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Evaluate the following integral: $$\frac{1}{a}\int_0^\infty{x^2}e^{-\large\frac{x^2}{2a}}\,dx.$$

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Hint: $~k=\dfrac1{2a}\quad=>\quad I(k)~=~\displaystyle\int_0^\infty e^{-kx^2}~dx\qquad=>\qquad I'(k)~=~-\int_0^\infty x^2e^{-kx^2}~dx$.

But the value of $~I(k)~$ can easily be computed using that of the Gaussian integral, with the help of one simple substitution. Then all that's left is to differentiate the result with respect to k, and replace it with its expression in terms of a.

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$$\frac{1}{a}\int_{0}^{\infty}x^{2}e^{-\frac{x^{2}}{2a}}dx=-\int_{0}^{\infty}x(e^{-\frac{x^{2}}{2a}})'dx$$

Use integration by parts and the value of Gaussian integral to evaluate:

$$=-(xe^{-\frac{x^{2}}{2a}})\vert_{0}^{\infty}+\int_{0}^{\infty}e^{-\frac{x^{2}}{2a}}dx=\sqrt{2a}\int_{0}^{\infty}e^{-(\frac{x}{\sqrt{2a}})^{2}}\frac{1}{\sqrt{2a}}dx=\sqrt{2a}\cdot\frac{\sqrt{\pi}}{2}$$

Alternatively notice that:

$$-\frac{1}{2}\sqrt{\frac{a\pi}{2}}=\frac{d}{db}(\sqrt{\frac{2a}{b}}\cdot\frac{\sqrt{\pi}}{2})\vert_{b=1}=\frac{d}{db}(\int_{0}^{\infty}e^{-\frac{bx^{2}}{2a}}dx)\vert_{b=1}=-\frac{1}{2a}\int_{0}^{\infty}x^{2}e^{-\frac{x^{2}}{2a}}dx$$

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  • $\begingroup$ please elaborate your answer , I cant understand it. I try to do by part but cant proceed $\endgroup$ – Zur Aug 9 '14 at 6:13
  • $\begingroup$ $\int_{0}^{\infty}x(e^{-\frac{x^{2}}{2a}})'dx=xe^{-\frac{x^{2}}{2a}}\vert_{0}^{ \infty}-\int_{0}^{\infty}e^{-\frac{x^{2}}{2a}}dx=-\sqrt{2a}\cdot\frac{\sqrt{\pi}}{2}$ $\endgroup$ – user71352 Aug 9 '14 at 6:20
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Without using the gamma function. With the same method usally used to compute the Gauss integral, but without using the Gauss integral itself :

enter image description here

(A typing mistake corrected)

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