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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?

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    $\begingroup$ Did you change arrangement of integrals.? $\endgroup$ – Nosrati Jan 31 '17 at 20:59
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    $\begingroup$ I would integrate with respect to $x$ first. $\endgroup$ – Simply Beautiful Art Jan 31 '17 at 20:59
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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 $$ \int_{-\infty}^\infty e^{-a^2x^2}\,\mathrm dx=\frac{\sqrt{\pi}}{a},\quad a>0. $$

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