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As the title, question 5 in this picture

https://i.stack.imgur.com/P52hf.jpg

thanks

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3 Answers 3

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If you know that $$ \int_0^\infty e^{-\alpha x^2} \: dx=\frac12\sqrt{\frac{\pi}{\alpha}}, \qquad \alpha>0. $$ Then you may deduce, using an integration by parts, that

$$ \begin{align} I_2(\alpha):= \int_0^\infty x^2e^{-\alpha x^2} \: dx&= \left.\left(-\frac{1}{2\alpha}\right)x \times\left( e^{-\alpha x^2}\right)\right|_0^{\infty}+\frac{1}{2\alpha}\int_0^\infty e^{-\alpha x^2} \: dx=\frac{1}{4\alpha}\sqrt{\frac{\pi}{\alpha}}. \end{align} $$

The other integral $$ I_1(\alpha):=\int_0^\infty xe^{-\alpha x^2} \: dx $$ is easily deduced by a change of variable, setting $u=x^2$, $du=2 xdx$, you get

$$ \begin{align} I_1(\alpha)= \int_0^\infty xe^{-\alpha x^2} \: dx&=\frac12\int_0^\infty e^{-\alpha u} \: du=\frac12\left.\left(-\frac{1}{\alpha}\right) \times\left( e^{-\alpha u}\right)\right|_0^{\infty}=\frac{1}{2\alpha}. \end{align} $$

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  • $\begingroup$ Thanks you very much,your calculus level is extremely high,haha $\endgroup$ Jul 11, 2015 at 14:04
  • $\begingroup$ @RichardTsai You are welcome! $\endgroup$ Jul 11, 2015 at 14:12
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you might generate arbritary even powers of $x$ by just noticing that

$$ \int_{0}^{\infty}x^{2n}e^{-\alpha x^2}=(-1)^n\partial^n_{\alpha}\int_{0}^{\infty}e^{-\alpha x^2}=(-1)^n \frac{\sqrt{\pi}}{2}\partial^n_{\alpha}\sqrt{\frac{1}{\alpha}} $$

But it is not possible to generate odd powers out of the knowledge that $\int_{0}^{\infty}e^{-\alpha x^2}=\frac{1}{2}\sqrt{\frac{\pi}{\alpha}}$!

You need another starting point like

$$ \int_{0}^{\infty}xe^{-\alpha x^2}=\frac{-1}{2\alpha}\int_{0}^{\infty}\partial_{x}e^{-\alpha x^2}=\frac{-1}{2\alpha}\left[e^{-\alpha x^2}\right]_{0}^{\infty}=\frac{1}{2\alpha} $$

The you might procced as above

$$ \int_{0}^{\infty}x^{2n+1}e^{-\alpha x^2}=(-1)^n\partial^n_{\alpha}\int_{0}^{\infty}xe^{-\alpha x^2}=(-1)^n\frac{1}{2}\partial^n_{\alpha}\frac{1}{\alpha} $$

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Check that if you let say $u=ax^2$ then $dx=du/(2ax)$ but $x=\sqrt{(u/a)}$ substituting in the integral and simplify it reduces to $(1/(2a^{1.5})) \gamma(3/2) $and further reduces to $(1/4)(1/a^{1.5})\sqrt{\pi}$. Or check integrals involving gamma functions.

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