Limit comparison

Limit comparison - Comparison theorem for improper...

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Comparison theorem for improper integrals ( with explicit examples ) 1 Intro This is a complement to the comparison theorem for improper integrals in the textbook. The vanilla version presented in the textbook is good enough to solve some very easy examples and it becomes exponentially gory with the complexity of the integral. Fortunately it is not hard to refine the statement in the book, and turn it into a powerful tool to estimate the convergence of arbitrarily complicated integrals. 2 Comparison ( vanilla version ) In words: suppose that a continuous positive function h ( x ) is smaller than another continuous function f ( x ) in a neighborhood of + ; if the integral of the bigger function is convergent, then the integral of the smaller function is convergent; vice versa, if the smaller function has a divergent integral then the integral of the bigger function must diverge too. More precisely: Proposition 2.1. Let f ( x ) be a non negative continuous function on [ a, + ) . Let h ( x ) be another continuous function on the same interval such that f ( x ) h ( x ) for all x a . The following implications hold: Z + a f ( x ) dx < + ∞ ⇒ Z + a h ( x ) dx < + and Z + a h ( x ) dx = + ∞ ⇒ Z + a f ( x ) dx = + and an analogous statement holds for the interval ( -∞ ,a ] and for negative functions. A similar statement holds in the neighborhood of a vertical asymptote. 3 Comparison ( enhanced version ) The idea that we try to implement is the following: the improper integral of a power function, an exponential, or a logarithm are very easy to study; we would like to reduce any given function f ( x ) to one of these functions or a product of them by comparison. This can be made precise with the definition of limit. Consider the example from the book: Z + 1 x + 1 x 4 - x 1
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To determine if the integral converges or not we need to study the behavior of the function at infinity and for x 1 where it has an asymptote. To study the behavior at 1 we rewrite the function in the following way: x + 1 x 4 - x = x + 1 p x ( x - 1)( x 2 + x + 1) = x + 1 p x ( x 2 + x + 1) · 1 x - 1 In this writing we have separated the part of this function responsible for the divergence 1 x - 1 from everything else. Using limits: lim x 1 + x + 1 p x ( x 2 + x + 1) = 2 3 lim x 1 + 1 x - 1 = + The enhanced version of the comparison theorem will tell us that in order study the integral in a neighborhood of 1 we can ignore everything that converges to a number or more precisely: Z 1+ ± 1 x + 1 x 4 - x < + ∞ ⇔ Z 1+ ± 1 1 x - 1 < + where ± is some positive number. We can repeat the same procedure at infinity, but in this case we want to isolate the part of the function which is responsible for the convergence to zero at infinity: x + 1 x 4 - x = x (1 + 1 x ) x 2 ( q 1 - 1 x 3 ) = 1 x · 1 + 1 x q 1 - 1 x 3 As in the previous case we can make this precise using limits: lim x + 1 + 1 x q
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This note was uploaded on 11/03/2011 for the course MATH 1090 taught by Professor Greenwood during the Spring '08 term at MIT.

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Limit comparison - Comparison theorem for improper...

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