1873_solutions

# N some exercises on cauchy products 1 suppose that f

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Unformatted text preview: , the integral Þ  1. x2 Ý Since the integral Þ j. Ý1 2 x2 1 exp 2 log x Þ3 Ý log x Since 1 log log x Ý1 2 x2 dx is convergent, the integral Þ Ý 2 1 exp 2 log x 1 log log log x log x  1. x2 dx is convergent. dx log log x 2 0 log x 1 log x whenever x is sufficiently large. For such x we have we know that log log x 2 2 1 1 1   1. x exp log log x 2 exp 1 log x log x log log x 2 lim xÝ Since the integral Þ l. Þ1 Ý exp Since 1 log x Ý 3 1 x dx is divergent, the integral Þ Ý 3 1 log x log log x dx 1 log x 2 log x whenever x is sufficiently large we have 1 exp log x 355 1 exp 1 2 log x  1 x dx is divergent. for x sufficiently large. Therefore, since the integral Þ 1 Ý 1 dx is divergent. Þ1 exp log x 2. Ý 1 x dx is divergent, the integral a. Prove that the integral Þ1 Ý x  1 e x dx converges for every number . Ý The convergence of this integral follows at once from the convergence of the integral Þ 1 and the fact that 1 x lim x e  0. xÝ 1 x2 dx 1 x2 b. Prove that the integral Þ0 1 x  1 e x dx converges if and only if   0 and deduce that the integral Þ0 Ý x  1 e x dx converges if and only if   0. The latter integral defines the value at  of the gamma function and is denoted as  . You will see more about this important function later. 1 The convergence of the integral Þ x  1 e x dx if and only if   0 follows at once from the 0 corresponding fact about the integral Þ 1 0 x  1 dx and the fact that 1 lim x  e x0 x1 x  1. 3. Prove that the integral Þ0 1 1 t 11 t dt converges if and only if both  and  are positive. This integral defines the value at the point ,  of the beta function and is denoted as B ,  . Some Further Exercises on Improper Integrals 1. Determine the convergence or divergence of the following integrals: a. Þ0 Ý sin x dx x This integral converges by Diriclet’s test. We can use exactly the same argument that was used in Subsection 13.4.7. b. Þ2 Ý sin 3 x dx x In order to use Dirichlet’s test for this integral we observe that if w 2 then w Þ 2 sin 3 xdx  cos w  1 cos 3 w  cos 2 1 cos 3 2 8 . 3 3 3 Ý sin 3 x dx is convergent. In this way we see that the integral Þ x 2 c. Þ1 Ý e x sin e x dx x 356 e x sin e x . x It is instructive to ask Scientific Notebook to sketch the graph y  20 10 0 1.5 2 2.5 x 3 3.5 4 4.5 -10 -20 As we move from left to right, the function oscillates with a rapidly expanding amplitude but the peaks, as they become higher, also become very narrow and, as we shall see in a moment, Ý e x sin e x the integral Þ dx is convergent. x 1 Given any number w 1 we have Þ 1 e x sin e x dx w ew Þe  sin udu and therefore Dirichlet’s test guarantees that the integral Þ 2 Ý 1 e x sin e x dx is convergent. x 2. Integrate by parts to obtain the identity w w 2 2 Þ 0 sin2 x dx  sin w  Þ 0 2 sin x cos x dx w x x and deduce that each of the following four improper integrals equals Ý Þ 0 sin x dx. x a. Þ0 Ý 2 sin x cos x dx x Þ0 Ý 2 sin x cos x dx  2 Þ Ý...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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