Re: [積分] 積分e積出拍

[Honor1984 (希望願望成真)]

※ 引述《suhorng ( )》之銘言：
: ※ 引述《andy05227900 (小黑)》之銘言：
: : 請問為什麼
: : 積分區間0到正無限大e^-u^2du=根號拍/2
: : 為什麼積分e可以積出拍??
: Apostol 書中一個很神的做法
: 7.19 Define
:             x                          1 e^{-x^2(t^2+1)}
:   f(x) = ( ∫e^{-t^2} dt )^2,  g(x) = ∫-----------------dt
:             0                          0     t^2 + 1
:                                                                           π
:   a) Show that g'(x) + f'(x) = 0 for all x and deduce that g(x) + f(x) = ---
:                                              __                           4
           x
f'(x) = 2[∫e^{-t^2} dt]exp(-x^2)
           0

         1
g'(x) = ∫-2xexp[-x^2(t^2+1)]dt
         0

         令 u=xt

           x
      = -2∫exp[-x^2-u^2]du
           0


      =  -f'(x)


=> g'(x) + f'(x) = 0


積分


g(x) + f(x) = C


g(0) = arctan(1) = π/4


f(0) = 0


所以g(x) + f(x) = π/4

:                             ∞             √π
:   b) Use (a) to prove that ∫e^{-t^2} dt = ----
:                             0                2

lim g(x) = 0
x→∞

=>lim √[f(x)] = √[π/4] = (√π)/2
  x→∞

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