If $X$ is normally distributed with mean $0$ and SD $1$, show that $$E(e^{sX^2}) = \dfrac{1}{\sqrt{1-2s}}$$ for $s < \dfrac{1}{2}$.
I obtain this from the paper 'Elementary proof of Johnson and Lindenstrauss',under the proof of Lemma $2.2(a)$.
I don't know how to obtain the equality above.
One way to go is to use LOTUS and solve this integral
$$E(e^{sX^2}) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{-x^2(\frac{1}{2} -s)}dx$$
We will try squaring both sides and solve using polar coordinates.
$$(E(e^{sX^2}))^2 = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(\frac{1}{2} - s)x^2} e^{-(\frac{1}{2} - s)y^2} dx \> dy $$
$$ = \frac{1}{2 \pi} \int_0^{2 \pi} \int_0^{\infty} e^{-(\frac{1}{2} - s)r^2} r \> dr \> d\theta$$
We let $u = (\frac{1}{2} - s)r^2$ and thus $\frac{du}{1 - 2s} = r dr$ .
$$ = \frac{1}{2 \pi (1 - 2s)} \int_0^{2 \pi} \int_0^{\infty} e^{-u} \> du \> d\theta$$
$$= \frac{1}{1- 2s}$$
Thus our initial expected value must equal
$$E(e^{sX^2}) = \frac{1}{\sqrt{1-2s}}$$
For reasons having to do with the convergence of the integral we must stipulate that $s < \frac{1}{2}$.