solutions for b

# Hence the integral above is also convergent 8 iii

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Unformatted text preview: = n=2 ∞ ζn n=2 = ∞ n=2 ∞ k=2 7 −1 1 kn +∞ . . = ∞ k=2 = ∞ k=2 = ∞ n=2 ∞ n=2 ∞ ∞ 2 1 k 1− 1 k ∞ k−1 1 k−1 k=2 = 1− 1 k − 1 2 + 1 1 − 2 3 + 1 1 − 3 4 + = n 1 k 1 k k=2 = swapped the summation symbols k=2 = 1 kn 1 1 − 4 5 1 = +∞. Hence the integral (∗) above is also convergent . 8 + ··· [III] (30pts) Recall that Γ s Γs = was deﬁned as +∞ ts−1 e−t dt s>0 . t=0 Know that for constant real numbers p and q with p > 0 and q > 0, the following holds: 1 u=0 (1) q −1 up−1 1 − u ΓpΓq Γ p+q = du We may use the above formula to evaluate Step 1. 1 2 p= Substitute 1 2 Γ as follows. q= 1 2 ·Γ and 1 2 formula, and obtain 1 u u=0 1 − 2 1−u 1 − 2 1 2 Γ du = 1 2 Γ 1 2 Γ = + 2 Γ1 = where we have used Γ1 Γ = 0 ! = 1. 9 1 2 2 , . 1 2 into the above 1 Step 2. We may evaluate the integral u − u=0 1 2 1−u − 1 2 du independently of Step 1 above . We rely on the following information: “An antiderivative of =u fu − 1 2 − 1−u 1 2 is = 2 arcsin Fu √ u. ” See “Review of Lectures – XXVII”, page 21, Exercise 3. It goes as follows: 1 u u=0 − 1 2 1−u − 1 2 du = 2 arcsin √ 1 u u=0 = 2 arcsin = 2· = √ √ π. π 2 1 − 2 arcsin − 0 2 ·0 Step 3. In Step 1 and Step 2, we have expressed the value of the same deﬁnite integral in two diﬀerent ways. Now equate them: Γ Taking into account the fact that Γ s Γ 2 1 2 = π. > 0 provided s > 0, we obtain 1 2 = 10 √ π. (2) By the deﬁnition of Γ s , Γ 1 2 +∞ = 1 t− 2 e−t dt. t=0 Hence the result of (1) above is paraphrazed as +∞ (#) 1 t− 2 e−t dt √ = π. t=0 +∞ We may utilize thi...
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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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