21 - Complex integration
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welcome to the first lecture in our
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fifth week of the course analysis of a
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complex kind today we'll learn about
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integration in the complex plane but to
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begin with let's review integration in
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are given a function f that's defined on
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some interval a b and is real valued and
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continuous we define the integral from A
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to B of F of T DT as the limit of these
00:31
left-hand sums so we take f at a point
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TJ multiplied it by the next point TJ
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plus 1 minus TJ and we add all these
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things up what are these TJ's well we
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take the integral from A to B and be
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divided up into these little chunks so a
00:55
is equal to t0 the next little point is
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t1 then t2 and so forth up to TN and NS
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on a large number and we're going to let
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that go to infinity later on so right
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now we're dividing up this interval
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duties and chunks and we evaluate F at
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the left hand point of each of these
01:13
integrals so for example f of T 0 is
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this value right here we take that value
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and multiply it by t1 minus t0 what is
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t1 minus t0 that's this length right
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here so this length is f of t0 t1 minus
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t0 is that length so when I multiply
01:31
those two I get the area of this little
01:34
rectangle similarly I take f of T 1
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which is that value right here so here's
01:44
F of T 1 and that's this length I
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multiply that beat by t2 minus t1 and
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get the area of the next rectangle and
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so by taking the sum of all these
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products I'm going to add up the values
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of all the areas of these rectangles but
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that's an approximation for the area
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under the curve so in the limit as n
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goes to infinity in these rectangles
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become very narrow they give a very good
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approximation for the area under the
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curve in other words if F happens to be
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above the x-axis just like I'm doing
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right here then this integral is
02:23
actually simply the area under the curve
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otherwise if F is sometimes above and
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sometimes below the x-axis we're going
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to take those portions that are above
02:34
the x-axis and subtract from them the
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portions that are below the x-axis so
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that's the meaning of this integral
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here's the fundamental theorem of
02:47
calculus if f is a function as we
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discussed above we can define a function
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uppercase f of X as the integral from a
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to X F of T DT so in other words if F is
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above the x-axis the integral from a to
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X is simply the area under the curve
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between a and X now X is the variables X
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varies so f of X gives you this area
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depending on that the value of X's of X
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gets bigger in my picture then f of X
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gets bigger if X is smaller f of X gets
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smaller because F happens to be above
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the x-axis but you can define a function
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upper case F like this also if F is
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above and below the x-axis it turns out
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this function that you defined is a
03:32
differentiable function and it's
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derivative is equal to little f of X
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that's the fundamental theorem of
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calculus and why is this so useful upper
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case F actually gives you an
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antiderivative of a function that when
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you differentiate it is little F in
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general we say that a function uppercase
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f that satisfies that a derivative is
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little F such a function is called an
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antiderivative of that another following
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suppose you found two different anti
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derivatives F and G so both of them have
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a property that F prime is equal to
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little F and also G prime is equal to
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the left suppose you found two different
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functions well what is G minus F prime
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what it's G prime minus F prime but both
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of these derivative
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the left so that's zero so G minus F is
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a function whose derivative is zero and
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that makes it a constant function so f
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and G to define anti derivatives of the
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same function f can only differ by a
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constant the conclusion is that we can
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use any anti derivative to evaluate an
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integral and you've probably seen this
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suppose little F is the function you
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want to integrate from A to B and you
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found an antiderivative upper case G and
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the integral from A to B F of T DT is
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the antiderivative evaluated at the
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upper bound B minus the antiderivative
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evaluated at the lower bound a the
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reason for this is if you wanted to
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prove such a fact you would notice that
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G this antiderivative that you found is
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related to the antiderivative F from the
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previous page by simply a constant where
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F was this function f of X is the
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integral from a to X little F of T DT
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which we know what the fundamental
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theorem of calculus that it is an
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antiderivative so what is f of a in that
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case F of a is the integral from a to a
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so that's nothing that's zero and what's
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F of B well that's the full integral of
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actually trying to find so we know this
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is equal to definitely F of B minus F of
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a because F of B is the integral and F
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of a is 0 but now F of B is the same as
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G of B minus this constant C and F of a
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is G of a minus this constant C the
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constants cancel out until you get G of
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B minus G of a so that's how easy that
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is to proof
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so how will we generalize this to see
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instead of having a real valued function
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of a complex valued function and the
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function doesn't have a real variable
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but a complex variable so instead of
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integrating over an interval we're now
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in C so what will we integrate over the
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answer it will integrate over curves
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recall that a curve is a smooth or
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piecewise smooth function gamma to find
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out an interval but into C we could
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write this function as gamma of T equals
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x of T plus I Y of T where X is the real
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part and Y is the imaginary part if F is
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a complex valued function that is
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defined on gamma then we define the
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integral over gamma f of zdz to be the
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limit of these sums we take F of C J
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times the J plus 1 minus CJ and the Z
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J's are the values of gamma at points TJ
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when we divide the integral from A to B
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up into these little chunks again as we
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did before so this is a direct
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generalization of the left hand son
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let's look at the picture and understand
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what the sum really does you can see the
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curve gamma gamma is actually the image
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of the interval from A to B under the
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function gamma again we divide this
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interval up into these little chunks
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where a is equal to t0 B is equal to TN
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and you know this is t1 t2 and so forth
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then we look at the image of these t1 t2
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t3 and so forth so for example the curve
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starts at Z 0 which is gamma of T 0 or
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gamma of a it ends at Zn which is gamma
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of B or gamma of T N and in between we
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have a Z J which is Kamath TJ in the
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next point Z J plus 1 which is gamma of
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TJ plus 1 so we have all these points
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here in order to form this sum we
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evaluate F at ZJ so what's the value of
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F or
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here and multiply that value by Z J plus
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1 minus CJ then we add up all these
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products and take the limit as n goes to
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infinity if that limit exists we call it
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the integral in fact we call it the path
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integral of F over gamma again it's the
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limit as n goes to infinity of the sum F
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of Z J times Z J plus 1 minus Z J when
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the CJS are gamma of T J's and the TJ's
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are those little points that divide the
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integral from A to B up into n pieces
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one can show the following if gamma is a
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smooth curve and F is continuous on
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gamma then this integral f of zdz over
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gamma can be found by taking F evaluated
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at gamma of T multiply that by the
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derivative can my friend of T and take
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the integral from A to B over that
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product why is that here's the idea of
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the proof this is the son that we used
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to define the integral to get the actual
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interval you would put the limit as n
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goes to infinity in front of it but the
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sum of the sum by itself ZJ is simply
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gamma of t j zj + 1 is gamma of t j plus
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1 zj is gamma of t j that makes up this
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sum but now we just divide by ki J plus
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1 minus TJ and then simply multiply by
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TJ plus 1 minus TJ so this will not
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affect the sum at all because we
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multiply and divide by the same number
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but now if you look at this term right
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here as you make TJ plus 1 and TJ be
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closer and closer to each other this
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approaches the derivative gamma prime of
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TJ this is
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approximately the derivative game a
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prime of TJ so the sound is over F of
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gamma TJ times gamma prime of teaching
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times TJ plus 1 minus TJ we often call
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this Delta Chi Chi in the limit this
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goes to the integral from A to B of F of
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gamma of t gamma prime of T DT note we
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haven't really defined how to integrate
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a complex valued function so let's
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quickly look at some examples suppose G
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is a function defined from A to B and
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maps into C so we can write G of T as U
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of T plus IV F T then if we write
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something like integral from A to B G of
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T DT what that really means is the
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interval over u plus I times the
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integral over V in other words we
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integrate the real part we integrate the
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imaginary part and then put them back
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together and that is how the integral
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over G is defined let's look at an
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example suppose I wanted to evaluate the
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integral from 0 to PI e to the I T DT
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you then break up e to the I T as cosine
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T plus I sine T and integrate the cosine
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function and separately integrate the
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sine function an antiderivative for
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cosine T is sine T because the
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derivative of science cousin so we need
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to take sine T and evaluate it at the
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upper bound and subtract from its value
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at the lower bound that's what this
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expresses right here next we need an
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antiderivative for sine T an
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antiderivative for sine T is minus
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cosine T because the derivative of minus
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cosine T is minus minus sine T which is
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sine T so we need to take minus cosine T
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and evaluated at PI and subtract from it
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the value it has at zero while sine of
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PI is actually 0 sine of 0 is also 0 so
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I put that all into this one 0 right
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here cosine of PI is negative 1 cosine
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of 0 is 1 what we're subtracting that
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value so minus another one and here's a
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minus I that's in front of it all
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together this gives you 2
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so the value of this integral is to I
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you actually don't have to break the
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integral up into real and imaginary part
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one can actually alternatively simply
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find an antiderivative right in the
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original form so if you can find an
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antiderivative of e to the I T then you
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can just use the fundamental theorem
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with that antiderivative Y turns out an
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antiderivative of e to the I T is - I
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eat the I T why is that true well let's
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take the derivative - I eat ooh the I T
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I so constant that goes to the side
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and trying to differentiate e to the I T
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the derivative e to the I TS e to the I
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T times the derivative of the inside
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function i T which is I when I multiply
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through I squared is minus 1 together
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with this - it all cancels out so it's e
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to the I T so indeed - I either the I T
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is an antiderivative of either the I T
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so it can take this antiderivative and
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evaluate it at PI and subtract its value
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at 0 the value at pi is minus ie to the
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I pi the value at 0 is minus I e to the
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0 together with the fact that we're
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subtracting it we're getting this plus
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right here e to the I PI is negative 1
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together with this negative sign it's
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plus so this whole expression evaluates
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to I even the 0 is 1 so this also
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evaluates to I so that I end up with 2 I
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so these are two alternative methods to
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find the same integral let's look at
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another example integral from 0 to 1 T
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plus I DT again you could break this up
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into the integral from 0 to 1 of the
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function T DT plus I times the integral
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from 0 to 1 of 1 DT and then find
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antiderivative separately and plug in
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the events or you could just find an
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antiderivative for the whole function
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instead of finding each part separate
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and had derivative 40 is 1/2 T squared
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and a derivative for I is I T and so 1/2
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T squared plus I T is an antiderivative
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for t plus I we need to evaluate that at
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1 and subtract from that the value at 0
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when I plug in T equals 1 I get 1/2 plus
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I when I plug in T equals 0 this all
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vanishes in other words the value of the
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integral is 1/2 plus I now let's look at
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some examples of actual path integrals
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so postive gamma of T is T plus I T for
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T between 0 and 1 what does that look
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like well when T is equal to 0 gamma of
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0 is 0 when T is equal to 1 game up 1 is
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1 plus I 1 and ice right here how about
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in between well I'm graphing X of T and
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y of T together but X of T is equal to Y
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of T so we could write this as X of T
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plus I Y of T and I see that X of T and
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Y of T are always the same value Y is
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equal to X well that's this line right
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here so the path gamma is really the
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line from the origin to 1 plus I we can
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find the derivative as 1 plus I just
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take the derivative of T which is 1 the
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derivative of ITSI and suppose I wanted
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to integrate the function f of Z equals
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Z squared over that path so I need to
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find the integral of f of Z DZ by our
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definition the integral of f of zdz
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because f is a continuous function that
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has smooth I can take f of gamma of t
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times gamma prime of T and evaluate that
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integral from 0 to 1 what is f of gamma
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of T I need to plug in gamma of T into
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my function f but the function f takes
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its input and squares it in other words
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this is gamma of T
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squared game of T squared is T plus I T
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that's gamma of T square so that's where
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this term comes from right here and I
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need to multiply by the derivative of
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gamma but we found that as 1 plus I and
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now I have a integral of a complex
17:56
valued function and I know how to
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evaluate that one way to evaluate that
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is to simply multiply through so this
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first term here becomes T squared plus
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2i times T times t minus T squared the
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minus T squared comes from the I squared
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right here in the second term and I need
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to multiply that by 1 plus I T squared
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and T squared cancel each other out so
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the first term is 2i T squared times 1
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plus I I multiply through I have 2 I
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teeth squared so this term right there
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plus I times 2i T squared but I squared
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is negative 1 so I end up with minus 2t
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squared I could break this up into the
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real part and the imaginary part so I
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get minus 2 times the integral over T
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squared that's from the second term here
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plus 2i times the integral over T
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squared that's the first term and
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antiderivative for T squared is 1/3 T
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cubed so I get minus 2/3 T cubed
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evaluated from 0 to 1 plus 2i and again
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an antiderivative of T squared is 1/3 T
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cubed I'm going to need to evaluate from
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0 to 1 when I plug in 1 for T I find
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minus 2/3 when I plug in 0 this term
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goes away for the second term when I
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plug in T equals 1 I get 2 I over 3 and
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again / 2 equals 0 there is no term
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altogether I can combine these into 2/3
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times negative 1 plus I will be using
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this fact later on again so I'll remind
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you of that when we need
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here's another example suppose I wanted
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to find the integral of 1 over Z DZ over
19:45
the curve for the absolute value of Z
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equals 1 and this is a notation that we
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haven't used so far the absolute value
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of Z being equal to 1 as a set the nose
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the circle of all those Z's were the
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absolute value of Z is once it's the
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circle of radius 1 when we write
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something like that absolute value of z
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equal to 1 we automatically assume that
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the circles oriented this way and most
20:15
of the time the parametrizations gamma
20:18
of T equals either the I T is used for T
20:24
runs from 0 to 2 pi so that's what's
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meant when I write something like this
20:30
integral over absolute value of Z equals
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1 we'll see later that other
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parametrizations feel the same result
20:37
with the integral and that's why this
20:39
notation is actually a reasonable
20:41
notation so how do I find this integral
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if gamma of T is e to the I T then gamma
20:47
prime of T it's the derivative is e to
20:50
the I T times the derivative of the
20:52
inside function which is the exponent so
20:54
times I now the integral over the curve
20:58
gamma of 1 over Z DZ is therefore the
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integral from 0 to 2 pi of 1 over gamma
21:05
of t times gamma prime of T DT so
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whatever we see is e we plug in gamma of
21:11
T and DZ is replaced with gamma prime of
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T DT next let's plug in what gamma of T
21:17
actually is it's either the I T and
21:19
gamma prime of T is ie either the I T
21:21
and we see that the e to the I T term
21:25
cancels out all we're left with this I
21:28
let's pull that outside of the integral
21:30
and were left with the integral from 0
21:32
to 2 pi of DT you could write a 1 right
21:36
here if you wanted to and antiderivative
21:38
of the function 1 is T because the
21:40
derivative of T is 1 we need to evaluate
21:43
that from 0 to 2 pi so we get I times 2
21:47
pi minus I times 0 which is 2 pi I so
21:52
integral of the function 1 over Z over
21:54
the circle of radius 1 is 2 pi here's
21:58
another example same curve is equals 1
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this time we're integrating over the
22:03
function Z instead of 1 over Z we take Z
22:05
so again gamma of T is e to the I T
22:07
gamma prime of T is I in the I T so the
22:12
integral over this curve gamma of Z DZ
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is therefore the integral from 0 to 2pi
22:18
we replace Z gamma and DZ with gamma
22:22
prime of T DT game of t's e to the I T
22:25
gamma prime of keys I eat the I T and we
22:29
can pull this I outside of the integral
22:31
and get the integral from 0 to 2 pi and
22:34
we can combine either the I T and either
22:36
the I T into e to the - I see so now
22:39
what we need is an antiderivative of e
22:42
to the 2i t in the claim is an
22:46
antiderivative of i times e to the 2i t
22:49
is 1/2 times e to the 2i t we better
22:53
check that that is true how do we check
22:56
it well we take the derivative of 1/2 e
22:58
to the 2i t and check said it really is
23:02
what we think it is so what is the
23:04
derivative of 1/2 e to the 2i t 1/2 is a
23:08
constant that goes to this side and the
23:11
derivative of e to the 2i t is e to the
23:14
2i t times the derivative of the
23:18
exponent which is 2i in other words
23:23
these twos cancel out nicely and we're
23:27
left with i times e to the 2i t so
23:30
indeed an antiderivative of i temped e
23:32
to the 2i t is 1/2 e to the 2i t
23:35
and we now take that antiderivative and
23:38
evaluate it at 2 pi subtract from it the
23:40
value at 0 when I plug in 2 pi party I
23:44
get 1/2 times e to the 2 times 2 pi I so
23:48
4 pi I when I plug in 0 I get e to the 0
23:52
but e to the 4 pi i is equal to 1 e to
23:57
the 0 is also equal to 1 and 1 1 is long
24:01
is equal to 0 so the integral evaluates
24:03
to 0
24:10
let's look at the same path gamma and
24:13
integrate 1 over Z squared DZ so so far
24:16
we integrated 1 over Z and we got that
24:23
was 2 pi then we integrated Z to Z and
24:30
we got that was 0 let's see what 1 over
24:32
Z squared gives us well
24:36
again 1 over Z squared means I ride one
24:40
over gamma squared of T or gamma of T
24:42
squared times gamma prime of T DT gamma
24:46
prime of T is this in the numerator and
24:48
it's ie to the I T that's this part
24:50
right here and we divide by the square
24:52
of gamma of G what does gamma of T
24:55
squared that is e to the I T square it's
25:01
either the it1000 I so e to the 2i T's
25:09
here in the denominator and I can
25:12
therefore cancel one of the e to the i
25:14
T's I'm left with 1 e to the I T in the
25:18
denominator or either - I T in the
25:20
numerator what is an antiderivative of I
25:23
II to the - I T the claim is it's minus
25:28
e to the minus I T again let's check how
25:32
do I check I take the derivative of
25:35
minus e to the minus I T the derivative
25:39
of that is minus e to the minus I T
25:42
times the derivative of the exponent
25:44
which is minus I the two negative signs
25:46
cancel out I'm left with items into the
25:49
I T which is exactly what I wanted to
25:51
get so minus e to the minus I t's indeed
25:54
an antiderivative of the function I'm
25:56
integrating and I need to evaluate the
25:58
antiderivative of 2 pi and at 0 at 2 pi
26:01
I get minus e to the minus 2 pi it's 0 I
26:05
get e to the 0 I'm subtracting that but
26:08
with this minus sign I get this plus
26:10
sign the first term is minus 1 either
26:14
the minus 2 pi is 1 with a negative sign
26:16
that gives me a minus 1 e to the 0 is 1
26:19
1 minus 1 0 and again the integral
26:22
it's 2-0 more generally one can see the
26:25
following the integral over Z to the M
26:29
is equal to 2pi when M is equal to minus
26:32
1 when M is equal to minus 1 what I'm
26:39
looking at is the integral over Z equals
26:43
1 of 1 over Z DZ and we showed that that
26:46
was 2pi
26:48
for all other exponents including M
26:51
equals 1 in which case I have the
26:53
integral over Z M equals -2 in which
26:56
case it's the integral 1 over Z squared
26:58
all other cases that integral is equal
27:00
to 0 so all powers of Z the integral
27:04
evaluates to 0 except for the function 1
27:08
over C then we'll see more on why that
27:11
is the case later in the next lecture
27:14
we'll look at more examples and some
27:16
first facts about complex path integrals