imaginary axis or the y axis in addition this natural log of the absolute value

of x becomes simply the natural log of X if I change the limits on my second integral to become zero to infinity so

for the second integral I can simply change it to going from zero to infinity because the integrals from a negative

infinity to zero is the same when I do that I can combine it with the first integral to get the following two times

that integral from 0 to infinity of lon x over x squared plus 1 now this integral of 1 over x squared plus 1 is

just the inverse tangent if you recall your identities from calculus 2 when I evaluate the inverse tangent from the

limit of X going to infinity and subtract the inverse tangent of 0 I get PI by 2 minus 0 which is just PI by 2 so

now if I isolate my integral of interest I end up with the following equation because the two constants cancel out all

that's left now is to evaluate these two integrals on the right but I won't really evaluate them I'll show that both

of them are 0 using the ml inequality theorem or the estimation lemma that I covered in a previous video links in the

description I'll start with this first integral the integral over C sub capital R because the function being integrated

is analytic over the C sub R I can use the ml inequality to say that the magnitude of this integral is less than

or equal to the maximum value on the contour of the function being integrated the capital M times the length of the

contour L and I'll call this inequality equation 2 the length of the contour L is just PI R because it's the

semicircular contour 2 pi R is the circumference of the full circle therefore PI R is the circumference of

the semicircle but what about the maximum value that's the real question well the

modulus of the whole integral is just the integral of the modulus which I can then split up into separate moduli of

the numerator and denominator so how do I find the maximum that possible value of the function that's being integrated

here well I can find the maximum possible value of the numerator and the minimum possible value of the

denominator if I expand out the numerator using the definition of natural log here's what I'll get now the

magnitude of the sums of the lon of the modulus of Z plus I times the argument of Z this magnitude is less than the sum

of the individual magnitudes for the triangle inequality of complex numbers now on the contour C sub R the modulus

of Z is just R in addition the maximum possible value of our Z on this contour is pi which is the left end of the CR

contour the eye goes away because it's magnitude is just 1 therefore the maximum possible value of the numerator

is just lon R plus pi for the denominator the minimum possible value is found by using expressions which the

denominator must be greater than or equal to using this logic the magnitude of Z squared plus 1 must be greater than

or equal to the magnitude of Z squared minus 1 this should make intuitive sense therefore since we're on the

semicircular contour CR the minimum possible value of the denominator is R squared minus 1 and as a result we can

write our M for this integral our maximum possible value as the following and if we substitute this into the ml

inequality and equation 2 we get the following equation if I now take the limit of this as capital R approaches

infinity I can show that the limit of this expression on the right-hand side approaches 0 using l'hopital's rule as

our approaches infinity as an exercise I invite you to apply l'hopital's rule to determine this limit to verify this I

won't do that here because that's not the primary goal of this lesson anyway since the magnitude is 0 we can conclude

that the integral itself is 0 as capital R approaches infinity let's now to turn

the integral over the smaller semicircular arc serum using the ml inequality the magnitude of this

integral again must be less than the maximum value on C row of the function being integrated times the length of C

row by the same logic of the previous calculation we can calculate the maximum possible value of the function on C row

to be the following the only difference here is that in the denominator I have 1 minus Rho squared because Rho is

approaching 0 so Rho is a small number so Rho must be less than 1 we can't have Rho squared minus 1 because that's

negative and you can verify this identity that the magnitude of Rho squared plus 1 must be greater than or

equal to 1 minus the magnitude of Rho squared anyway the length of the contour here is

pi times Rho and if I take the limit as Rho approaches 0 I can once again show that the answer is 0 for the Rho times

lawns Eero term I can just use l'hopital's rule to prove this and I invite you to verify that this is indeed

the case we can therefore conclude that the integral over C Rho is 0 as Rho approaches 0 finally when we plug all

this back into the equation for the integral we ultimately want we find that since both of our integrals on the right

are 0 as we take the limits the integral from 0 to infinity of lon x over x squared plus 1 is 0

all that work for nothing but the result doesn't matter it's the journey that really counts and in this journey we

learn how to integrate complex logs using branch cuts anyway that should do it for this video I'd like to thank the

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