Expectation of $E[e^{-X^2}]$ of a Gaussian distribution

Question

Given Gaussian distribution $X \sim N(0, t)$, what is $E\left[e^{-X^2}\right]$?

I used the Taylor expansion and Moment Generating Function to get this far:

$E(e^{-X^2}) = 1 - E(X^2) + \frac{E(X^4)}{2} - \frac{E(X^6)}{3!} + \ldots$

$=1 - t + \frac{3t^2}{2} - \frac{5t^3}{6} + \frac{7t^4}{24} + \ldots$

At this point I have two questions:

1. Can I claim that $E(X^{2k}) = (2k-1)t^k$ for all $k \in \{1, 2, 3, \ldots\}$?
2. If so, how might I simplify this series?

Answer 1

Write it as an integral,$$\int_{\Bbb R}\frac{1}{\sqrt{2\pi t}}\exp\left[-\frac{(1+2t)x^2}{2t}\right]dx=(1+2t)^{-1/2}.$$

Answer 2

Hint :

Define $Y = \frac{X^2}{t} $

$Y \sim {\chi^2}_{(1)}$

Now use it’s MGF to compute the expectation.

$E(e^{-X^2}) = E(e^{-tY}) = M_Y(-t) = (1 + 2t)^{\frac{-1}{2}}$

PS: The MGF exists since $t>0$ (as it’s the variance)