By considering the integral of $\displaystyle e^{-yx^2}$ over a suitable region of $\displaystyle \mathbb{R}^2$, show that for $\displaystyle 0<a<b<\infty$,
$\displaystyle \int_{0}^{\infty} \frac{e^{-ax^2} - e^{-bx^2}}{x^2} dx = \sqrt{\pi} ( \sqrt{b} - \sqrt {a} )$
2. the hint was already given, we have that $\displaystyle \frac{e^{-ax^2}-e^{-bx^2}}{x^2}=\int_a^b e^{-tx^2}\,dt,$ now reverse integration order (justified by Tonelli's Theorem since the integrand is non negative, becuase of $\displaystyle b>a$) and use the fact that $\displaystyle \int_{x\ge0}e^{-x^2}\,dx=\frac{\sqrt\pi}2,$ hence, the result.