正态分布概率密度函数的积分 I = ∫ − ∞ ∞ 1 2 π σ e − ( x − μ ) 2 2 σ 2 d x \begin{aligned} I = \int _{- \infty} ^{\infty} \frac{1}{\sqrt{2 \pi} \sigma} e ^ {- \frac{(x - \mu) ^2}{2 \sigma ^ 2}} dx \end{aligned} I=∫−∞∞2π σ1e−2σ2(x−μ)2dx
指数中包含二次项，无法直接求解，需要做一些变换。
夹逼定理
F ( x ) , G ( x ) F(x), G(x) F(x),G(x)在 x 0 x_0 x0连续且极限相同,即 lim x → x 0 F ( x ) = lim x → x 0 G ( x ) = A \begin{aligned} \lim_{x \to x_0}F(x) = \lim_{x \to x_0}G(x) = A \end{aligned} x→x0limF(x)=x→x0limG(x)=A
f ( x ) f(x) f(x)在 x 0 x_0 x0的某领域内恒有 F ( x ) ≤ f ( x ) ≤ G ( x ) F(x) \le f(x) \le G(x) F(x)≤f(x)≤G(x)
则 lim x → x 0 f ( x ) = A \begin{aligned} \lim_{x \to x_0}f(x) = A \end{aligned} x→x0limf(x)=A
二重积分的换元法
∬ x 2 + y 2 ≤ R 2 f ( x , y ) d x d y = ∫ 0 R ∫ 0 2 π f ( r s i n θ , r c o s θ ) r d r d θ \begin{aligned} &\iint_{x^2+y^2 \le R^2} f(x, y)dxdy \\ = &\int_0^R \int_0^{2\pi} f(rsin\theta, rcos\theta) \ r \ dr d\theta \end{aligned} =∬x2+y2≤R2f(x,y)dxdy∫0R∫02πf(rsinθ,rcosθ) r drdθ
正态分布概率密度函数的积分
(1) 令 y = x − μ 2 σ \begin{aligned} y = \frac{x - \mu}{\sqrt{2} \sigma} \end{aligned} y=2 σx−μ, 得到 I = 1 π ∫ − ∞ ∞ e − y 2 d y \begin{aligned} I = \frac {1}{\sqrt{\pi}} \int _{-\infty} ^{\infty} e ^ {- y^2} dy \end{aligned} I=π 1∫−∞∞e−y2dy
(2) 令 U = ∫ − ∞ ∞ e − y 2 d y \begin{aligned} U = \int _{-\infty} ^{\infty} e ^ {- y^2} dy \end{aligned} U=∫−∞∞e−y2dy, 则 I = 1 π U \begin{aligned} I = \frac {1}{\sqrt{\pi}} U \end{aligned} I=π 1U
(3) U = lim R → ∞ ∫ − R R e − y 2 d y \begin{aligned} U = \lim_{R \to \infty} \int _{-R} ^{R} e ^ {- y^2} dy \end{aligned} U=R→∞lim∫−RRe−y2dy
(4) 转换为正方形区域内的二重积分 U 2 = lim R → ∞ ∫ − R R e − x 2 d x ∫ − R R e − y 2 d y = lim R → ∞ ∫ − R R ∫ − R R e − x 2 − y 2 d x d y = lim R → ∞ ∬ − R ≤ x ≤ R , − R ≤ y ≤ R e − x 2 − y 2 d x d y \begin{aligned} U^2 &= \lim_{R \to \infty} \int _{-R} ^{R} e ^ {- x^2} dx \int _{-R} ^{R} e ^ {- y^2} dy \\ &= \lim_{R \to \infty} \int _{-R} ^{R}\int _{-R} ^{R} e ^ {- x^2-y^2} dx dy \\ &= \lim_{R \to \infty} \iint _{-R \le x \le R, -R \le y \le R} e ^ {- x^2-y^2} dx dy \end{aligned} U2=R→∞lim∫−RRe−x2dx∫−RRe−y2dy=R→∞lim∫−RR∫−RRe−x2−y2dxdy=R→∞lim∬−R≤x≤R,−R≤y≤Re−x2−y2dxdy
(5) 内切圆积分 U 1 = lim R → ∞ ∬ x 2 + y 2 ≤ R 2 e − x 2 − y 2 d x d y = lim R → ∞ ∫ 0 R ∫ 0 2 π e − r 2 r d r d θ = ∫ 0 2 π d θ × lim R → ∞ ∫ 0 R e − r 2 r d r = 2 π ∫ 0 ∞ e − r 2 r d r = 2 π ( − 1 2 e − r 2 ∣ 0 ∞ ) = π \begin{aligned} U1 &= \lim_{R \to \infty} \iint _{x^2+y^2 \le R^2} e ^ {- x^2-y^2} dx dy \\ &= \lim_{R \to \infty} \int _{0}^{R} \int_{0}^{2 \pi} e ^ {- r^2} r \ dr d\theta \\ &= \int_{0}^{2 \pi} d\theta \times \lim_{R \to \infty} \int _{0}^{R} e ^ {- r^2} r \ dr \\ &= 2 \pi \int _{0}^{\infty} e ^ {- r^2} r \ dr \\ &= 2 \pi (- \frac{1}{2} e ^ {- r^2} |_0^\infty) \\ &= \pi \end{aligned} U1=R→∞lim∬x2+y2≤R2e−x2−y2dxdy=R→∞lim∫0R∫02πe−r2r drdθ=∫02πdθ×R→∞lim∫0Re−r2r dr=2π∫0∞e−r2r dr=2π(−21e−r2∣0∞)=π
(5) 外接圆积分 U 2 = lim R → ∞ ∬ x 2 + y 2 ≤ 2 R 2 e − x 2 − y 2 d x d y = lim R → ∞ ∫ 0 2 R ∫ 0 2 π e − r 2 r d r d θ = ∫ 0 2 π d θ × lim R → ∞ ∫ 0 2 R e − r 2 r d r = 2 π ∫ 0 ∞ e − r 2 r d r = 2 π ( − 1 2 e − r 2 ∣ 0 ∞ ) = π \begin{aligned} U2 &= \lim_{R \to \infty} \iint _{x^2+y^2 \le 2R^2} e ^ {- x^2-y^2} dx dy \\ &= \lim_{R \to \infty} \int _{0}^{\sqrt{2}R} \int_{0}^{2 \pi} e ^ {- r^2} r \ dr d\theta \\ &= \int_{0}^{2 \pi} d\theta \times \lim_{R \to \infty} \int _{0}^{\sqrt{2}R} e ^ {- r^2} r \ dr \\ &= 2 \pi \int _{0}^{\infty} e ^ {- r^2} r \ dr \\ &= 2 \pi (- \frac{1}{2} e ^ {- r^2} |_0^\infty) \\ &= \pi \end{aligned} U2=R→∞lim∬x2+y2≤2R2e−x2−y2dxdy=R→∞lim∫02 R∫02πe−r2r drdθ=∫02πdθ×R→∞lim∫02 Re−r2r dr=2π∫0∞e−r2r dr=2π(−21e−r2∣0∞)=π
(6) e − x 2 − y 2 e ^ {- x^2-y^2} e−x2−y2在整个平面上都大于0，因此有 ∬ x 2 + y 2 ≤ R 2 e − x 2 − y 2 d x d y < ∬ − R ≤ x ≤ R , − R ≤ y ≤ R e − x 2 − y 2 d x d y < ∬ x 2 + y 2 ≤ 2 R 2 e − x 2 − y 2 d x d y \begin{aligned} &\iint _{x^2+y^2 \le R^2} e ^ {- x^2-y^2} dx dy \\ \lt &\iint _{-R \le x \le R, -R \le y \le R} e ^ {- x^2-y^2} dx dy \\ \lt &\iint _{x^2+y^2 \le 2R^2} e ^ {- x^2-y^2} dx dy \end{aligned} <<∬x2+y2≤R2e−x2−y2dxdy∬−R≤x≤R,−R≤y≤Re−x2−y2dxdy∬x2+y2≤2R2e−x2−y2dxdy
夹逼定理可得 U 2 = U 1 = U 2 = π U^2 = U1 = U2 = \pi U2=U1=U2=π 因此 U = π , I = 1 π U = 1 \begin{aligned}U=\sqrt{\pi}, I = \frac {1}{\sqrt{\pi}} U = 1 \end{aligned} U=π ,I=π 1U=1
