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I need find $x^2\cdot e^{{-x^2}/2} * e^{{-x^2}/2}$. I used statements, that $\widehat{xf}=i \widehat{f}'$ and $\widehat{f*g}=\sqrt{2\pi}\widehat{f}\cdot \widehat{g}$. So, $\widehat{x^2\cdot e^{{-x^2}/2}}(\lambda)={e^{{{-\lambda^2}/2}} \cdot(1-\lambda^2)}$ and $\widehat{e^{{-x^2}/2}}(\lambda)=e^{-\lambda^2/2}$. Then, $\widehat{{(x^2\cdot e^{{-x^2}/2})*e^{{-x^2}/2}}}(\lambda)=\sqrt{2\pi}\cdot e^{-\lambda^2}\cdot(1-\lambda^2)$.

So, $(x^2\cdot e^{{-x^2}/2} * e^{{-x^2}/2})(y)= \int_{-\infty}^\infty e^{-\lambda^2+i\lambda y} \cdot (1-\lambda^2) \, d\lambda$, but I don't know how solve this integral, perhaps, I did some mistake in counting. Please help me solve this problem.

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    I see, wish you good luck!2012-05-18

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$\begin{align*} \left(x^2e^{-x^2/2}\right)*\left(e^{-x^2/2}\right) &= \int_{-\infty}^{\infty}t^2e^{-t^2/2}e^{-(x-t)^2/2}\;\mathrm dt\\ &= e^{-x^2/2}\int_{-\infty}^{\infty}t^2e^{-t^2+xt}\;\mathrm dt\\ &= e^{-x^2/4}\int_{-\infty}^{\infty}t^2e^{-t^2+xt -(x/2)^2}\;\mathrm dt\\ &= e^{-x^2/4}\int_{-\infty}^{\infty}t^2e^{-(t-x/2)^2}\;\mathrm dt\\ &= \sqrt{\pi} e^{-x^2/4}\int_{-\infty}^{\infty}\frac{1}{\frac{1}{\sqrt{2}}\sqrt{2\pi}}t^2e^{-(t-x/2)^2/2(1/\sqrt{2})^2}\;\mathrm dt \end{align*}$ where the integral can be recognized as the value of $E[X^2]$ for a normal random variable $X$ with mean $\frac{x}{2}$ and variance $\frac{1}{2}$. Since $E[X^2] = \frac{1}{2}+\left(\frac{x}{2}\right)^2$, we get $\left(x^2e^{-x^2/2}\right)*\left(e^{-x^2/2}\right) = \sqrt{\pi}\left(\frac{1}{2}+\left(\frac{x}{2}\right)^2\right)e^{-x^2/4}$