证明 $\dps{\int_0^\infty f\sez{\sex{Ax-\frac{B}{x}}^2}\rd x=\frac{1}{A}\int_0^\infty f(y^2)\rd y}$ (其中左、右积分存在, 且 $A,B>0$). 

证明: $$\beex \bea I&\equiv \int_0^\infty f\sez{\sex{Ax-\frac{B}{x}}^2}\rd x\\ &=\int_0^\infty f\sez{\sex{\frac{B}{t}-At}^2}\frac{B}{At^2}\rd t\quad\sex{t=\frac{B}{Ax}}\\ &=\frac{1}{2}\int_0^\infty f\sez{\sex{Ax-\frac{B}{x}}^2} \sex{1+\frac{B}{Ax^2}}\rd x\\ &=\frac{1}{2A}\int_0^\infty f\sez{\sex{Ax-\frac{B}{x}}^2} \rd \sex{Ax-\frac{B}{x}}\\ &=\frac{1}{2A}\int_{-\infty}^{+\infty}f(y^2)\rd y\quad\sex{y=Ax-\frac{B}{x}}\\ &=\frac{1}{A}\int_0^\infty f(y^2)\rd y. \eea \eeex$$