Asymptotic behavior of $\mathbb{E}\left[\exp\left(-|X|^\nu\right)\right]$ wrt the mean of the normal rv $X$ Here is a statement that I would like to prove:

Let $X \sim \mathcal{N}(\mu, 1)$. Let $\nu >0$, show that $$
-\log \mathbb{E}\left[\exp\left(-|X|^\nu\right)\right] \quad\underset{\mu \to +\infty}{\sim}\quad \mu^\nu $$
  where $f(x) \sim g(x)$ means $f(x) = g(x) + o(g(x))$.

I have checked numerically that it is always verified whatever $\nu>0$, but I cannot prove it.
Attempt of proof:
Here is what I have tried:
\begin{align}
\mathbb{E}\left[\exp\left(-|X|^\nu\right)\right]
=
\sum_{k=0}^\infty \frac{(-1)^k}{k!}\mathbb{E}\left[|X|^{\nu k}\right]
\end{align}
We have, for any $p>0$
[see here]
$$
\mathbb{E}\left[|X|^{p}\right]
=
\frac{2^{\frac{p}{2}} \Gamma[\frac{1}2 +\frac{p}{2}]}{\sqrt{\pi}}
M\left(-\frac{p}2, \frac12, -\frac{\mu^2}2\right)
$$
where M is the Kummer function.
And we have [see here]
$$
M\left(-\frac{p}2, \frac12, -\frac{\mu^2}2\right)
\quad\underset{|\mu| \to \infty}{\sim}\quad
\frac{\Gamma(1/2) \left(\frac{\mu^2}2\right)^{\frac{p}2}}{\Gamma(\frac12 + \frac{p}2)}
$$
Pluging these two together give
$$
\mathbb{E}\left[|X|^{p}\right]
\quad\underset{\mu \to +\infty}{\sim}\quad
\mu^p
$$
Now I would like to conclude that
\begin{align}
\mathbb{E}\left[\exp\left(-|X|^\nu\right)\right]
&\quad\underset{\mu \to +\infty}{\sim}\quad
\sum_{k=0}^\infty \frac{(-1)^k}{k!}\mu^{k \nu}
=
\exp\left(-\mu^{\nu}\right)
\end{align}
but I cannot because I have no control on the error term with respect to $k$ (in order to use dominated or monotone convergence arguments). Maybe this reference could help: here.
Note 1: I have also tried using the Delta method, but I did not succeed.
Note 2: In fact the statement that I need is a bit weaker:
$$
\log\left[C_\nu-\log \mathbb{E}\left[\exp\left(-|X|^\nu\right)\right] \right]\quad\underset{\mu \to \infty}{\sim}\quad \nu \log \mu
$$
where $C_\nu$ is a constant ensuring the quantity inside the outer log to be positive when $|\mu|>0$.
 A: Here is an argument for $\leq$: 
Fix $v>0$. Define $f:[0,\infty)\rightarrow\mathbb{R}$ by $f(y) = \exp(-y^v)$.  Then: 
\begin{align}
f'(y) &= -vy^{v-1} \exp(-y^v)\\
f''(y) &= [(vy^{v-1})^2  + -v(v-1)y^{v-2}]\exp(-y^v)
\end{align}
Notice that if $0 < v \leq 1$ then $f$ is a convex function over the domain $y \geq 0$. Further, if $v>1$, there is a threshold $\theta>0$ such that $f$ is convex over $[\theta, \infty)$. 
Case $0 < v \leq 1$:
By convexity of $f$ for this case we get by Jensen's inequality: 
$$ E[\exp(-|X|^v)] = E[f(|X|)] \geq f(E[|X|]) = \exp(-E[|X|]^v) $$
Taking $-\log()$ of both sides gives: 
$$ \boxed{-\log(E[\exp(-|X|^v)]) \leq E[|X|]^v} $$ 
In fact this holds for any random variable $X$ provided that $E[|X|]$ is finite. Notice that if $X$ is $N(\mu,1)$ and $\mu$ is large then $E[|X|]\approx \mu$.  This is because: 
$$ |X| = X - 2X1\{X<0\} \implies E[|X|] = \mu - 2E[X1\{X<0\}] $$
and $E[X1\{X<0\}]$ is very small when $\mu\rightarrow\infty$. 
Case $v>1$:
Recall that $f(y)$ is convex over $y \in [\theta, \infty)$.  Define the event $A=\{|X|\geq \theta\}$. Then taking expectations conditioned on $A$ we get by a similar argument: 
$$ \boxed{-\log(E[\exp(-|X|^v) \:  |A]) \leq E[|X| \:  | A]^v} $$ 
Now when $\mu\rightarrow \infty$ we get $P[A]\rightarrow 1$ and also $E[|X|] \approx \mu$ for large $\mu$. 
A: Here is an approach to the lower bound (see my other answer for the upper bound). 
Fix $v>0$, $\mu>1$ and let $X$ be $N(\mu, 1)$.  Define the event $B_{\mu} = \{\mu - \sqrt{\mu} \leq X \leq \mu + \sqrt{\mu}\}$ and note that $\lim_{\mu\rightarrow\infty} P[B_{\mu}] =1$. Since $\mu>1$ we know $\mu - \sqrt{\mu}>0$. By Taylor's theorem we have, if $x \in [\mu - \sqrt{\mu}, \mu + \sqrt{\mu}]$: 
$$ \exp(-x^v) \leq \exp(-\mu^v) + (x-\mu)(-v\mu^{v-1}) \exp(-\mu^v) + \frac{(x-\mu)^2}{2}\exp(-\mu^v)h(\mu) $$
where $h(\mu)$ also depends on $v$ and I will not write it out since I'm lazy...
Hence, conditioning on $B_{\mu}$: 
$$ E[\exp(-X^v) | B_{\mu}] \leq \exp(-\mu^v) +  \underbrace{E[X-\mu | B_{\mu}]}_{=0}(-v\mu^{v-1})\exp(\mu^v) + \underbrace{E\left[\frac{(X-\mu)^2}{2} | B_{\mu} \right]}_{\leq 1/2}\exp(-\mu^v)h(\mu)$$
where the underbrace values hold because the distribution of $X$ is symmetric about $\mu$, and conditioning on being in an interval about $\mu$ reduces the variance. We get: 
$$ E[\exp(-X^v)|B_{\mu}] \leq \exp(-\mu^v)[ 1 + (1/2)h(\mu)] $$
Taking the $-\log(\cdot)$ of both sides gives
$$ \boxed{-\log(E[\exp(-X^v)|B_{\mu}])\geq \mu^v - \log(1 + (1/2)h(\mu))} $$
and if someone is less lazy than me, that person could likely show $\log(1+(1/2)h(\mu))$ is asymptotically negligible in comparison to $\mu^v$. 
Then note that for large $\mu$ we have $P[B_{\mu}]\approx 1$ and $X=|X|$ with high probability, leading to the desired result.
A: Below I have completed Michael's proof and raise an issue.
1. Recall that
$$
  f''(\mu) =
  \left[
    \mu^{\nu}
    + (1 - \nu)
    \right]
  \nu \mu^{\nu-2} e^{-\mu^\nu}
$$
I have chosen to focus at $x \in [\mu - \mu^\rho, \mu + \mu^\rho]$ for $0 < \rho <1$. And I have define
$$h(\mu) = 2 \nu (\mu - \mu^{\rho})^{2\nu-2} \exp(\mu^\nu - (\mu - \mu^{\rho})^{\nu})$$
such that for $\mu$ large enough, $\exp(-\mu^\nu)h(\mu) \geq f''(x)$ for all $x \in [\mu - \mu^\rho, \mu + \mu^\rho]$, and then the targeted inequality holds
$$
\exp(-x^v) \leq \exp(-\mu^v) + (x-\mu)(-v\mu^{v-1}) \exp(-\mu^v) + \frac{(x-\mu)^2}{2}\exp(-\mu^v)h(\mu)
$$
It remains to show that $\log(1+h(\mu)/2)$ is negligeable for all $\nu >0$ in comparison to of $\mu^\nu$. We have
$$
    \log(1 + h(\mu)/2)
    \sim
    \log(h(\mu)/2)
    =
    \log \nu + (2\nu-2)\log(\mu - \mu^{\rho})
    +
    \mu^\nu - (\mu - \mu^{\rho})^{\nu}
$$
And the result follows as
$$  (\mu - \mu^{\rho})^{\nu}
  =
  \mu^\nu (1-\mu^{\rho-1})^{\nu}
  =
  \mu^\nu (1+o(1))
  =
  \mu^\nu + o(\mu^\nu)
$$
Then following Michael's arguments, the result holds true for all $\nu >0$.
2. Nevertheless, I have an issue. For $\nu=2$, I know that $$
-\log \mathbb{E}\left[\exp\left(-|X|^2\right)\right] =
\frac13 \mu^2 + \frac12 \log 3 
$$
which contradicts the result because of the factor $1/3$ (convolution of a gaussian of variance 1, with a gaussian of variance 2). See with Maple
-log(int(1/sqrt(2*Pi)*exp(-x^2)*exp(-(x-mu)^2/2), x=-infinity..infinity));

My numerical simulations show that the asymptotical behavior is $\mu^\nu$ for $\nu < 2$, but nor for $\nu =2$. For $\nu > 2$, my simulations are too unstable to make any conclusion. Where is in the proof the reason to discard $\nu = 2$?
We can check with Maple that it works for $\nu=1$ (Maple cannot solve it otherwise)
> limit(-log(int(1/sqrt(2*Pi)*exp(-abs(x)^1)*exp(-(x-mu)^2/2), x=-infinity..infinity))/mu^1, mu=infinity);
1

> limit(-log(int(1/sqrt(2*Pi)*exp(-abs(x)^2)*exp(-(x-mu)^2/2), x=-infinity..infinity))/mu^2, mu=infinity);                                                                          
1/3

Thanks 
