36 - Evaluating an Improper Integral via the Residue Theorem
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welcome to this last lecture in the
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seventh week of our course analysis of a
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complex kind in this lecture I'll show
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you how to evaluate an improper integral
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via the rezident there this is a
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real-valued integral it doesn't look
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complex at all and it's very hard to
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tackle using methods from real analysis
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but it turns out by turning to complex
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analysis we can actually find this
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integral here is the integral we want to
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evaluate the integral from 0 to infinity
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of cosine x over 1 plus x squared DX you
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can try all kinds of substitutions and
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take sneaks you know from calculus but
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it's really really hard to crack this
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integral let me quickly explain what all
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this notation means first of all when we
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write the integral from 0 to infinity
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what that really means this would take
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will be integral from 0 to R or R some
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large number and we're letting our go to
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infinity and so we need to consider the
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integral from 0 to R of cosine x over 1
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plus x squared DX
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again this is a real valued integral
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integral on the real axis and then we
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need to let our go to infinity here's
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the next idea the integral from 0 to R
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of this function cosine x over 1 plus x
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squared is half of the integral when we
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go from negative R to R the reason is
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that the function cosine x over 1 plus x
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squared is an even function look at this
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picture here you can see the function
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cosine x over 1 plus x squared from 0 to
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10 the integral of this function is the
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area under the curve where areas above
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the x axis count is positive in areas
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below the x axis count is negative now
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look at what happens when we take the
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integral from negative R to R instead
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you see the function is symmetric and so
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exactly the same areas are added up to
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the left of the origin as they are to
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the right of the origin so that when I
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integrate from negative R to R instead
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of from 0 to R they get twice the value
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of the integral so instead of
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integrating from 0 to R and I just as
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well integrate from negative R to R
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to make things more symmetrical and then
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divide the result by two here's the next
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idea I can add an extra term in the
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numerator the extra term I just added is
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PI times sine X and the claim is that
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didn't change anything why is that let's
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look at the function sine X divided by
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one plus x squared that function is an
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odd function look at this picture for
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the function sine x over one plus x
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squared when I integrated from zero to R
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I get the opposite value of the aterial
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that I get when I integrated from
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negative R to zero because the function
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is odd but ever it used to be above the
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x-axis is now below the x-axis and
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vice-versa areas to the left cancel out
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with areas to the right so that the
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integral from negative part R of sine x
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over one plus x squared DX is equal to
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zero now if I multiply this in addition
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by I that doesn't change anything
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multiplied by I simply multiply zero by
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I which is still zero that's why I can
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simply add I times sine X and the
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integral here but cosine X plus I sine X
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that's equal to e to the i-x and all of
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a sudden I get my exponential function
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into the picture here so instead of
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finding the original integral from zero
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to R of cosine x over 1 plus x squared
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DX I might just as well find 1/2 times
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the integral from negative part r e to
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the i-x over one plus x squared here's
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the next idea this integral from
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negative R to R of e to the i-x over 1
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plus x squared DX can be written as a
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complex contour integral namely as the
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interval over the line segment from
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negative R to R of the function e to the
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eysie over 1 plus Z square DZ why is
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that this line segment from negative
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part are in the complex plane can be
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parameterize as gamma of x equals x
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where X goes from negative R to R then
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the derivative of this parameterization
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is simply 1 and so if I plug into this
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contour
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integral apart parametrizations this
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becomes 1/2 times the integral from
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negative R R of e to the I and now a
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plugin gamma of X grizzy but gamma of
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axis with X divided that's why of 1 plus
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and again for Z a plugin gamma of X by
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gamma fixes X then to multiply with a
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derivative game I prime of X which is 1
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and take the integral DX but that's
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exactly the integral that we actually
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wanted to find so the integral want to
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find can be written as a complex contour
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integral and that's what we want because
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ultimately we really want to use the
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residue theorem currently or simply
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integrating over a line segment we're
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going to have to integrate over a closed
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curve in order to be able to use the
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residue theorem so here's the next idea
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this is the closed curve over which
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we're going to integrate we're going to
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first integrate over the line segment
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from negative R to R and then go around
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the semicircle from R to negative 1 so
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that's our curve CR CR is the line
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segment together with that semicircle
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gamma R but that's clearly not the same
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thing as what we want before we really
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only wanted to integrate over the line
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segment so if we integrate over all of C
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R which is the line segment Plus this
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semicircle then we need to subtract the
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integral over the semicircle afterwards
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in order to make things right and that's
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what we're doing right here
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now the great thing is what is the
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integral over the curve CR of the
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function e to the I Z over 1 plus Z
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squared e to the I Z over 1 plus Z
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squared has two singularities that
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isolated and the two singularities
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happen when the denominator of this
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function is equal to zero
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when is the denominator equal to zero
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when Z is plus or minus I only the
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singularity it eyes enclosed by the
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curve C are the other singularity at
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negative I is down here and we don't
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really care about it now by the residue
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theorem the interval receipt R of the
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function e to the i z over 1 plus z
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squared is 2 pi i times the residue of
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my function at the singular point so
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here we have 2 pi i times the residue of
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my function at i and the 1/2 of course
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had to stay there and then we need to
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subtract this integral over the
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semicircle so we have two things to do
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we need to find that residue of the
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function e to the i z over 1 plus z
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squared' pet i and then we need to take
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care of the integral over the semicircle
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first we're going to find the residue of
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e to the I Z over 1 plus Z squared at
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the singular point I we notice that we
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have a simple pole at Z equals I because
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1 plus Z squared factors into Z plus I
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times Z minus 5
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therefore the residue can simply be
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found by multiplying F by Z minus I and
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then taking the limit as Z approaches I
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when I multiply F by Z minus I that Z
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minus I will cancel out with this others
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you might as I in the denominator and
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all that is left is then the Z plus I so
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I need to find the limit as C approaches
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I of e to the I Z over Z plus I I find
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that limit but simply plugging in PI for
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Z then i plug in I trizzie I have an I
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squared in the exponent of the
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exponential function so that's e to the
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negative 1 and the denominator becomes
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I so that also is equal to 1 over 2 e
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I therefore the integral over the closed
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curve C R of e to the I Z over 1 plus Z
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squared DZ by the residue theorem is 2
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pi I times the residue the 1/2 you need
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to carry around right here here's 2 pi I
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and here's the residue 1 over 2 ie I can
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cancel a few things out it's with this 2
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and that 2 for example this I and that I
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and I'm left with PI over 2 e next I
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need to estimate the integral over the
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semicircle gamma R of my function f
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remember gamma R was simply the semi
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circle of radius R centered at the
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origin plus were only really interested
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for what happens when R goes to infinity
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what we want to show is that the
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integral actually goes to 0 as paarvo so
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infinity so it doesn't really matter in
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order to show that some complicated
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expression goes to 0 as R goes to
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infinity it really surprises to show
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that the absolute value of that
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complicated expression is bounded above
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by some constant for which it is easier
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for us to show that it goes to 0 so as
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long as this constant goes to 0 this
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integral which is stuck between 0 and
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this constant that goes to 0 then is
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sandwiched in between 0 and a constant
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and therefore has to go to 0 as well so
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we're going to try to find an upper
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bound with the absolute value of the
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integral and show that our bound goes to
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0 then the integral is forced to go to 0
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as well we need to show that the
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absolute value of the integral is
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bounded above by a constant or the
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constant goes to 0 how do we estimate
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the absolute value of an integral
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remember what we learned a while ago the
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absolute value of a contour integral is
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bounded above by the length of the curve
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over which we're integrating times the
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maximum of the absolute value of F for
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Z's that are on the curve let's use that
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so therefore an our case the absolute
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value of the integral over gamma R of
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the function e to the IG over 1 plus Z
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squared is bounded above by the length
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of the curve gamma R and that's pretty
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easy to find the length of Kirk gamma R
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it's a semi circle of radius R so it's
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PI R times the maximum of e the Isay
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over 1 plus Z squared so how do i
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estimate that first i need to look at
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the absolute value of e to the eysie
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let's write Z that's X plus iy then e to
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the I Z becomes either that I X plus I
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squared Y so minus y now remember the
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absolute value of e to the something is
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e to the real part of the exponent but
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what is the real part of pi x minus y
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the real part of that is simply minus y
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so the absolute value of e to the I Z is
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e to the minus y the denominator is
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simply the absolute value of 1 plus Z
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squared and here we use the reverse
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triangle inequality the absolute value
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of 1 plus Z squared is greater than or
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equal to the absolute value of the Z
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squared minus the absolute value 1 the
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absolute value of Z squared and that
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semicircle is almost equal to R squared
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and so the denominator is found that
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below by R squared minus 1 since it's in
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the denominator we found that 104 with
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the absolute value of 1 plus Z squared
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then is bounded above by 1 over R
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squared minus 1 and that's what you see
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right here so the absolute value of my
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integrand is bounded above by e to the
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minus y divided by r squared minus 1 now
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remember that we're only looking at the
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upper half of the circle of radius R
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centered at the origin so all my Z's are
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from this upper half of the circle which
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means the Y value of all of these heats
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it's greater than or equal to zero
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because the Y value is greater than or
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equal to zero all these exponents minus
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y are less than or equal to 0 which
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means e to such an exponent is never
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greater than e to the 0 which is 1
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therefore on the curve gamma r the
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integrand is bounded above by 1 divided
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by r squared minus 1
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so we have seen that the ox ability of
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the interval is bounded above by the
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length of the curve which is PI R times
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the maximum of the integral which was 1
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over R squared minus 1 and what happens
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to this product as R goes to infinity
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well the denominator goes to infinity
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much faster than the numerator R squared
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wins over R and therefore this whole
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fraction PI R over R squared minus 1
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goes to 0 as R goes to infinity and
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that's what we wanted to show therefore
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the integral of e to the eysie over 1
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plus Z squared over the curve an arm
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goes to 0 as R goes to infinity now it's
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time to put it all together remember we
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wanted to find the integral from the 0
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to infinity of cosine x over 1 plus x
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squared here's what we have we first
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realized that instead of taking the
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integral from 0 to infinity we need to
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take the integral from 0 to R and then
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that our go to infinity
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next we realized that instead of taking
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the integral from zero to R of cosine X
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over 1 plus x squared DX we can turn
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this into something complex namely
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integrate over the function e to the
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eysie over one plus Z squared we
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integrated over the whole curve C R
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which consisted of the line segment from
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negative R to R and the semicircle with
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gamma R once we showed that the integral
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over the closed curve C R of e to the I
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Z over 1 plus Z squared DZ is 1/2 times
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2 pi I times the residue of the function
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at I and that turned out to be PI over 2
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e so this integral right here that
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evaluated to PI over 2 e next we showed
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that the integral over gamma R of e to
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the I Z over 1 plus Z squared Z goes to
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0 as R goes to infinity so this integral
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right here that goes to 0 as R goes to
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infinity therefore as R
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to infinity this integral here converges
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to PI over 2 e the value of the first
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interval over CR because the second
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interval simply goes to 0 so we have
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just shown that the integral of cosine x
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over 1 plus x squared from 0 to infinity
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is equal to PI over 2 e this is a
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fabulous result as the outcome of the
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equation includes some of the most
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important mathematical constants namely
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pi - only even prime number and feet I
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hope you enjoyed this course it's been a
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pleasure teaching you all