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By squeeze theorem, the result follows.
t=0 +∞ tx e−t dt − x tx−1 e−t dt We may rewrite this last identity using Γ x
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 +
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
2 , Γ 2+x
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.
- Fall '08