Problem Evaluate the following definite integral
$$\int_0^{\infty} \left(x – \frac{x^3}{2} + \frac{x^5}{2\cdot 4} – \frac{x^7}{2\cdot4\cdot 6} + \cdot \cdot \cdot \right) \left(1 + \frac{x^2}{2^2} + \frac{x^4}{2^2 \cdot 4^2 } +\frac{x^6}{2^2 \cdot 4^2 \cdot 6^2 } + \cdot \cdot \cdot \right) $$
I figured the first term in the expansion is just equal to
$$ xe^{-x^2 / 2} = x\left(1 – \frac{x^2}{2} + \frac{x^4}{(2\cdot 2)(2\cdot 1)} – \frac{x^6}{(3\cdot 2)(2\cdot 2)(2 \cdot 1)} + \cdot \cdot \cdot \right)$$
Thus the integral is reduced to
$$\int_0^{\infty} xe^{-x^2 / 2} \left(1 + \frac{x^2}{2^2} + \frac{x^4}{2^2 \cdot 4^2 } +\frac{x^6}{2^2 \cdot 4^2 \cdot 6^2 } + \cdot \cdot \cdot \right) $$
A similar expression for the second term does not seem plausible to me. Any hints would be appreciated.
Attempting to integrate term by term, we obtain the series.
$$ \sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{(t)^{n} e^{-t}}{2^n (n!)^2}$$
We may integrate each term in the series by parts, to obtain the following
$$ \sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{(t)^{n} e^{-t}}{2^n (n!)^2} = \sum_{n=0}^{\infty} \frac{-e^{-t}}{2^n (n!)} \sum_{i=0}^{n} \left(\frac{t^{n-i}}{(n-i)!} |_{0}^{\infty} \right) $$
However I do not know how to proceed, nor do I think this approach is correct since the problem was given in an high school exam.
Best Answer
We may integrate term by term. Call the integral in question $I$ and let $X\sim \mathcal N(0,1)$. It is an easy exercise in using Stein's Lemma to show $$E|X^{2p+1}| = \sqrt{\frac{2}{\pi}}(2p)!!$$ Thus $$\frac{I}{\sqrt{2\pi}} = \frac{1}{2}\sum_{p=0}^\infty \frac{1}{[(2p)!!]^2}E|X^{2p+1}| = \frac{1}{\sqrt{2\pi}}\sum_{p=0}^\infty \frac{1}{(2p)!!} = \frac{1}{\sqrt{2\pi}}\sum_{p=0}^\infty \frac{(1/2)^n}{n!} = \sqrt{\frac{e}{2\pi}}$$