Laplaces Method not in standard form

Question

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?

Answer 1

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

$$\bbox[5px,border:2px solid #C0A000]{I(x)\sim -\sqrt{x}\sqrt{\frac{2\pi}{\sqrt{x}|f''(-1)|}}e^{xf(-1)}=-\sqrt{x}\sqrt{\frac{\pi}{\sqrt{x}}}e^{-2\sqrt{x}}=-\sqrt{\pi}x^{1/4}e^{-2\sqrt{x}}}$$

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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.

Answer 2

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.