Eufisky - The lost book

误差函数Erf专题

1.int e^(-x^2-2y^2), y=0..x, x=0..\infty用WolframAlpha计算二重积分.

$\int_0^\infty {\int_0^x {{e^{ - {x^2} - 2{y^2}}}} } dydx = \frac{{\arctan \left( {\sqrt 2 } \right)}}{{2\sqrt 2 }}.$

求 $I\left( b \right) = \int_0^\infty {\int_0^x {{e^{ - {a^2}{x^2} - {b^2}{y^2}}}} } dydx,\qquad a,b\geq 0.$

解.事实上,我们有

$\begin{align*}I\left( b \right) &= \int_0^\infty {\int_0^x {{e^{ - {a^2}{x^2} - {b^2}{y^2}}}} } dydx = \int_0^\infty {{e^{ - {a^2}{x^2}}}dx\int_0^x {{e^{ - {b^2}{y^2}}}} } dy\\&= \frac{1}{b}\int_0^\infty {{e^{ - {a^2}{x^2}}}dx\int_0^{bx} {{e^{ - {y^2}}}} } dy = \frac{{\sqrt \pi }}{{2b}}\int_0^\infty {{e^{ - {a^2}{x^2}}} \mathrm{erf}\left( {bx} \right)dx}. \end{align*}$

对

$b$ 求导我们有

$\begin{align*}I'\left( b \right) &= - \frac{1}{{{b^2}}}\int_0^\infty {{e^{ - {a^2}{x^2}}}dx\int_0^{bx} {{e^{ - {y^2}}}} } dy + \frac{1}{b}\int_0^\infty {x{e^{ - \left( {{a^2} + {b^2}} \right){x^2}}}dx} \\&= - \frac{1}{b}I\left( b \right) + \frac{1}{{2b\left( {{a^2} + {b^2}} \right)}}.\end{align*}$

又

$I(0)=\frac1{2a^2}$ ,解得

$I\left( b \right) = \frac{1}{{2ab}}\arctan \frac{b}{a},$ 即

$\int_0^\infty {{e^{ - {a^2}{x^2}}}\mathrm{erf}\left( {bx} \right)dx} = \frac{1}{{a\sqrt \pi }}\arctan \frac{b}{a}.$

法医秦明，完美案件