Laplaces Method not in standard form I am trying to figure out the following integral to leading order
$$I(x)=\int_0^{-\infty}dk\, \frac{e^{kx}e^{1/k}}{k^2}$$
I have thought about 2 different methods.  The first is to consider $h(k)= e^{1/k}/k^2,g(k)=k$ So that $I(x)=\int dk\,h(k)e^{xg(k)}$ However g(k) is not twice differentiable, so that is no good.
The second thought I had was to just re express this as
$$\int dk\,\frac{\exp[\frac{k^2x+1}{k}]}{k^2}$$
However, I do not see how to write this as $\exp(g(k)*x)$ and additionally, the maximum point occurs at $k=\sqrt x$ Which is moving.
Is there a simple way I can reexpress the initial integrand or can I proceed with either of the methods I thought of?
 A: Following the suggestions from @tired, we first let $k\to k/\sqrt{x}$.  Then, the integral of interest becomes 
$$\begin{align}
I(x)&=\int_0^{-\infty}\frac{e^{kx+1/k}}{k^2}\,dk\\\\
&=\sqrt{x}\int_0^{-\infty}\frac{e^{\sqrt{x}\left(k+\frac1k\right)}}{k^2}\,dk \tag 1
\end{align}$$
Next, we enforce the substitution $k\to 1/k$ in $(1)$ to reveal 
$$I(x)=\sqrt{x}\int_0^{-\infty}e^{\sqrt{x}\left(k+\frac1k\right)}\,dk \tag 2$$
Now, we directly apply Laplace's method to $(2)$.  To that end, we let $f(k)=k+\frac1k$.  Then, $f'(k)=1-\frac1{k^2}$ and $f''(k)=2/k^3$.  
Then $k=-1$, we have $f(-1)=-2$, $f'(-1)=0$ and $f''(-1)=-2$.  
Therefore, the integral of interest becomes
$$\bbox[5px,border:2px solid #C0A000]{I(x)\sim -\sqrt{x}\sqrt{\frac{2\pi}{\sqrt{x}|f''(-1)|}}e^{xf(-1)}=-\sqrt{x}\sqrt{\frac{\pi}{\sqrt{x}}}e^{-2\sqrt{x}}=-\sqrt{\pi}x^{1/4}e^{-2\sqrt{x}}}$$

As an alternative approach, we note that
$$I''(x)-\frac1x I(x)=0$$
Then, $I(x)=A\sqrt{x}K_1(2\sqrt{x})$.  Note that the coefficient on the modified Bessel function $I_1(2\sqrt{x})$ is zero since $I_1$ exhibits exponential growth as $x\to \infty$ whereas the integral $I(x)$ goes to zero as $x\to \infty$.
Using $I(0)=-1$ along with the small argument approximation for $\sqrt{x}K_1(2\sqrt{x})\sim \frac12$ reveals that $A=-2$ and therefore
$$I(x)=-2\sqrt{x}K_1(2\sqrt{x})$$
and we can apply the large argument asymptotic expansion for the modified Bessel function as reported by @jackd'aurizio.
A: Assuming $x>0$,
$$J(x)=\int_{0}^{+\infty}\exp\left(-kx-\frac{1}{k}\right)\frac{dk}{k^2} = 2\sqrt{x}\cdot K_1(2\sqrt{x})$$
where $K_1$ is a modified Bessel function of the second kind. For moderately large values of $x$,
$$ J(x) \approx e^{-2\sqrt{x}}\sqrt{\pi}\left( x^{\frac{1}{4}}+\frac{3}{16}\,x^{-\frac{1}{4}}\right) $$
where the previous asymptotics comes from an application of Laplace's method.
