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$\begin{gathered} \int {x{{\text{e}}^{ - {x^2}}}{\kern 1pt} {\text{d}}x} \hfill \\ u = - {x^2},\frac{{{\text{d}}u}}{{{\text{d}}x}} = - 2x, \hfill \\ = - \frac{1}{2}\int {{{\text{e}}^u}{\kern 1pt} {\text{d}}u} = - \frac{{{{\text{e}}^u}}}{2} \hfill \\ = - \frac{{{{\text{e}}^{ - {x^2}}}}}{2} + C \hfill \\ = - \frac{1}{{2{{\text{e}}^{{x^2}}}}} + C \hfill \\ \int_0^\infty {x{{\text{e}}^{ - {x^2}}}{\kern 1pt} {\text{d}}x} = - \mathop {\lim }\limits_{x \to \infty } \frac{1}{{2{{\text{e}}^{{x^2}}}}} + \frac{1}{2} = \frac{1}{2} \hfill \\ \end{gathered}$

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$\int_{-\infty}^{+\infty}x e^{-x^2/2}dx\\ =\int_{-\infty}^{+\infty} e^{-x^2/2} d \frac{x^2}{2}\\ =(\int_{-\infty}^{0}+\int_{0}^{+\infty}) e^{-x^2/2} d \frac{x^2}{2}\\ =\int_{-\infty}^{0} e^{-x^2/2} d \frac{x^2}{2}+\int^{+\infty}_{0} e^{-x^2/2} d \frac{x^2}{2}\\ =\int_{+\infty}^{0} e^{-u} d u+\int^{+\infty}_{0} e^{-u} d u\\ =(\int_{+\infty}^{0} +\int^{+\infty}_{0}) e^{-u} d u\\ =(-\int^{+\infty}_{0} +\int^{+\infty}_{0}) e^{-u} d u\\ =0$

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