7.8 Improper Integrals

Transcription

1 CHAPTER 7. TECHNIQUES OF INTEGRATION Improper Integrals Eample. Find Solution. Z Z e d. e d = lim F () F (), where F () = e! = lim e! = (e ) = Eample. Find the volume of the shape known as Gabriel s Horn. This is defined by rotating f() = around the -ais, from =to =. Solution. We start by picturing the original region, and the volume generated by rotating it. graphics/_over for_rotating-eps-converted-to.pdf

2 CHAPTER 7. TECHNIQUES OF INTEGRATION 66 graphics/gabriels_horn_rotated-eps-converted-to.pdf Ihopethepicturesstimulateyoutothink volumebydisks. Weslice the volume into disks, moving along the -ais. Thus, the volume is given by Z V = d. Using the techniques for improper integrals that we have learned, this integral is fairly straight forward, Z Z d = d = lim F () F ()! = lim =! + = ( + ) =, where F () = Here s one way to think about this eample. The shape we get by rotatation is a physical solid, that is infinitely long, but has a finite volume. Here s a nice way to think about this: if you turned this horn vertically, you could fill it with liquid, using a finite amount of liquid. Eample Z 3. Figure out which of the following converge: (a) d Z (b) d.5

3 CHAPTER 7. TECHNIQUES OF INTEGRATION 67 (c) Z Solution. d.7 (a) Z d = lim ln() ln()! = = So in this case, the integral is divergent. (b) Z d =.5 Z.5 d = lim F () F () where F () =.5!.5 = lim.5!.5 =.5 = ( ).5 =.5 So in this case, the integral is convergent. (c) Z d =.7 Z.7 d So in this case the integral is divergent. = lim F () F () where F () =.3!.3 = lim.3!.3 = ( ).3 = Z Eample 4. For which values of p does d converge? p Solution. We saw in the previous eample that the integral diverges for p =. For p>welookatcase(b)andseethesamecalculationswould work in general: the anti-derivative will come out to be,with p

4 CHAPTER 7. TECHNIQUES OF INTEGRATION 68 p >, and this function has a horizontal asymptote. So the limit will eist, and the integral converges. For p<welookatcase(c)andseethatthesamecalculations will work in general: the anti-derivative will come out to be p with p>, and this function does not have an horizontal asymptote. So the limit will not eist, and the integral diverges. In conclusion: Z ( d = convergent if p> p divergent if p apple Z Eample 5. Find d. Solution. We do this problem twice: once the WRONG way, and once the correct way. WRONG WAY Z d = = = ( + /) = 3/ Can you see how we can tell that this answer must be wrong? Here is the graph of this function: graphics/area_under over squared-eps-converted-to.pdf This graph is always positive, so the area under the curve has to be positive, and so there s no way the integral should be negative. If this

5 CHAPTER 7. TECHNIQUES OF INTEGRATION 69 integral is defined, it has to be positive, but we ll see in a moment that it s not defined. Here s the right way to do this. CORRECT WAY Z d = Z d + Z d = lim F () F ( ) + F () lim F (), where F () =! a! + = lim lim!! + = ( + ) ( )) = = + Since the integral comes out to be infinite, we say that the integral diverges. Challenge. Can you come up with an eample, like the previous one but where the integral, when done correctly, converges? Eample Z 6. Figure out which of the following converge. (a) d Z (b) d.3 Z (c) d.6 Solution. (a) Z d =lim a! a =ln() lim ln(a) a! = ( ) = So in this case, the integral is divergent. (b) Z d =.3 Z.3 d.3 =lim a!.3 a = lim a.3.3 a!

6 CHAPTER 7. TECHNIQUES OF INTEGRATION 7 = lim.3 =.3 = ( ).3 = ± So in this case, the integral is divergent. (c) Z d =.6 Z.6 d.4 =lim a!.4 =.4 a a! a.3 lim a.4 a! = ( ).4 =.4 So in this case the integral is convergent. Eample 7. Figure out which values of p make the following integral converge. Z p d Solution. We saw in the previous eample that d is divergent. For p>welookatcase(b)andseethesamecalculationswould work in general: the anti-derivative will come out to be,with p p >, and this function has a vertical asymptote. So the limit will not eist, and the integral diverges. For p<welookatcase(c)andseethatthesamecalculations will work in general: the anti-derivative will come out to be p with p>, and this function can be evaluated at =. Sothelimit will eist and the integral converges. In conclusion: Z Z p d = ( convergent if p> divergent if p

7 CHAPTER 7. TECHNIQUES OF INTEGRATION 7 Eample 8. Find the following, or show that it diverges Z 3 ( ) /3 d Solution. Note: you have to spot what the problem is with this eample, it s not made eplicit. You have to remember that the negative power means that you have,andthenyou have to see that ( ) /3 this could give division by when you plug in =. Thismeansthat there is a vertical asymptote at =,andsowehavetosplitthe integral up there: Z 3 Z Z 3 d = d + d ( ) /3 ( ) /3 ( ) /3 Now we do a very simple u-substitution, u = = = 3 ( )/3 3 andget ( )/3 + /3 /3 lim (! )/3 ( ) /3 +(3 ) /3 lim( ) /3! = 3 ( + ) = Note, this integral comes out to be zero since ( ) /3 is odd, and the area on the left, below the -ais cancels the area on the right, above the -ais.