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# 7.8 - 7.8 Improper Integrals(Dated In conclusion IMPROPER...

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7.8: Improper Integrals (Dated: October 29, 2011) IMPROPER INTEGRALS OF TYPE 1 (a) If R t a f ( x ) dx exists for all t a , then Z a f ( x ) dx : = lim t →∞ Z t a f ( x ) dx provided the limit on the right exists as a finite num- ber. Here and hereafter ‘ : = ’ is an abbreviation for ‘is defined as’. (b) If R b t f ( x ) dx exists for all t a , then Z b -∞ f ( x ) dx : = lim t →-∞ Z b t f ( x ) dx provided the limit on the right exists as a finite num- ber. The improper integrals R a f ( x ) dx and R b -∞ f ( x ) dx are called convergent provided the corresponding limits exist and divergent otherwise. (c) If both R c f ( x ) dx and R c -∞ f ( x ) dx are convergent, then Z -∞ f ( x ) dx : = Z c -∞ f ( x ) dx + Z c f ( x ) dx and is called convergent. Note that any real number c can be used in the definition of R -∞ f ( x ) dx. The im- proper integral R -∞ f ( x ) dx is called divergent if it is not convergent. Example 1 Consider the improper integral R 1 1 x p dx. For p = 1 , Z 1 1 x p dx = lim t →∞ Z t 1 1 x dx = lim t →∞ [ln | x | ] t 1 = lim t →∞ ln( t ) does not exist as a finite number. In other words, R 1 1 x dx is divergent. For p 6= 1 , Z 1 1 x p dx = lim t →∞ Z t 1 1 x p dx = lim t →∞ Z t 1 x - p dx = lim t →∞ x 1 - p 1 - p t 1 = lim t →∞ t 1 - p 1 - p - 1 1 - p = if p < 1; 1 p - 1 if p > 1. In conclusion, Z 1 1 x p dx is ( divergent if p 1; convergent if p < 1. Example 2 [Exercise 7.8.7 in the text] Since Z - 1 -∞ 1 p 2 - w dw = lim t →-∞ Z - 1 t 1 p 2 - w dw = lim t →-∞ h - 2 p 2 - w i - 1 t = lim t →-∞ 2 + 2 p 2 - t · = ∞ , R - 1 -∞ 1 p 2 - w dw is divergent. Example 3 [Exercise 7.8.8 in the text] Letting u = x 2 + 2 yields du 2 = x dx and Z x ( x 2 + 2) 2 dx = 1 2 Z 1 u 2 du = - 1 2 u + C = - 1 2( x 2 + 2) + C . Thus for t 0 , Z t 0 x ( x 2 + 2) 2 dx = - 1 2( t 2 + 2) + 1 4 and Z 0 x ( x 2 + 2) 2 dx = lim t →∞ - 1 2( t 2 + 2) + 1 4 = 1 4 is convergent. Example 4 [Exercise 7.8.10 in the text] Since Z - 1 -∞ e - 2 t dt = lim t →-∞ Z - 1 t e - 2 x dx = lim t →-∞ - e - 2 x 2 - 1 t = lim t →-∞ e - 2 t 2 - e 2 2 = ∞ , R - 1 -∞ e - 2 t dt is divergent.

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2 Example 5 [Exercise 7.8.11 in the text] Since Z 0 x 1 + x 2 dx = lim t →∞ Z t 0 x 1 + x 2 dx = lim t →∞ 1 2 Z 1 + t 2 1 du u where u = 1 + x 2 , du = 2 x dx = lim t
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7.8 - 7.8 Improper Integrals(Dated In conclusion IMPROPER...

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