asymptoticsintegrationlaplace-method
I'm still having a little trouble applying Laplace's method to find the leading asymptotic behavior of an integral. Could someone help me understand this? How about with an example, like:
$$\int_0^{\infty} t^{3/4}e^{-x(t^2+2t^4)}dt$$ for $x>0$, as $x\rightarrow\infty$.
Plot integrand for few values of $x$:
It is apparent that the maximum shifts closer to the origin as $x$ grows.
Let's rewrite the integrand as follows:
$$
\int_0^{\pi/2} \sqrt{\sin(t)} \exp\left(-x \sin^4(t)\right) \mathrm{d}t = \int_0^{\pi/2} \exp\left(\frac{1}{2} \log(\sin(t))-x \sin^4(t)\right) \mathrm{d}t
$$
The maximum of the integrand is determined by
$$
0 = \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{1}{2} \log(\sin(t))-x \sin^4(t)\right) = \cot(t) \left( \frac{1}{2} - 4 x \sin^4(t)\right)
$$
that is at $t_\ast = \arcsin\left((8 x)^{-1/4}\right)$. Then using Laplace's method:
$$
\int_0^{\pi/2} \sqrt{\sin(t)} \exp\left(-x \sin^4(t)\right) \mathrm{d}t \approx \int_{0}^{\pi/2} \exp\left(\phi(t_\ast) + \frac{1}{2} \phi^{\prime\prime}(t_\ast) (t-t_\ast)^2 \right) \mathrm{d}t = \exp\left(\phi(t_\ast)\right) \sqrt{\frac{2\pi}{-\phi^{\prime\prime}(t_\ast)}}
$$
Easy algebra gives $\exp\left(\phi(t_\ast)\right) = (8 \mathrm{e} x)^{-1/8}$, $-\phi^{\prime\prime}(t_\ast) = 4 \sqrt{2 x} - 2$, giving
$$
\int_0^{\pi/2} \sqrt{\sin(t)} \exp\left(-x \sin^4(t)\right) \mathrm{d}t \approx (8 \mathrm{e} x)^{-1/8} \sqrt{ \frac{\pi}{2 \sqrt{2 x} -1}}
$$
In the present case, there is a shortcut to Watson's lemma (mentioned in the comments), which is to scale properly the variable of integration of the integral $I(x)$ to be evaluated.
Let $x=1/z^2$ with $z\gt0$, hence $z\to0^+$. Using the change of variable $t\to z^2t$, one gets $$ I(x)=z^2\int_0^{+\infty}\mathrm e^{-t}\log(1+z\sqrt{t})\,\mathrm dt. $$ For every $N\geqslant0$, an expansion of $\log(1+s)$ up to order $s^N$ when $s\to0$ is $$ \log(1+s)=\sum_{n=1}^N(-1)^{n-1}\frac1ns^n+o(s^{N}). $$ This yields $$ I(x)=z^2\sum_{n=1}^N(-1)^{n-1}I_nz^n+o(z^{N+2})=\sum_{n=1}^N(-1)^{n-1}I_nx^{-1-n/2}+o(x^{-1-N/2}), $$ where, for every $n\geqslant1$, $$ I_n=\frac1n\int_0^{+\infty}\mathrm e^{-t}t^{n/2}\,\mathrm dt=\frac1n\Gamma\left(\frac{n}2+1\right)=\frac12\Gamma\left(\frac{n}2\right). $$ Finally, for every $N\geqslant0$, $$ I(x)=\sum_{n=1}^N(-1)^{n-1}\frac12\Gamma\left(\frac{n}2\right)x^{-1-n/2}+o(x^{-1-N/2}). $$ Note that the radius of convergence of the series $\sum\limits_n\Gamma\left(\frac{n}2\right)t^n$ being zero, the formulas above are indeed asymptotic expansions.
Best Answer
The basic idea is that the maximum contribution of the integral comes from a neighborhood of $t=0$, and near there we have $t^2+2t^4 \approx t^2$. This problem is particularly nice because we can do everything explicitly. I'll do the calculation in two steps (two changes of variables) to illustrate what's going on.
Start with the change of variables $t^2+2t^4 = s^2$, where $s \geq 0$. This gives
$$ t=\frac{1}{2}\sqrt{-1+\sqrt{1+8s^2}}, $$
so that the integral becomes
$$ \int_0^{\infty} t^{3/4}e^{-x(t^2+2t^4)}\,dt = \int_0^\infty s^{3/4} f(s) e^{-xs^2}\,ds, $$
where
$$ f(s) = \frac{2^{1/4}s^{1/4}}{\sqrt{1+8s^2}\left(-1+\sqrt{1+8s^2}\right)^{1/8}} = 1 - \frac{15}{4}s^2 + \frac{713}{32}s^4 + \cdots. $$
Now we can make the second change of variables $s^2 = r$ to put the integral into a form where we can directly apply Watson's lemma. Indeed, this gives
$$ \int_0^\infty s^{3/4} f(s) e^{-xs^2}\,ds = \int_0^\infty r^{-1/8} g(r) e^{-xr}\,dr, $$
where
$$ g(r) = \frac{1}{2}f\left(\sqrt{r}\right) = \frac{1}{2} - \frac{15}{8}r + \frac{713}{64}r^2 + \cdots. $$
Finally
$$ \begin{align*} \int_0^\infty r^{-1/8} g(r) e^{-xr}\,dr &\approx \sum_{n=0}^{\infty} \frac{g^{(n)}(0) \Gamma(n+7/8)}{n! x^{n+7/8}} \\ &= \frac{1}{2}\Gamma\left(\frac{7}{8}\right) x^{-7/8} - \frac{15}{8}\Gamma\left(\frac{15}{8}\right)x^{-15/8} + O\left(x^{-23/8}\right) \end{align*} $$
as $x \to \infty$, by Watson's lemma.