Solutions to Problem Set 5

Transcription

1 Solutions to Problem Set Problem 4.6. f () ( )( 4) For this simle rational function, we see immediately that the function has simle oles at the two indicated oints, and so we use equation (6.) to nd the residues: we multily f () with ( ), where is the osition of the ole, then take the limit as goes to. R f (); lim! f () lim! ( ( ) )( 4) ( 4) : Note the minus sign - it comes from switching the terms in the rst arenthesis in the denominator. Now for the residue at the second ole: R f (); lim f () lim! 4! 4 4 ( )( ) Problem 4.6.' ( 4 ) : n this roblem, we will use the residue theorem to solve this integral: f ()d; where f : jj g, that is, a circle with radius and center in the origin. f () is the function from roblem The residue theorem states that if f () is meromorhic, the integral around a simle, closed curve is just i times the sum of all the residues of at oles inside. We see clearly that f () has two oles, at and at 4, and we found the residues at these oles above. We see that the residues are the same excet for the sign, which means that the sum of the residues inside, and then also our contour integral, is ero by the residue theorem: Problem f ()d i : f () sin : To nd the residue at the ole, we use the same method as above: sin Rf (); lim! () lim f! ( ) sin :

2 Problem f () + ( + 9)( + ) : We are to nd the residue of f () at the oint i. We can see that there is a ole at i since the rst factor in the denominator is ero there. Then it is also ero at i, and this factor is easily factoried. Then it is straight forward to nd the residue: + R(f (); i) lim ( i)!i ( i)( + i)( + ) i + 6i 8 i 48 : Problem 4.6.8' Here, we will integrate the function from roblem around, which is a circle of radius and center in the origin (as in roblem 4.6.'). We see that f () is a meromorhic function with two oles inside : at i. We need to nd the residues at these oints. (The oles we investigated above are not inside, so we do not need those residues.) We factorie the second factor in the denominator, and nd the residue at i: + R(f (); i) lim( i)!i ( + 9)( + i)( i) i 6 : By the same method, we nd the residue at i: R(f (); i) The integral then becomes Problem f ()d i i 6 Here, we want to solve the integral lim ( + i)f () i +! i 6 : ( + cos ) d; + i + i 6 4 : by using the residue theorem. To do so, we need to construct a contour integral whose value is equal to the value of the given integral. The rst thing we do is to notice that the integrand is an even function, meaning that f () f ( ). A consequence of this is that the integral from to a is half the integral from a to a: ( + cos ) d Now we do a substitution e i. Then we have cos ei + e i ( + cos ) d: +

3 We also need d d(i). nserting this for the integrand, we get ( + cos ) d d + ( + ) i f ()d: We then notice that when runs over the interval [ ; ], runs along the unit circle f : jj g. This means that the integral on the right hand side above is equal to the corresonding function f () integrated around the unit circle: ( + cos ) f ()d To solve the contour integral, we need to locate the oles that are inside our integration curve, and nd the residues at those oles. We can see at once that there is a ole at, but if we look at the integrand f (), we susect that there will be a lot of work to calculate the residue directly because of the term inside the arenthesis. So we look at the inside of the arenthesis rst. There is a second order ole where this is ero: + + ) ) : So here we nd two oles, but only one of them, at +, is inside our contour. We now factorie this: + + ( + + )( + ) : nserted into our integrand, we get this: f () (++ )(+ ) 4i ( + + )( + i ) : Now it is easy to nd the two residues we need. First at : R(f (); ) lim f () lim!! 4i ( + + )( + ) : Next, we look at the ole at +. This is a second order ole, so it involves more work to comute the residue. We will use the method near the bottom of age 68 in Boas, with m : R(f (); + d d ( + ) 4i ( + + )( + ) 4i( + + ) + 8i( + + ) ( + + ) 4! ) ( + ) ( + ) 8i( + ) 4i ( ) i :

4 Short summary: we multily f () with (+ ), then dierentiate, divide by (m )! (which is just here), and then evaluate the exression at +. Now we have the residues, and can calculate the contour integral: i f ()d i 4 : Now, remember that we had a factor of / when we set u our contour integral above. With this, the integral we wanted to solve becomes Problem ( + cos ) d n this roblem, we will solve this integral: f ()d sin cos d: : We will use the same method for this integral as for the revious one (this method is the same for integrals of rational exressions with trigonometrical functions). First, we notice yet again that the integrand is an even function, so we can multily by / and integrate over [ ; ]. Doing the same substitution e i, we see that when varies over [ ; ], varies along the unit circle, and so we get a contour integral that we can solve with the residue theorem. We rst insert our substitution into the arts of the integrand: sin i (ei e i ) cos i + So we have this relation between our integrals: sin cos d sin : d d i : cos d 4 + : + 6( + ) 4i d: We need to nd the oles of the integrand that are inside, and then we must nd the residues at those oles. We see immediately that there is a ole at, which is inside. There are two more oles, which we nd by setting the denominator to ero. Multilying in, and setting the total factor i8 outside the integral, we get this: i d: The denominator is now a second order olynomial with roots f ; g. The root is inside. We factor the denominator, giving us this integrand: f () + 6( )( ) : 4

5 alculating the residue at is straight forward: R f (); lim! + 6( )( ) 6 : alculating the residue at gives a little more work, as it is a second order ole: R(f (); ) d + d ( ) (4 4)( 6 6) ( 4 + )( ) ( ) ( 6 6) 6 : Notice that have gone back to the olynomial form in the denominator, since that is easier to dierentiate. Now that we have the residues, it is straight forward to comute the integral. Remember that we had a re-factor of i8 as well, so we get sin cos d i + 8 ( 6( + )) d i i : Problem 4.7. n this roblem, we will solve this integral: x + 4x + dx: To solve this integral by use of the residue theorem, we must nd a contour integral that has the same value as this integral. This time, we rst construct the corresonding comlex function, and factor the denominator by use of the quadratic formula: f () Now construct the following contour: ( + + i)( + i) : r f : jj r; m g [ f : jj r; m > g [ : This is just a semi-circle in the uer half lane with radius r (do not worry if you do not understand the set notation, it is not imortant). Let us integrate f () around this contour, and slit the integral into two arts: one along the real axis (the rst art of the set r, which we call ), and one along the circumference of the semi circle (the second art of r, which we call ): ( + + i)( + i) d ( + + i)( + i) d + ( + + i)( + i) d: When integrating along, we can substitute x + iy, and since y when, we get just an integral over x, from r to r, with d dx. For the

6 integration along, we substitute re i. When, r is a constant, and varies between and. Also, d rie i d here. So we have f ()d r r (x + + i)(x + i) dx + rie i (re i + + i)(re i + i) d: Now we let r go to innity. Then the rst integral becomes the integral we started with, and the second integral becomes ero since the integrand goes to ero (since the numerator in the second integral is roortional to r, and the denominator is roortional to r ). So: lim r! f ()d x + 4x + dx: The integral to the left is a contour integral, which we can solve with the residue theorem. We have two oles, at i. The ole at + i is inside our contour when r goes to innity. The residue at + i is R(f (); + i) lim ( + i)! +i ( + + i)( + i) i : So our contour integral is f ()d i i ; which is indeendent of r, as long as the ole is inside, which it is when r goes to innity. This is then our nal answer as well: Problem 4.7. dx : x + 4x + n this roblem, we will solve this integral: x x dx: We will use exactly the same methods as in the revious roblem, so will not go through all the details here. But one thing is dierent: this integral is only from to. Luckily, the integrand is an even function (f (x) f ( x)), so we can rewrite it as x x dx x x dx: We need to nd the roots of the olynomial in the denominator, which are simly the comlex fourth roots of 6. The fourth roots of 6 are ( i). These will be the oles where we will nd residues. Factoring, and writing in terms of, we get f () ( + + i )( + i )( + i )( i ) : 6

7 We use the same contour as in roblem Again we get x lim f ()d r! x dx by the same argument: the integral along the arc of the semi-circle goes to ero as the radius goes to innity. Now there are two oles inside when r is large, at (i ). The residues at these oles are straight forward to nd, since both are simle oles (but there are some factors of to kee track of): R(f (); ( + i)) lim! ( (i+) R(f (); (i )) (i + ))f () ; 4 ( + i) 4 (i ) : The sum of the residues is i(4 ), and by the residue theorem, the value of the integral is i f ()d i 4 : But we had a factor / in front of the contour integral because of our integration limits. So our nal answer is x x dx : 4 7