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

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

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

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

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

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

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

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

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

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