22 - Complex Integration: examples and first facts
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welcome back to our second lecture in
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the fifth week of our course analysis of
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a complex kind today we'll learn more
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about complex integration but look at
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some examples and we'll learn some first
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facts remember this is how we define the
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complex path integral given a smooth
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curve gamma and a complex valued
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function f that is defined on gamma we
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define the integral over gamma F of a
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ctz to be the integral from A to B F
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gamma of T times gamma prime of T DT so
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this second integral can be broken up
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into its real and imaginary parts and
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then integrated according to the rules
00:45
of calculus let's look at some more
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examples suppose gamma of T is given by
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1 minus T times 1 minus I where T runs
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from 0 to 1 let's get a quick idea of
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what this path looks like when T is
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equal to 0 gamma of T equals 1 and then
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when T is equal to 1 gamma of 1 is equal
01:09
to 1 minus 1 minus I in other words I so
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that is gamma of 1 and what happens to
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the path in between in between there's a
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linear relationship between X of T and y
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of T who you write gamma of T as X of T
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plus iy of T then the real part is 1
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minus T and the imaginary part is simply
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T so Y is equal to T X is equal to 1
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minus T they're linearly related so we
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just get this line segment from 1 to I
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let's see what the integral that's the
01:56
integral or for gamma f of zdz by
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definition is the integral from 0 to 1
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these are the bounds for the T values of
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the function f the function f of C is
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given by the real part of Z so we have
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to take the real part
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of gamma of T and multiply that by gamma
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prime of T what is gamma prime of T
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gamma prime of T is what the derivative
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of 1 is 0 derivative of minus T times 1
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minus I is minus 1 minus I so remember
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the path integral integral over gamma f
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of zdz is defined to be the integral
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from A to B f of gamma of T gamma prime
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of T DT that's what we're using right
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here so the integral over gamma F of C
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DC is the integral from 0 to 1 F is the
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function takes the real part of whatever
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is put into it we're putting gamma of T
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into it that's right here and then we
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multiply by gamma prime of T so this is
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the integral we have to evaluate let me
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clear this screen here to evaluate this
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integral we need to find the real part
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of 1 minus T times 1 minus I but the
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real part is everything that's real in
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here
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so what's real 1 is real minus T is real
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but then if we multiply through we have
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a plus T times I that's the imaginary
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part so the real part is 1 minus T and
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we're multiplying by minus 1 times 1
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minus I which is the same as I minus 1
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but that's a constant we pull that out
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of the integral so here's the I minus 1
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and then the integral of 1 minus T DT
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antiderivative of 1 minus T is T minus
03:55
1/2 T squared we evaluate that from 0 to
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1 so in the end we get I minus 1 times 1
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minus 1/2 times 1 squared that's the
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integral evaluated at the upper bound
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and if you evaluate it at the lower
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bound we get a 0 so all together 1 minus
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1/2 is 1/2 the value of the integral is
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I minus 1 over 2 let's look at a second
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example
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let's Gayla of t BR e to the I T where T
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runs from 0 to 2 pi we know that that
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parameterize is a circle of radius R
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gamma prime of T we also know what that
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is that is our I either the I T and the
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function f we're looking at is f of Z is
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the complex conjugate of Z let's find
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the integral over gamma f of ZD C F of C
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is the complex conjugate so it's
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integral over gamma of the complex
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conjugate of Z DZ and we know what we
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have to do is we have to look at F of
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gamma of T times gamma prime of T and
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integrate that over the balance of from
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0 to 2 pi so we get the integral from 0
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to 2 pi this is f of gamma of T and
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since gamma of T its are either the I
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see we have to take the complex
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conjugate of our identity so you get a
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complex conjugate of gamma of T and then
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we have to multiply by gamma prime of T
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the complex conjugate of our even though
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I t we looked at that a while ago that's
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our either the minus I T and the
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derivative of gamma is our I into I T so
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I have an R in another art which gives
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me this R squared then there's this I we
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can also pull that up front and what's
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left inside is either the minus I T
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times e to the I T those two cancel each
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other out so the integral is just over
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if you want put the 1 there you could DT
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an antiderivative of 1 its T and we need
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to plug in the upper bound and subtract
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them that the value at the lower bound
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so if the aqua problem we get 2pi at the
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lower bound 0 so the value of the
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integral is 2 pi times R squared I if we
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rewrite that we could write that as 2 I
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times PI R squared and PI R squared is
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the area of this disk so turns out this
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integral is the area of the region that
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is surrounded by the curve and there's
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absolutely a more general fact that says
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if gamma surrounds and I have simply
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connected region then the integral over
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gamma Z bar you see is the area of the
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region it surrounds
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I want to remind you of an integration
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tool from calculus that will come in
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handy for our complex integrals so both
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a B and C B are integrals in R and H is
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a smooth function from C D to a B so a B
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and there C D and H is a function from C
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D to a B F is a continuous function
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defined on a V then integration by
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substitution sets that you can integrate
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f of T DT from H of C to H of D so H of
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C and H of do some points in this
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interval so where F is defined or
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alternatively you can integrate from C
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to D the function f of H of s multiplied
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by H prime of SDS and these two
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integrals are the same thing don't look
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at an example to remind you how this
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goes suppose you wanted to integrate
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from 2 to 4 the function s squared times
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s Q plus 1 to the fourth power D s we
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recognize that that is an integral off
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the form on the right where this is my
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function f of H of s if I said H of s to
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be s cubed plus 1 so this right here is
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my H of s then here I see H of s to the
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fourth power and over here I see almost
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H prime of s H prime of s is 3 s squared
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so I need an extra 3 there and then that
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is H prime of s but I can't just put a 3
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there and you to make up for that so put
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a 1/3 in front of the integral and all
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of a sudden this integral here is of the
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form f of H of s times H prime of SDS
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where F is the function that raises its
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input to the fourth power by integration
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by substitution
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this integral is the same thing as the
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integral from H of 2 to H of 4 H of 2 to
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H of 4 of F of TDT
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is the function that raises its input to
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the fourth power so f of T is T to the
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fourth and I integrate DT and this 1/3
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needs to remain there because that's
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outside of the integral that doesn't
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affect what's happening with my
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transitions on the inside so now I need
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to find the integral of H of 2 to H of 4
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T to the 4th DT what is H of 2 I need to
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plug in 2 for s right here that is 2
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cubed plus 1 that's 9 what is H of 4
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what's 4 cubed plus 1 that's 65 I need
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to find 1/3 times the integral from 9
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and 65 left T to the 4th DT and it had
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derivative of T to the 4th is 1 v 2 v so
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we need to evaluate that from 9 to 65 so
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the result is 115 times 65 to the fifth
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minus 9 to the fifth we can use
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integration by substitution to find out
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that the complex path integral is
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independent of the parametrizations that
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we choose to make process what I mean by
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that let gamma be a smooth curve defined
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on an interval a B and let beta be
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another smooth parametrizations of the
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same curve given by beta of s is equal
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to gamma of H of s where H is a smooth
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ejection so let's look at this picture
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here's the interval from A to B and
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here's the intro from C to D and aged is
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a smooth project in between these two
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intervals gamma is a curve to find them
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a B so here's that curve gamma but I'm
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also looking at a curve beta that's
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given by beta of s it's the same thing
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ass going over with H and then applying
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gamma so gamma H of s is the same as
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beta of s now suppose I have a complex
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valued function that's defined on gamma
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then what
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the integral over beta F of CDC what by
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definition that the integral from C to D
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F of beta of s beta prime of STS what is
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beta of s beta FS is gamma of H of s and
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what is beta prime of s I see the
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composition of two functions so by the
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chain will let scam a prime of H of s
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times H prime of s so that's what you
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see down here now we use our integration
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by substitution factor H of s is our T
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so this is also our team and there's our
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H prime of SDS which will become our DT
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so by integration by substitution this
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is the same thing as the integral from A
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to B f of gamma of t gamma prime of T DT
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and that by definition is the integral
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over gamma f of zdz so the integral over
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beta is the same thing as the integral
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over gamma
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therefore the complex path integral is
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what we say independent of the chosen
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hair motorisation
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now so far we've been talking about
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smooth curves only what if you had a
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curve that was almost smooth except
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every now and then there's a little
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corner like the one I drew down here
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well for a piecewise smooth curve so
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that's a curve where it's put together
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from finally many smooth pieces where
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each piece starts where the previous one
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ends then one can show that the integral
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over gamma f of zdz is the same thing as
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integrating over gamma one adding to
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that the integral over gamma two adding
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to that the integral over gamma three
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and so forth up to the integral over
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gamma and I also want to introduce you
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to reversed paths given a curve gamma
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defined in an integral from A to B
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there's a curve minus gamma and this is
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a confusing notation because we don't
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mean to take the negative of gamma of T
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it's literally a new curve minus gamma
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so if you don't like this notation call
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this gamma tilde or gamma star or
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something like that so this is a new
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curve for : even beta so there's a new
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curve also defines an AV and it's given
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by taking the original curve gamma but
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instead of evaluating at T we evaluated
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at a plus B minus T so if you take minus
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gamma and evaluate at its initial point
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a what you actually get is gamma of a
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plus B minus a which is gamma of B so
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the initial point of the curve minus
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gamma is actually the point with the
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original curve gamma in it so in this
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picture down here gamma ends at gamma P
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but that is the starting point of the
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curve minus gamma furthermore minus
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gamma of B is gamma of a plus B minus B
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so that's gamma
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II so minus gamma ends where gamma used
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to start the curve minus gamma passes
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through all the points that gamma went
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through but in Reverse orientation
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that's why it's called the reverse path
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now what's the derivative of minus gamma
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minus gamma prime of T is the derivative
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of this function gamma of a plus B minus
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T that's a composition of two functions
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so we could gamma prime of a plus B
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minus T times the derivative of what's
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inside but the derivative of a plus B
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minus T is minus one if that is a
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continuous function that's complex
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valued on gamma what happens when I
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integrated over minus gamma the integral
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over a minus gamma a folksy dizzy by
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definition is the integral from A to B F
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of minus gamma of s minus gamma prime of
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s yes and then you can go through what I
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wrote down here to find out it's
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actually the negative of the integral
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over gamma F of C DC so if you integrate
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a function over a reverse path the
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integral flips its sign as compared to
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the integral over the original path here
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are some facts about complex curved
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integrals so as always gamma is a curve
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C is a complex constant and F and G are
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continuous and complex valued and gamma
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the integral over gamma of F plus G can
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be pulled apart just like in regular
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calculus you can pull the integral apart
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along with some so this equals to
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integral over gamma f of zdz plus the
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integral over gamma G of CT see further
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more complex constants can be pulled out
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and we've been doing this so the
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integral C times F is C times the
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integral of F and this one we just
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showed the integral over the reverse
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path is the same as the negative of the
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integral over the original path
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now let's figure out how we could find
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the length of a curve given a curve
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gamma how do we find how long it is
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normally you would take maybe a piece of
16:53
yarn lay it along the curve then
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straighten it out and measure its length
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how do you actually do that
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well suppose we take this interval from
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A to B and subdivided again to its
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little pieces and look at these
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intermediate points on the curve and we
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can approximate the length of the curve
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by just measuring straight between all
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those points and the closer the points
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are together the better the
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approximation seems to be so the length
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of gamma can be approximated by taking
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gamma of tj+ along minus gamma of T J
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and the absolute value of that if this
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sum has a limit as n goes to infinity
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that is called the length of gamma and
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if those limit exists we say that the
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curve gamma is rectifiable or it has a
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length
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note that not every curve has a length
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you could imagine that you know even
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though it seemed that this piece was a
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good approximation of this curve here as
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you zoom in really far you you know zoom
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into little little piece right here if
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you zoom into that maybe there's a lot
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more going on than you actually thought
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and it's a whole lot longer than you
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thought and then if you zoom into
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another little piece that happens again
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so none of your approximations will ever
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be any good if that is the case the
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curve won't be rectifiable
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but for us most of the curves we deal
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with are rectifiable and have a length
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my question is how do we find that
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length so we know it's given by the
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limit of the sums but that doesn't
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really help because you can't really go
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measure all these little distances and
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add them up so is there a way to
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actually calculate the length of a curve
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given its parameterization and there is
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the idea comes by looking at this some a
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little bit more carefully and applying a
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trick that we applied before so we look
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at gamma of TJ plus 1 minus gamma of TJ
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that's the line segment between
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consecutive points and divide that by TJ
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plus 1 minus TJ and immediate
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multiplied by TJ plus one minus TJ and
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we observe that this term here if the
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TJ's are close to each other is roughly
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the absolute value of the derivative
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gamma prime of TJ and this is my delta
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TJ as before as n goes to infinity this
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sum goes to the integral from A to B of
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gamma prime of T DT in other words the
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length of gamma can be found as the
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integral from A to B the absolute value
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of gamma prime of T DT this is true for
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any smooth or piecewise smooth curve
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gamma for smooth or piecewise smooth
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curves gamma you don't have to worry
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about the length not existing those all
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have a length and it can be found in
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this way let's look at some examples
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let's go back to our curve gamma of T
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equals R e to the I T we all know what
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that looks like that's simply a circle
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of radius R and we even know how long
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that curve should be this is the
19:59
circumference of this circle but that's
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actually calculated with our formula we
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know that gamma prime of T is our either
20:08
the I T and so the length of gamma is
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given by the integral from 0 to 2 pi of
20:12
the absolute value of RI e to the I T
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that's my gamma prime of T right here DT
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but the absolute value of either the I T
20:23
is 1 I has absolute value 1 so the
20:26
absolute value of gamma prime is simply
20:28
R so we're integrating our from 0 to 2pi
20:31
our two constant and antiderivative is R
20:34
times T we plug in 2 pi we get 2 pi R we
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plug in the 0 that's nothing so the
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length of this curve is 2 PI R and we
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knew that the circumference of a circle
20:45
of radius R is indeed 2 pi R let's look
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at another example let gamma of T be the
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curve T plus I 2 we looked at this curve
20:55
before here's what it looks like when T
20:58
is equal to 0 it's at the origin when t
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is equal to 1 it is had 1 plus I and in
21:07
between
21:08
goes linearly so here is my curve gamma
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and I want to find out how long in this
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and again by looking at this picture I
21:15
can calculate its length because it is
21:17
the hypotenuse of a triangle both of its
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legs have length 1 so the hypotenuse has
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length square root of 2 let's see if our
21:25
formula gives us the same result well
21:28
first of all gamma prime of T is 1 plus
21:31
I and so the length of gamma is found by
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integrating from 0 to 1 the absolute
21:35
value of gamma prime of T so the
21:37
absolute value of 1 plus I DT what is
21:40
the absolute value of 1 plus I the
21:43
absolute value of a complex number can
21:45
be found by taking the square root of
21:47
the real part squared plus the imaginary
21:50
part squared both real part and
21:53
imaginary part are one together adds up
21:55
to two as the square root of 2 is the
21:57
length of 1 plus I so we're integrating
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from 0 to 1 square root of 2 DT square
22:04
root of 2 as an antiderivative which is
22:05
square root of two attempts T we're
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plugging in 1 and 0 and so we find
22:10
square root of 2 as the answer and
22:13
that's exactly what we expected this
22:14
length or T is indeed square root of 2
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now this prompts and new definition
22:25
suppose gamma is a smooth curve F
22:27
complex-valued and continuous on gamma
22:29
we defined the integral over gamma f of
22:32
Z DZ and the only way that's differed
22:35
from the previous integral is that we
22:37
always haven't put these absolute value
22:39
signs around you see we're defining that
22:42
to be the integral from A to B F of
22:44
gamma of teeth times the absolute value
22:46
of gamma prime of T DT so that's the
22:49
only way in which this new integral that
22:52
were defining differs from the complex
22:54
path integral we call this integral of F
22:57
over gamma with respect to arc length
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because this absolute value of gamma
23:04
prime of T was related to finding the
23:06
length of a curve let's look at some
23:09
examples the length of a curve gamma we
23:13
just found that that can be found by
23:14
taking the integral from A to B of gamma
23:18
prime of T absolute value TT so in my
23:22
notation the function f of gamma T is
23:25
just the function 1 so there's F
23:28
identically equal to 1 and then this
23:32
length interval agrees with the integral
23:34
on the right but by definition that is
23:37
then the interval of 1 times absolute
23:39
value dizzi
23:40
so the length of gamma is the integral
23:43
over gamma of the absolute value of T C
23:45
let's look at another example suppose we
23:48
wanted to integrate over the circle of
23:50
radius 1 and remember when we use this
23:52
notation absolute value of C equals to 1
23:54
we automatically assume the circle is
23:57
oriented counterclockwise and typically
24:00
we choose the parametrization gamma of T
24:03
equals e to the I T which he runs from 0
24:07
to 2pi so this is the integral from 0 to
24:10
2pi f of gamma of t but f of z is the
24:14
function z so f of gamma of t is simply
24:17
either the I T times the absolute value
24:20
of the derivative gamma prime of T in
24:23
this case is I either the I T but the
24:26
absolute value of gamma prime of t is
24:30
equal to 1 so that's where this one
24:32
right here comes from so we're
24:34
integrating from
24:35
to 2pi either the IT and antiderivative
24:38
either the IT we found that last time as
24:41
minus items evil IT we integrate that
24:44
from 0 to 2pi
24:45
and find minus items either the 2pi
24:50
minus minus plus i times e to the 0 and
24:54
those terms cancel each other out and we
24:56
end up with 0 note that we could have
24:58
also used the piecewise smooth curves in
25:00
all of the above we would have broken up
25:03
the integral into the sum over smooth
25:05
pieces as before he is a great estimate
25:10
sometimes it's impossible to find the
25:12
actual value of an integral but all we
25:14
need is an upper bound is no bigger than
25:16
some certain number and this is called
25:18
the ML estimate given curve gamma and a
25:22
continuous function on gamma it can be
25:24
shown that the integral over gamma f of
25:26
zdz the absolute value of that integral
25:29
is bounded above but the integral over
25:31
gamma absolute value of f of z absolute
25:34
value TZ in other words the absolute
25:37
value can kind of be pulled to the
25:38
inside this reminds us a little bit of
25:41
the triangle inequality remember a plus
25:44
B absolute value is bounded above by the
25:46
absolute value of a plus the absolute
25:48
value of B this can be viewed in the
25:51
similar manner actually proofs in a
25:53
similar manner in particular if you
25:56
happen to know that your function f is
25:58
bounded by some constant M along gamma
26:01
then this f of C would be less than or
26:05
equal to M so you could go one step
26:08
further it's less than equal to the
26:09
integral over gamma M DZ you could then
26:14
pull the M outside of the integral and
26:16
you're left with the integral over gamma
26:18
DZ which is the length of gamma so if F
26:22
is bounded by some constant M on gamma
26:24
then the absolute value of this path
26:26
integral is bounded above by M times the
26:29
length of gamma which length L would be
26:32
a good approximation put that that's why
26:34
this is called the ML estimate let's
26:38
look at some more examples suppose we
26:40
wanted to find the integral over the
26:44
circles equals 1 of 1 over the absolute
26:47
value of T Z so the integral with
26:49
respect to arc length again we know the
26:52
parametrization we're using is gamma of
26:54
T equals e to the I T and we already
26:57
showed that the absolute value of gamma
26:59
prime of T is 1 so the integral of 1
27:03
over Z ops of value DZ by definition is
27:06
the integral from 0 to 2pi this is my F
27:09
of gamma of T so 1 over gamma of T so 1
27:13
over e to the I T times gamma prime of T
27:16
which is 1 DT an antiderivative of
27:19
either the minus I T it's I times e to
27:22
the minus I T evaluated from 0 to 2 pi
27:25
again the two terms that you get cancel
27:28
each other out
27:28
the integral evaluates to 0 next let's
27:34
look again at our path gamma of T equals
27:36
T plus I T so again that was the path
27:38
from the origin to 1 plus I we'd like to
27:45
find an upper bound for the integral
27:47
over gamma of the function Z squared DZ
27:50
let's first use the ML estimate so the
27:54
second part of our theorem which said
27:56
that the integral over gamma f of zdz
27:58
absolute value is bounded above by M
28:00
times the length of gamma where m is
28:03
abound than F on this path gamma so for
28:07
us F of Z is the function Z squared
28:09
what can I bend do we have for F for Z
28:12
values that are from this path gamma
28:15
what F of Z is in absolute value the
28:18
absolute value of Z squared and the
28:21
absolute value of Z on this entire path
28:23
gamma never gets bigger than this
28:26
absolute value of 1 plus I which is the
28:29
biggest it gets an absolute value but
28:31
one of us is absolute value is square
28:32
root of 2 so the absolute value of Z
28:35
never gets bigger than the square root
28:36
of 2 and so the absolute value of Z
28:38
squared is founded about Phi 2 on gamma
28:42
so we can use M equals 2 on gamma we
28:45
also know that the length of gamma is
28:47
root 2 we calculated that earlier and
28:49
therefore using the ML estimate the
28:52
absolute value of the path integral of Z
28:54
squared TC is bounded above by M which
28:57
is 2 times the length of gamma which is
29:00
square root of 2 so it's 2 square root
29:01
of 2
29:09
let's try to also use the first part of
29:12
that theorem to find an estimate maybe
29:15
even a better estimate for the integral
29:17
of Z squared is Z over gamma
29:18
so again gamma of T is T plus I T gamma
29:21
prime of T is 1 plus I and so the
29:25
absolute value of gamma prime of T is
29:27
square root of 2 which is the absolute
29:30
value of 1 plus I F of Z is the function
29:32
Z squared the first part of the theorem
29:35
said that the absolute value of the
29:37
integral over gamma f of zdz is bounded
29:40
above by just pulling the absolute
29:42
values inside so the integral over gamma
29:45
absolute value of f of c absolute value
29:47
of TC let's see if we can calculate that
29:50
so the integral over Z squared DZ is
29:55
bounded above by the integral over the
29:57
absolute value of C squared absolute
29:59
value of G Z by definition that's the
30:03
integral from 0 to 1 we look at gamma of
30:07
T instead of Z squared and then we need
30:09
to fill in absolute value of gamma prime
30:12
of T DT but gamma of T is T plus I T so
30:17
we need to take the absolute value of
30:18
that and square it and then multiply
30:21
with the absolute value of gamma prime
30:23
of T which is square root of 2 what is
30:27
the absolute value of T plus I T squared
30:34
well we take the real part and squared
30:36
the real part is T and then we take the
30:39
imaginary part in squared the imaginary
30:41
part is also T so all together the
30:44
absolute value is 2t squared and that's
30:46
what you see right here and then we
30:48
multiply it with square root of 2 which
30:49
was the absolute value of the derivative
30:51
the 2 on that squared of 2 can be pulled
30:54
outside of the integral we're left with
30:56
the integral from 0 to 1 of T squared an
30:58
antiderivative of T squared is 1/3 T
31:01
cubed and that's what you see right here
31:03
evaluate that from 0 to 1 when you plug
31:06
in 1 for T you get 2 root 2 over 3 when
31:09
you plug in 0 the term vanishes and so
31:12
the integral has value 2 root 2 over 3
31:15
we found that the integral over gamma Z
31:17
square DZ it's bounded above by 2/3
31:21
rrah - now let me remind you that we
31:25
actually calculated this integral
31:27
earlier we calculated its actual value
31:30
as 2/3 times the minus 1 plus I in the
31:34
last lecture minus 1 plus I has absolute
31:38
value of square root of 2 so if you put
31:42
absolute values around this this
31:45
actually equals 2/3 times Row two so the
31:50
estimate we got was as good as it gets
31:53
it's a sharp estimate it doesn't get any
31:56
better the estimate is actually
31:58
inequality in this particular case you
32:00
will not always get inequality but this
32:02
example is set up to yield an equality
32:05
here so this doesn't get any better you
32:07
cannot improve this estimate because we
32:09
found an example in which case equality
32:12
is actually true next up is the
32:16
fundamental theorem of calculus for
32:18
analytic functions we already saw it for
32:20
real valued functions and will now be
32:23
able to prove a similar fact for
32:26
analytic crunches