Re: [微積] 一題積分..
※ 引述《ihha (王建民!!!)》之銘言:
: 請問一題積分證明
: 証:
: 積 e^(-x^2) 從負無限大到無限大 = 根號 pi
: 謝謝大家^^||||
∞ ∞
證明: ∫ e^(-x^2) dx = 2∫ e^(-x^2) dx
-∞ 0
∞ ∞
令 I = ∫ e^(-x^2) dx = ∫ e^(-y^2) dy
0 0
∞ ∞
則 I^2 = ∫ e^(-x^2) dx ∫ e^(-y^2) dy
0 0
∞ ∞
= ∫ ∫ e^(-(x^2 + y^2)) dxdy
0 0
令 x = rcosΘ , y = rsinΘ 則 |J| = r
∞ ∞
所以 I^2 = ∫ ∫ e^(-(x^2 + y^2)) dxdy
0 0
∞ π/2
= ∫ ∫ r*e^(-r^2) dΘdr
0 0
π ∞
= (---)*∫ r*e^(-r^2) dr
2 0
π 1 |∞ π
= (---)*(- ---e^(-r^2)) | = ---
2 2 |0 4
∞ π^(1/2)
因此 I = ∫ e^(-x^2) dx = ----------
0 2
∞ ∞
所以 ∫ e^(-x^2) dx = 2I = 2∫ e^(-x^2) dx = π^(1/2)
-∞ 0
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 61.66.173.21
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