35 - Evaluating Integrals via the Residue Theorem

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welcome to the fifth lecture in the

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seventh week of our course analysis of a

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complex kind in today's lecture also

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your number of examples that demonstrate

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how the residue theorem can be used to

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evaluate complex integrals let me start

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by reminding you of the resident there

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suppose DS is simply connected domain

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and F is analytic in D but the

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intersection of isolated singularities

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and the nodal feels isolated

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similarities across below suppose

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furthermore that sees a simple closed

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curve and D that doesn't pass through

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any of the singularities of F and let C

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1 through Zn B those singularities of F

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that are surrounded by C then the

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integral of F over the curve C can be

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found as 2 pi I times the sum of the

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residues have those funds that are

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surrounded by sir let's look at an

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example suppose we wanted to find the

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integral of the function each of the 3

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over C over the circle of radius 1

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centered at the origin

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where does the function f have

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singularities even the 3 over zis

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undefined when Z is equal to 0 because

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we would be dividing by 0 at that point

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but it's actually analytic everywhere

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else so we have an isolated singularity

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at 0 and it turns out that isolated

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singularities of essential singularity

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the integral can therefore be found by

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the residue theorem as 2 pi I times the

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residue of F and 0 how do we find that

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residue first we remember that the

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exponential function has a Taylor series

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representation centered at the origin s1

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was W thus following the squared over 2

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factorial and so forth if we write that

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in summation notation that is the sum

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came from 0 to infinity of 1 over K

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factorial and W to the K now I'm going

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to find Laurent series expansion for e

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to the 3 do regime so that we can find

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the residue of e to the 3 over Z at the

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origin because 0 is an essential

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singularity for the function f we can

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use any of the tricks we learned last

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class rather we actually have to find

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the Laurent series expansion but that's

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no easy because we simply plug in 3 over

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C

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for W we find that e to the three or

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Izzy's the sum 1 over K factorial times

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3 over Z to the K in order to find the

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residue of this function at 0 we need to

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find the 1 over Z term what does 1 over

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Z get multiplied with well we get a 1

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over is d exactly when this K is equal

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to 1 when that K is equal to 1 we have a

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1 over 1 factorial and the 3 from the

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numerator that must be might a negative

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1 term so therefore the residue which is

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a negative 1 is equal to 3 and that

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means that the integral that we're

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interested in the integral of the

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function e to the 3 over Z integrated

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over the circle of radius 1 is equal to

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3 times 2 pi and that equals 6 pi I

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here's another example suppose we wanted

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to integrate the function tangency what

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with a circle of radius two where are

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the isolated singularities of the

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tangent function tangency assigns the

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over cosine Z which is analytic

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everywhere except for a cosine Z is

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equal to zero but cosine Z is equal to

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zero at PI over two negative PI over 2 3

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PI over 2 negative 3 PI over 2 and so

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forth those singularities that are

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surrounded by that circle if we're used

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to all right PI over 2 and negative PI

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over 2 in order to evaluate the integral

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and therefore need to find the residue

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of the tangent function at PI over 2 and

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the negative PI over 2 we need to add

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those two residues up and multiply the

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result by 2 pi how do we find the

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residue of F at PI over 2 we notice that

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f is of the form an analytic function

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divided by another analytic function

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sine same cosine T and those two

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analytic functions satisfied that the

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denominator function is equal to 0 and

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has a simple zero at PI over 2 but

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theorem that we end up with in last

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class we remember that we can find the

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residue of F at PI over 2 by simply

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taking the numerator function evaluating

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it at PI over 2 and dividing it by the

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derivative of the denominator at PI over

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2 the numerator is sine Z so we need to

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find sine of PI over 2 the denominator

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is cosine Z its derivative is negative

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sine Z if PI over 2 that gives us

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negative sine of PI over 2 sine of PI

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over 2 is 1 negative sine of PI over 2

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is negative 1 the quotient evaluates to

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negative 1 and therefore the residue of

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F at PI over 2 equals negative 1

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similarly the residue of F at negative

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PI over 2 is the numerator at negative

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PI over 2 divided by the derivative of

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the denominator at negative PI over 2

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because the denominator function cosine

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Z again has a simple 0 at negative PI

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over 2 therefore we can evaluate G of

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negative PI over 2 just sine of negative

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PI over 2 divide that by

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the derivative of the denominator

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negative sine of negative PI over two

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then we find ourselves with negative one

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which is signed up negative PI over 2

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and negative negative one which

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evaluates all together to a quotient of

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negative one

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both residues are actually equal to

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negative one so that the integral of

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tangent cbz over the circle of radius

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two evaluates to two pi I times negative

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1 plus negative 1 which equals negative

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4 pi it's a third example this is a

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function we've looked at before f of Z

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equals 1 over C minus 1 squared C minus

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30 this function has isolated

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singularities x equals 1 and Z equals 3

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a-drinkin curves here curve C one that

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surrounds only the super there gave one

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and occur c2 that surrounds both of

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these isolated singularities we'll find

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both integrals first we'll start with

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the integral over the curve C 1 which by

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the residue theorem is to point items

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the residue of F at 1 to find the

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residue of FN want to notice that and

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one of the function has a double pull to

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find the residue the double pull I

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simply multiply the function by Z minus

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the double we'll squared and then take

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the derivative and in the limit as Z

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approaches default that's the formula we

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developed last class

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conveniently we notice that Z minus 1

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squared in the numerator is U n is 1

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squared in the denominator cancelled out

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so I'm relaxed with having to find the

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derivative of 1 over C minus 3 and then

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the limit as C approaches 1 but the

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derivative of 1 over Z minus 3 is

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negative 1 over Z minus 3 squared when

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we take the limit as Z approaches 1 we

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can simply replace this is u 1 1 minus 3

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equals negative 2 we squared that we get

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for the residue is equal to negative 1/4

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therefore the interval of my function

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over the curve C 1 is 2 pi I times

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negative 1/4 and that evaluates the

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negative PI over 2 next let's look at

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the other curve the curve C 2 this time

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we need to find the residue

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one and the residue of F F 3 and add

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those two things up we already found the

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residue of F at 1 and that is equal to

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negative 1/4 now we need to find the

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residue of F n 3 we notice that the

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function has a simple pole that's free

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therefore we multiply by Z minus 3 and

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then take the limit as Z approaches 3 no

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need to take a derivative or a simple we

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notice the Z minus 3 term cancels out

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conveniently we're left with having to

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find the limit as Z approaches 3 of 1

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over Z minus 1 squared to find that

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limit we simply replace the zenith 3 3

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minus 1 is 2 2 squared is 4 the residue

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turns out to be 1/4 now we notice when

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we add those two residues negative 1/4

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plus 1/4 adds up to 0 so the integral

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comes out to be 0 when we integrate it

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over the curve CT

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so far all the integrals we evaluated

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were in drills in the complex plane the

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residue theorem can actually also be

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used to evaluate real intervals for

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example of the following forms an

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integral for a rational function cosine

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T and sine T do you see a rational

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function of x and y both numerator and

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denominator are polynomials and X and

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water when I replace X and y by cosine T

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and sine T and find that R cosine T sine

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T equals cosine Square t minus science

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worked over 1 minus cosine T plus 1/4

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and we can actually find this integral

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he was in the resident theory it turns

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out is integral equals PI over 3

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other examples are integrals of rational

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functions integrals of a rational

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function times a cosine of alpha X

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integral of a rational function times

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sine of alpha X in the next and last

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lecture I will demonstrate for you how

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to find an integral of this last time I

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will evaluate an integral from minus

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infinity to infinity

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for rational function times cosine of X