各种积分

$Guass$积分

$$\int_{0}^{+\infty}e^{-x^{2}}dx$$

设$I(t)={(\int_{0}^{t} e^{-x^{2}}dx)}^2$, 则求

$$\sqrt{\lim\limits_{t \to +\infty}I(t)}$$

$$I^{‘}(t) = 2(\int_{0}^{t}e^{-x^{2}}dx)e^{-t^2} = 2\int_{0}^{t}e^{-x^2-t^2}dx = 2\int_{0}^{1}2te^{(y^2+1)t^2}dy$$

$$\frac{dI(t)}{dt} = \frac{\int_0^1\frac{e^{-(y^2+1)t^2}}{-(y^2+1)}dy}{dt}$$

$$I(t) = \int_0^1\frac{e^{-(y^2+1)t^2}}{-(y^2+1)}+C$$

$$I(0)=\int_0^1\frac{-1}{y^2+1}dy+C = -\frac{\pi}{4} + C=0,C=\frac{\pi}{4}$$

$$I(t) = \int_0^1\frac{e^{-(y^2+1)t^2}}{-(y^2+1)}+\frac{\pi}{4}$$

$$\lim\limits_{t \to +\infty}I(t)=0+\frac{\pi}{4} = \frac{\pi}{4}$$

$$\therefore\int_{0}^{+\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}$$

$Euler$积分

$Beta$函数

$$B(p,q)=\int_0^1x^{p-1}(1-x)^{q-1}dx\ (p > 0, q > 0)$$

$$let\ x = \sin^2\theta,B(p,q)=\int_0^{\frac{\pi}{2}}{(\sin\theta)^{2p-2} (\cos\theta)^{2q-2} d(\sin^2\theta)} = \int_0^{\frac{\pi}{2}}{\sin^{2p-1}\theta \cos^{2q-1}\theta d\theta}$$

$$let\ x = \frac{y}{1+y}, B(p,q) = \int_0^1\frac{y^{p-1}}{(y+1)^{p+q}}dy.$$

(1) 对偶性$B(p,q)=B(q,p)$

由区间再现，$B(p,q)=\int_0^1x^{p-1}(1-x)^{q-1}dx = \int_0^1(1-x)^{p-1}x^{q-1}dx = B(p, q).$

(2) $B(p,q)$在定义域上连续，且有任意阶连续偏导数.

(3) 递推公式

$$B(p,q+1)=\frac{q}{p+q}B(p,q),\ B(p+1,q)=\frac{p}{p+q}$$

PROOF:

$$B(p,q)=\int_0^1x^{p-1}(1-x)^{q-1}dx=\frac{1}{p}\int_0^1(1-x)^{q-1}dx^p$$

$$=\frac{1}{p}(1-x)^{q-1}x^p|_0^1+\frac{q-1}{p}\int_0^1x^p(1-x)^{q-2}dx$$

$$=\frac{q-1}{p}\int_0^1x^{p-1}(1-(1-x))(1-x)^{q-2}dx$$

$$=\frac{q-1}{p}[\int_0^1{x^{p-1}(1-x)^{(q-2)}dx}-{\int_0^1x^{p-1}(1-x)^{q-1}dx}]$$

$$=\frac{q-1}{p}[{B(p,q-1)-B(p,q)}]$$

$$B(p,q+1) = \frac{q}{p}[(B(p,q)-B(p,q+1))]$$

$$B(p,q+1)=\frac{q}{p+q}B(p,q)$$

由对偶性, $B(p+1,q) = \frac{p}{p+q}B(p,q).$

(4) 设$p>0$

$$B(p,p)=\frac{1}{2^{2p-1}}B(\frac{1}{2},p)$$

PROOF:

$$B(p,p)=2\int_0^{\frac{\pi}{2}}{(\sin^{2p-1}\theta) (\cos^{2p-1}\theta) d\theta}$$

$$=\frac{2}{2^{2p-1}}\int_0^{\frac{\pi}{2}}(2\sin\theta \cos\theta)^{2p-1}d\theta$$

$$=\frac{1}{2^{2p-1}}\int_0^{\frac{\pi}{2}}(\sin(2\theta))^{(2p-1)}d(2\theta)$$

$$=\frac{1}{2^{2p-1}}\int_0^{\pi}\sin^{(2p-1)}\theta d\theta$$

$$=\frac{1}{2^{2p-1}}2\int_0^{\frac{\pi}{2}}\sin^{(2p-1)}\theta d\theta$$

$$=\frac{1}{2^{2p-1}}B(\frac{1}{2},p).$$