Unformatted text preview: e 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 = − e−t u − e−t v x tx−1 dt v − du . In other words,
tx e−t dt − e−t + x tx−1 dt =
= tx · − e−t +C − tx e−t + C . The corresponding deﬁnite integral: tx e−t dt
t=0 +∞ +∞ +∞ +
t=0 − e−t 10 x tx−1 dt = − tx e−t t=0 . That is,
t=0 tx e−t dt − x tx−1 e−t dt
+∞ − tx e−t = = − t=0 tx e−t lim − t − +∞
→ − 0x e−0 0
= 0 0. Here, the reason why the limit
lim tx e−t
t − +∞
equals 0 is as follows: First, when x is a positive integer, then tx is a polynomial
in t, and hence this is “Review of Lectures – XIV”, Formula A. Second, when x is
not a positive integer, then choose any positive integer m which is bigger than x.
Then 0 < tx e−t < tm e−t holds for an arbitrary real number t with t > 1...
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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