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?

  • 1
    $\begingroup$ Did you change arrangement of integrals.? $\endgroup$ – Nosrati Jan 31 '17 at 20:59
  • 2
    $\begingroup$ I would integrate with respect to $x$ first. $\endgroup$ – 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 $$ \int_{-\infty}^\infty e^{-a^2x^2}\,\mathrm dx=\frac{\sqrt{\pi}}{a},\quad a>0. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.