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# By integration by parts we have tx et dt u tx dv u

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Unformatted text preview: 4) : Start with the indeﬁnite integral tx e−t dt . u dv Here, treat x as a constant. By “integration by parts”, we have tx e−t dt u tx · dv = u − e−t − v x tx−1 dt v − e−t du . In other words, tx e−t dt x tx−1 dt − e−t + = = tx · − e−t +C − tx e−t + C . The corresponding deﬁnite integral: t t=0 +∞ +∞ +∞ x −t e dt −t + t=0 −e 10 xt x−1 dt = x −t −t e t=0 . That is, (♥) +∞ +∞ +∞ tx e−t dt − x t=0 tx−1 e−t dt − tx e−t = t=0 t=0 . Explain why the right-hand side of (♥) equals 0. Your explanation must involve the limit lim tx e−t . t− ∞ → Hence +∞ t=0 +∞ tx e−t dt − x tx−1 e−t dt = 0. t=0 Finally, rewrite this last identity using Γ x and Γ x + 1 . (5) (Extra 20pts) Let Γ x ln Γ x be as in (3). Know that ln Γ x = lim − γ + n− ∞ → − ln x 1+x n 1 x k k=1 1+ has the following expression: x 2 1+ x 3 · ··· · 1+ x n where γ is the Euler’s constant . One way to deﬁne the Euler’s constant γ is 11 , +∞ γ = − Γ′ 1 = t=0 e−t dt. − ln t Let f x and g x be as before page 4 . From the above expression of ln Γ x and the result of (4), one can prove −f x +g x = 1−γ x− ln Γ 2 + x , f x +g x = 1−γ x+ ln Γ 2 − x . Or, equivalently, fx = 1 2 gx = 1−γ ln Γ 2−x Γ 2+x x− 1 2 , Γ 2+x ln . Γ 2−x Use the above information to evaluate each of the following: ∞ (5a) m=2 −1 ∞ (5c) m=2 −1 m−1 m−1 ζm −1 m ζm −1 m · 2m ∞ , (5b) m=2 ∞ , (5d) m=2 The answers may involve γ . 12 ζm −1 m , ζm −1 m · 2m ....
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