sol for exam 2 take home

T0 x x as follows 0 to the right hand side and we

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Unformatted text preview: . By squeeze theorem, the result follows. In short, +∞ t=0 +∞ tx e−t dt − x tx−1 e−t dt We may rewrite this last identity using Γ x Γ x+1 Shift x Γ x = 0. t=0 − xΓ x as follows: = 0. to the right-hand side and we obtain the required identity. 11 (5) (Extra 20pts) Let Γ x be as in (3). Know that ln Γ x ln Γ x = − ln x n lim − γ + n− ∞ → 1+x 1 x k k=1 x 2 1+ has the following expression: 1+ x 3 · ··· · 1+ x n , where γ is the Euler’s constant . One way to define the Euler’s constant γ is +∞ 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− 12 1 2 , Γ 2+x ln . Γ 2−x Use the above information to evaluate each of the following: ∞ (5a) −1 m=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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