# How to evaluate $\int_{-\infty}^\infty\int_1^2\frac{y}{\sqrt{2\pi}}e^{-x^2y^2/2}\,\mathrm dy\,\mathrm dx.$

How to evaluate the following integral: $$\int_{-\infty}^\infty\int_1^2\frac{y}{\sqrt{2\pi}}e^{-x^2y^2/2}\,\mathrm dy\,\mathrm dx.$$

What I did was integrate with respect to $y$ first, but then I get $$\int_{-\infty}^\infty\left[-\frac{2}{x^2\sqrt{2\pi}}e^{-x^2}+\frac{2}{x^2\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\right]\,\mathrm dx.$$ From here on I wouldn't know how to continue. I've evaluated the integral in Mathematica, which yields to 1. So it should be integrable. Could someone help me out?

• Did you change arrangement of integrals.? – Nosrati Jan 31 '17 at 20:59
• I would integrate with respect to $x$ first. – Simply Beautiful Art Jan 31 '17 at 20:59

Hint. Integrating with respect to $x$, one gets $$\int_{-\infty}^\infty\int_1^2\frac{y}{\sqrt{2\pi}}e^{-x^2y^2/2}\,\mathrm dy\,\mathrm dx=\int_1^2\frac{y}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-x^2y^2/2}\,\mathrm dx\,\mathrm dy=\int_1^2\frac{y}{\sqrt{2\pi}}\cdot \frac{\sqrt{2\pi}}{y}\,\mathrm dy$$ where we have used the gaussian result $$\int_{-\infty}^\infty e^{-a^2x^2}\,\mathrm dx=\frac{\sqrt{\pi}}{a},\quad a>0.$$