solutions for c

# 2 s1 ln 5 du 2 1 1 c 3a substitute s0 in

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Unformatted text preview: +λ ln u · us−1 · with respect to ln u sin π s − cos π s π As it turns out, C equals +∞ du = λs−1 π 1 . 1−λ 2 + s λs−1 ln u π2 Accordingly, you have evaluated ln u sin π s − cos π s π ln u du = . 2 + λs−1 = ln λ 5 du π2 + 1 . 1−λ : + C. (3a) Substitute s=0 in your answer in (2), then negate the whole quantity, and we may thereby evaluate +∞ u=0 (3b) 1 u u+λ Substitute · ln u s=1 +∞ u+λ : −1 = λ + π2 −1 . 1−λ + ln λ in your answer in (2), and we may thereby evaluate 1 u=0 Note 2 du · ln u 2 du 1 ln λ = + π2 The integrals in (3a) and (3b) for λ=1 1 . 1−λ + both exist. In order to ﬁnd their values, take the limit as λ − 1 of the outcomes of (3a) and (3b): → +∞ u=0 1 u u+1 · = and +∞ u=0 2 ln u du + π2 −1 lim λ− 1 → λ ln λ 1 u+1 · ln u = 2 + −1 1−λ = 1 , 2 du + π2 1 lim ln λ λ− 1 → respectively. 6 + 1 1−λ = 1 , 2...
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## This note was uploaded on 01/12/2014 for the course MATH 116 taught by Professor Scholle,minho during the Fall '08 term at Kansas.

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