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Show that $ {1 \over {a \sqrt{2\pi}}}\int_{-\infty}^{\infty}{x^n}e^{-{{x^2}\over{2a^2}}}dx = 1.3.5....(n-1)a^n$ for even n, and 0 for odd n.
It is specifically asked to use the standard mathematical table.
But I can't find any equation in the following standard table.
Please help me how to prove the above.

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2 Answers

up vote 3 down vote accepted

You will find the integral for $n=0$ in the table. Afterwards, substitute $\mu=2/a^2$ and differentiate a sufficient amount of time with respect to $\mu$, i.e, use $$ \frac{d^m}{d\mu^m}\int e^{-\mu x^2} = (-1)^m\int x^{2m} e^{-\mu x^2} dx $$

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We must assume, of course, that $a > 0$. Use an exponential generating function:

$$ \eqalign{\sum_{j=0}^\infty \frac{t^{j} }{\sqrt{2\pi} a j!} \int_{-\infty}^\infty x^{j} \; e^{-x^2/(2 a^2)}\ dx &= \frac{1}{\sqrt{2\pi} a} \int_{-\infty}^{\infty} \sum_{j=0}^\infty \frac{t^j x^{j}}{ j!} e^{-x^2/(2a^2)}\ dx\cr &= \frac{1}{\sqrt{2\pi} a} \int_{-\infty}^{\infty} e^{tx} e^{-x^2/(2a^2)}\ dx\cr &= \frac{1}{\sqrt{2\pi} a} e^{a^2 t^2/2} \int_{-\infty}^{\infty} e^{-(x/a - at)^2/2}\ dx \cr \text{(using the substitution $s=x - a^2 t$)} &= \frac{1}{\sqrt{2\pi} a} e^{a^2 t^2/2} \int_{-\infty}^{\infty} e^{-s^2/(2a^2)}\ ds\cr &= e^{a^2 t^2/2} = \sum_{k=0}^\infty \frac{a^{2k} t^{2k}}{2^k k!}}$$

Now match up terms in each power of $t$ on both sides.

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