# Need integrating e^(−x^2) x^2

Integrating ${e}^{-{x}^{2}}{x}^{2}$
Show that $\mathcal{L}\left\{{t}^{1/2}\right\}=\sqrt{\pi }/\left(2{s}^{3/2}\right),\phantom{\rule{mediummathspace}{0ex}}s>0$
By the definition of Laplace transform we get:
$\mathcal{L}\left\{{t}^{1/2}\right\}={\int }_{0}^{\mathrm{\infty }}{t}^{1/2}{e}^{-st}\phantom{\rule{thinmathspace}{0ex}}dt=\left\{x=\sqrt{st}\right\}=\frac{2}{{s}^{3/2}}{\int }_{0}^{\mathrm{\infty }}{e}^{-{x}^{2}}{x}^{2}\phantom{\rule{thinmathspace}{0ex}}dx.$
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scranna0o
Rewrite the integral as
$\underset{0}{\overset{\mathrm{\infty }}{\int }}-\frac{x}{2}\left(-2x{e}^{-{x}^{2}}\right)dx$
and use integration by parts with $f\left(x\right)=-\frac{x}{2}$ and $g\left(x\right)={e}^{-{x}^{2}}$ to obtain
${-\frac{x}{2}{e}^{-{x}^{2}}|}_{0}^{\mathrm{\infty }}+\frac{1}{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-{x}^{2}}dx=\frac{\sqrt{\pi }}{4}$
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Cyrus Travis
${\int }_{0}^{\mathrm{\infty }}{x}^{2}{e}^{-{x}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{0}^{\mathrm{\infty }}\frac{x}{2}{e}^{-{x}^{2}}\left(2x\phantom{\rule{thinmathspace}{0ex}}dx\right)={\int }_{0}^{\mathrm{\infty }}\frac{\sqrt{u}}{2}{e}^{-u}\phantom{\rule{thinmathspace}{0ex}}du=\frac{1}{2}\mathrm{\Gamma }\left(\frac{3}{2}\right).$
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