I am struggling with this problem where we're asked to use Laplace's method:
Find the leading term of the asymptotics of the following intergral for $\lambda\to\infty$:
$$\int_0^{\infty} \exp\left(-\lambda x-\frac{4}{x^2}\right)dx.$$
I'm not particularly sure how to approach this one. A walk through would be very helpful.
Let $x=\lambda^{-1/3} y$. Then the integral is
$$\lambda^{-1/3} \int_0^{\infty} dy \, \exp{\left [-\lambda^{2/3} \left (y+\frac{4}{y^2} \right )\right ]} $$
By rescaling, we now may use Laplace's method, i.e., find the min of the exponential, Taylor expand about that min and evaluate the resulting integral.
The min is at $1 - 8/y^3 = 0 \implies y=2$. The integral is then, about this min:
$$\lambda^{-1/3} \int_0^{\infty} dy \, \exp{\left [-\lambda^{2/3} \left ((y-2)+2+\frac{1}{(1+(y-2)/2)^2} \right )\right ]} \\ = \lambda^{-1/3} \int_0^{\infty} dy \, \exp{\left [-\lambda^{2/3} \left ((y-2)+2+1-(y-2)+\frac{3}{4} (y-2)^2 + O[(y-2)^3] \right )\right ]}\\ = \lambda^{-1/3} e^{-3 \lambda^{2/3}} \int_0^{\infty} dy \, \exp{\left [-\frac{3}{4} \lambda^{2/3}(y-2)^2\right ] } \left (1+O[(y-2)^3] \right )$$
Keep in mind that when $\lambda^{2/3} (y-2)^2 < \epsilon$, then $|y-2| \lt \lambda^{-1/3} \sqrt{\epsilon}$. Thus, with exponential error, we may simply integrate over the entire real line to get the asymptotic term sought:
$$\begin{align}\int_0^{\infty} dx \, \exp{\left (-\lambda x - \frac{4}{x^2} \right )} &= \lambda^{-1/3} e^{-3 \lambda^{2/3}} \int_{-\infty}^{\infty} dy \, \exp{\left (-\frac{3}{4} \lambda^{2/3}y^2\right ) } \left (1+O[y^3] \right )\\ &= e^{-3 \lambda^{2/3}}\left [ 2 \sqrt{\frac{\pi}{3}} \lambda^{-2/3} + O\left (\lambda^{-5/3} \right ) \right ] \end{align}$$