Calculus 1120, Spring 2012 Dan Barbasch October 11, 2012 Dan - PowerPoint PPT Presentation

Calculus 1120, Spring 2012 Dan Barbasch October 11, 2012 Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 1 / 6

Improper Integrals Integration via Riemann sums assumes functions are continuous over a bounded interval (or at least piecewise continuous over finitely many bounded intervals). Deviating from this leads to improper integrals. � ∞ e − x dx . infinite intervals 0 � 1 1 unbounded functions, vertical asymptotes √ x dx . 3 0 The functions in the two examples are > 0 , so you can think of the integrals as representing areas below the curves and above the x − axis. Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 2 / 6

Improper Integrals � A � ∞ Type I: y = f ( x ) continuous on [ a , ∞ ) f ( x ) dx = lim f ( x ) dx . A →∞ a a � a Similarly for f ( x ) dx . −∞ � R � ∞ e − x dx := lim e − x dx . Example: R →∞ 0 0 Type II: y = f ( x ) continuous on [ a , b ) then � b � A f ( x ) dx = lim f ( x ) dx . A → b − a a Similarly if y = f ( x ) is continuous on ( a , b ] , � b � b f ( x ) dx = lim f ( x ) dx . A → a + a A � 1 � 1 1 1 √ x dx := lim √ x dx . Example: 3 3 δ → 0 + 0 δ Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 2 / 6

Improper Integrals Important: If more than one type occurs in the integral, MUST divide into a sum and treat each term separately. If the interval is infinite in both directions, and there are asymptotes, we need to separate into a sum, and examine each of them. � 0 � b � ∞ � − a � ∞ 1 1 1 1 1 √ x dx = √ x dx + √ x dx + √ x dx + √ x dx . 3 3 3 3 3 − a 0 b −∞ −∞ It does not matter what a and b are, except − a < 0 and b > 0. This integral is improper because there are two type I and two type II issues! Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 2 / 6

� 1 � 1 � A � � � � 1 1 1 x dx = lim x dx + lim x dx B → 0 + A → 0 − − 2 − 2 B � 1 �� − R 1 1 � Contrast this with lim x dx + x dx = − ln 2 . R → 0 + − 2 R � 1 � − R 1 1 − 2 + ln | x | | 1 x dx = ln | x | | − R x dx + R = − 2 R =(ln R − ln 2) + (ln 1 − ln R ) = − ln 2 . � 1 1 This DOES NOT SAY that x dx converges and equals − ln 2 . − 2 The order in which you take the integrals and the limit makes a difference! Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 3 / 6

R → 0 + [ln R − ln 1] + lim lim R → 0 + [ln 1 − ln R ] = −∞ + ∞ . � 1 1 Since each of the limits are infinite separately, x dx diverges. − 1 We need to know lim A → 0 + ln A = −∞ and other limits of this type. Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 3 / 6

Comparison Tests Direct Comparison: If 0 ≤ f ( x ) ≤ g ( x ) for x ≥ a , then � ∞ � ∞ g ( x ) dx converges ⇒ f ( x ) dx converges. 1 a a � ∞ � ∞ f ( x ) dx diverges ⇒ g ( x ) dx diverges. 2 a a � ∞ � ∞ LHS = f ( x ) dx ≤ RHS = g ( x ) dx . a a If the RHS is finite, so is the smaller one. If the LHS is infinite, so is the bigger one. Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 4 / 6

Comparison Tests � ∞ 1 1 1 √ Example: 1 + x 4 dx . For example, 1+ x 4 ≤ 1+ x 2 . √ √ 0 � 1 + x 2 dx = 1 1 � √ 2 ln | x + 1 + x 2 | + C from the tables. � A √ 1 + x 2 = 1 dx � � � √ 1 + A 2 | − ln | 0 + ln | A + 1 | − → ∞ as A → ∞ . 2 0 � 1 � ∞ � ∞ dx dx dx √ √ √ 1 + x 4 = 1 + x 4 + 1 + x 4 ≤ 0 0 1 � 1 � ∞ dx dx ≤ √ 1 + x 4 + √ 1 + x 2 = ∞ 0 1 DOES NOT GIVE ANY CONCLUSION. Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 4 / 6

Comparison Tests 1 We need a better sense of how 1+ x 4 behaves as x → ∞ . √ 1 1 1 + 1 / x 4 = 1 1 1 √ 1 + x 4 = = x 2 · � x 2 � � x 4 (1 + 1 / x 4 ) 1 + 1 / x 4 1 + x 4 ≈ 1 1 1 √ But lim x →∞ 1+1 / x 4 = 1 , so √ x 2 for x → ∞ . In fact 1 √ 1+1 / x 4 ≤ 1, so � 1 � ∞ � ∞ dx dx dx √ 1 + x 4 = √ 1 + x 4 + √ 1 + x 4 ≤ 0 0 1 � 1 � ∞ dx dx ≤ √ 1 + x 4 + x 2 < ∞ . 0 1 The integral converges. We had to split the intergral into a sum � 1 � ∞ because 1 / x 2 has a vertical asymptote at x = 0 . 0 + 1 � 1 � A A � dx x 2 = − 1 � � = − A − 1 − → 1 as A → ∞ . � x � 1 1 Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 4 / 6

Important Special Cases: 1 � 1 1 x p dx converges for 0 < p < 1 and diverges for p ≥ 1 . 0 2 � ∞ 1 x p dx diverges for 0 < p ≤ 1 and converges for p > 1 . 1 By Direct Computation, see the text. The calculations we just did, are encoded in what is called the Limit Comparison Test. Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

Limit Comparison Test: f ( x ) 0 ≤ f ( x ) , g ( x ) for x ≥ a , and lim g ( x ) = L , with 0 ≤ L ≤ ∞ . x →∞ 1 0 < L < ∞ , ( f ( x ) ≈ g ( x )) � ∞ � ∞ g ( x ) dx converges ⇔ f ( x ) dx converges. a a 2 L = 0 , (eventually f ( x ) << g ( x ) ) � ∞ � ∞ g ( x ) dx converges ⇒ f ( x ) dx converges a a � ∞ � ∞ f ( x ) dx diverges ⇒ g ( x ) dx diverges a a 3 If L = ∞ , (eventually g ( x ) << f ( x ) ) � ∞ � ∞ f ( x ) dx converges ⇒ g ( x ) dx converges a a � ∞ � ∞ g ( x ) dx diverges ⇒ g ( x ) dx diverges. a a Case (1) is the more common one. Remark: These tests also work for integrals over finite intervals for functions with vertical asymptotes. Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

lim f ( x ) g ( x ) = L , 0 < L < ∞ ⇔ L 2 g ( x ) ≤ f ( x ) ≤ L 1 f ( x ). You can apply the comparison test both ways, f ( x ) ≤ L 2 g ( x ) AND g ( x ) ≤ f ( x ) L 1 . � b � b f ( x ) dx converges ⇔ g ( x ) dx converges . a a lim f ( x ) g ( x ) = 0 means f ( x ) << g ( x ) eventually. � b � b g ( x ) dx diverges ⇒ g ( x ) dx diverges a a lim f ( x ) g ( x ) = ∞ means g ( x ) << f ( x ) eventually. � b � b f ( x ) dx converges ⇒ g ( x ) dx converges a a Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

1+ x 4 and g ( x ) = 1 1 f ( x ) = x 2 . √ f ( x ) 1 g ( x ) = 1 + 1 / x 4 − → 1 as x → ∞ . � f ( x ) = e − x 2 / 2 and g ( x ) = e − x / 2 . f ( x ) g ( x ) = e ( − x 2 + x ) / 2 − → 0 as x → ∞ because x − x 2 → −∞ as x → ∞ . (Other) Important Special Cases: 1 � 1 1 x p dx converges for 0 < p < 1 and diverges for p ≥ 1 . 0 2 � ∞ 1 x p dx diverges for 0 < p ≤ 1 and converges for p > 1 . 1 3 � ∞ e − x dx converges a 4 � a 0 ln x dx diverges 5 � ∞ 1 ln x dx diverges. 2 We write f ( x ) ≈ g ( x ) in case (1) of the limit comparison test. Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

More Examples: � ∞ dt (a) t 2 + sin t 0 � ∞ e x dx (b) e 2 x − 1 , 0 � ∞ dx (c) (ln x ) p . 2 Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

� 1 � ∞ � ∞ dt dt dt t 2 + sin t = t 2 + sin t + t 2 + sin t . 0 0 1 t 2 + sin t = 1 1 1 + sin t / t 2 ≈ 1 1 t 2 · t 2 for t → ∞ t 2 + sin t = 1 1 t + sin t / t ≈ 1 1 t for t → 0 + t · lim t → 0 sin t Necessary Facts: = 1 . | sin t | ≤ 1 . t Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

Remark: The solution is incomplete. We need to check that t 2 + sin t � = 0 for any t > 0 . This is true because on the one hand for t > 1 , t 2 + sin t > 1 + sin t ≥ 0 , so no zeroes. On the other hand, for 0 < t ≤ 1 , t 2 + sin t is increasing by using the 1 st derivative test; it has derivative 2 t + cos t which is positive because both 2 t > 0 and cos t > 0 for 0 < t ≤ 1 . Since t 2 + sin t equals 0 for t = 0 , being increasing, it is > 0 for 0 < t ≤ 1 . This is NOT the case for the analogous problem from last semester, � ∞ dt t 2 − sin t . 0 t 2 − sin t = 0 for t = 0 , and its derivative is 2 t − cos t equal to − 1 at t = 0 . Thus the function is decreasing so < 0 for t > 0 . But at t = 1 , the value is 1 2 − sin 1 > 0 . So by the intermediate value theorem t 2 − sin t must be = 0 for at least one value 0 < c < 1 . BONUS CREDIT is awarded for a complete careful analysis of all the � ∞ dt integrals involved in determining whether t 2 − sin t converges or not. 0 You should attempt this only after having done all the homework problems which are easier. Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

� 1 � ∞ e x dx e x dx � ∞ e x dx e 2 x − 1 = e 2 x − 1 + e 2 x − 1 . 0 0 1 e 2 x − 1 = e x e x 1 − e − 2 x = 1 1 1 − e − 2 x ≈ 1 1 e 2 x · e x · for x → ∞ e x Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6

For x → 0 + , e x → 1 but e 2 x − 1 → 0 . So e x e 2 x − 1 → ∞ , but how fast? The integral is improper. e 2 x − 1 lim = 2 . x x → 0 e x e 2 x − 1 = 1 e 2 x − 1 · e x ≈ 1 x as x → 0 + . x · 2 x Dan Barbasch () Calculus 1120, Spring 2012 October 11, 2012 5 / 6