Is it possible to find an upper bound on the moment generating function of $\sqrt{|X|}$, where $X\sim \mathcal{N}(0,1)$? So basically we need to find an upper bound on the integral $$\int_0^\infty \exp(f(x)) dx,$$ where $f(x) = \lambda \sqrt{x} - x^2/2$ with $\lambda>0$. I used the Laplace's method of approximation and ended up with the following result, $$\lim_{\lambda \to \infty}\frac{\int_0^\infty e^{f(x)} dx}{g(\lambda)}=1,$$ for some function $g$, exact form of which is not relevant here. However, I am not sure how to use this method further to produce an upper bound of the integral of the following form $$\int_0^\infty \exp(f(x)) dx\le h(\lambda),\ \lambda>0.$$ Can anyone give any relevant suggestions or probably some reference that can help me progress in this direction? Thanks in advance.
 A: We'll first consider the case when $\lambda > 0$.
Begin by writing
$$
\int_0^\infty \exp\!\left\{\lambda\sqrt{x}-x^2/2\right\}dx = \lambda^{2/3} \int_0^\infty \exp\!\left\{ \lambda^{4/3} \left(\sqrt{y}-y^2/2\right) \right\}dy.
$$
Near the maximum at $y = 2^{-2/3}$ we have
$$
\sqrt{y} - \frac{1}{2}y^2 = 3 \cdot 2^{-7/3} - \frac{3}{4}\left( y - 2^{-2/3} \right)^2 + O\left(\left( y - 2^{-2/3} \right)^3\right).
$$
Calculus (check the second derivative) shows that
$$
\sqrt{y} - \frac{1}{2}y^2 \leq 3 \cdot 2^{-7/3} - \frac{1}{2}\left( y - 2^{-2/3} \right)^2
$$
for all $y > 0$, and so
$$
\begin{align}
\int_0^\infty \exp\!\left\{\lambda\sqrt{x}-x^2/2\right\}dx &< \lambda^{2/3} \int_0^\infty \exp\!\left\{ \lambda^{4/3} \left[ 3 \cdot 2^{-7/3} - \frac{1}{2}\left( y - 2^{-2/3} \right)^2 \right] \right\}dy \\
&< \lambda^{2/3} \int_{-\infty}^\infty \exp\!\left\{ \lambda^{4/3} \left[ 3 \cdot 2^{-7/3} - \frac{1}{2}\left( y - 2^{-2/3} \right)^2 \right] \right\}dy \\
&= \sqrt{2\pi} \exp\!\left\{ 3 \cdot 2^{-7/3} \lambda^{4/3} \right\}.
\end{align}
$$
This upper bound agrees with the asymptotic of the integral up to the constant factor $\sqrt{2\pi}$. The correct asymptotic constant is $2\sqrt{\pi/3}$.
Now we'll suppose $\lambda < 0$.
In this case
$$
\int_0^\infty \exp\!\left\{\lambda\sqrt{x}-x^2/2\right\}dx < \int_0^\infty \exp\!\left\{\lambda\sqrt{x}\right\}dx = \frac{2}{\lambda^2},
$$
and this bound is asymptotically tight in the sense that
$$
\int_0^\infty \exp\!\left\{\lambda\sqrt{x}-x^2/2\right\}dx \sim \frac{2}{\lambda^2} \qquad \text{as } \lambda \to -\infty
$$
by Watson's lemma.
A: Since this is about distributions, let's estimate the function
$$M(\lambda)=\sqrt{\frac2{\pi}}\int^\infty_0\exp(\lambda\sqrt{x}-\frac{x^2}2)\,dx.\tag1$$ In the spirit of Laplace's method, we're interested in the maximum of the integrand, which is attained at $$x_0=\mu(\lambda)=\left(\dfrac{\lambda}2\right)^{2/3}\tag2$$ 
But we won't approximate the exponent by a second degree polynomial, we'll estimate it from above. No polynomial with negative curvature will stay above $\sqrt{x}$, so we'll have to take the tangent ($\sqrt{x}$ is concave), and we take the tangent at $x=\mu$, of course, to be as accurate as possible near the maximum. Taking into account (2), we get $$\lambda\sqrt{x}-\frac{x^2}2\le\mu x+\mu^2-\frac{x^2}2=\frac32\mu^2-\frac{(x-\mu)^2}2,$$ so the final result will be $$M(\lambda)\le2\Phi(\mu)\,\exp\left(\frac32\mu^2\right),$$ where $\Phi$ is the cdf of the standard normal distribution, and $\mu$ is defined by (2). Since we had to ignore the non-linearity of $\sqrt{x}$, this will not be asymptotically optimal, but an overestimate by a factor of $\sqrt{7}/2<1.33$, I think. Of course, $\Phi(\mu)$ is almost $1$ for large $\lambda$.
A: Because we have $$\lim_{\lambda \to \infty}\frac{\int_0^\infty e^{f(x)} dx}{g(\lambda)}=1$$ there is by definition of the limit an $N \in \Bbb N$ s.t. for all $\lambda \ge N$ we have $$\frac{\int_0^\infty e^{f(x)} dx}{g(\lambda)} \le 2$$ what gives us 
$$\int_0^\infty e^{f(x)} dx \le 2g(\lambda), \lambda \ge N$$
Additionally $$\int_0^\infty e^{f(x)} dx$$ is monotonous increasing in $\lambda$ and so $$\int_0^\infty e^{f(x)} dx \le 2g(N), \lambda < N$$  so we if we choose
$$h(\lambda) = \begin{cases} 2g(N), & \lambda < N \\ 2g(\lambda), &\lambda \ge N\end{cases}$$ we are done.
