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Unformatted text preview: p then 1 (1 + x 2 ) p ≤ 1 x 2 p . This means that the integral ∞ R 1 1 (1+ x 2 ) p dx converges when the integral ∞ R 1 1 x 2 p dx which happens when 2 p > 1. To show that it doesn’t converge for other values, we will use comparison theorem again. This time we can write 1 + x 2 ≤ 2 x 2 , since ,x ≥ 1 . Then doing similar manipulations we arrive at the conclusion that 1 (1 + x 2 ) p ≥ 1 2 p x 2 p . Therefore the integral ∞ R 1 1 (1+ x 2 ) p dx diverges whenever the integral ∞ R 1 1 2 p x 2 p dx = 1 2 p ∞ R 1 1 x 2 p dx diverges. Which happens for p ≤ 1 / 2. 1 2 Question 2 See your lecture notes from 2/5....
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This note was uploaded on 07/10/2008 for the course MATH 1B taught by Professor Reshetiken during the Spring '08 term at Berkeley.
 Spring '08
 Reshetiken
 Math, Calculus, Continuity

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