I have the integral
$$I(x)=\int_0^{\pi/2}\exp(-xt^3\cos t)dt$$
and I want to derive the first two terms in the asymptotic expansion for $x\rightarrow \infty$, which should give me
$$\frac{1}{3x^{1/3}}\Gamma(1/3)+\left(\frac{1}{6}+\frac{8}{\pi^3} \right)\frac{1}{x}+\dots$$
Using the Laplace Method as stated here,
I get only one term so my idea was to use partial integration, which gives me $$\pi/2-\int_0^{\pi/2} \left[ \exp(-xt^3\cos t)\,\, t\,\,(xt^3\sin(t)-3xt^2\cos(t)) \right] dt$$
Best Answer
Well, there are two contributions because the exponent is zero at $t=0$ and $t=\pi/2$. Let's consider $t=0$ first. In the immediate neighborhood of $t=0$ (we'll get to what that means in a bit), the exponent behaves as $x t^3$ so that as $x\to\infty$, we have
$$I(x) \sim \int_0^{\infty} dt \, e^{-x t^3} = \frac{\Gamma\left ( \frac{4}{3}\right )}{x^{1/3}} = \frac{\Gamma\left ( \frac{1}{3}\right )}{3 x^{1/3}} \quad (x\to\infty)$$
What is the size of the neighborhood? Well, we want $0 \lt x t^3 \lt \epsilon$ for some small $\epsilon$, so we have $0 \lt t \lt (\epsilon/x)^{1/3}$.
We also have a contribution in a neighborhood near $t=\pi/2$; we Taylor expand and get that
$$t^3 \cos{t} = -\frac{\pi^3}{8} \left ( t-\frac{\pi}{2}\right ) + O\left [ \left ( t-\frac{\pi}{2}\right )^2\right ]$$
Really, we are only interested in positive values of this expansion, as the exponent is positive through the integration region. If we look at only an immediate neighborhood near $t=\pi/2$, but with $t \lt \pi/2$, then we may approximate the contribution to the integral there as
$$\int_0^{\infty} dy \, e^{-\pi^3 x y/8} = \frac{8}{\pi^3 x}$$
It doesn't make sense to simply add these two terms together and declare them the leading behavior of $I(x)$ until we investigate the next leading behavior of the contribution at $x=0$. Note that the next contribution in the exponent is $x t^5/2$; within the interval of interest, this is $O(x^{-2/3})$, so we may Taylor expand this exponential term separately. The result is the following integral for the next contribution at $t=0$:
$$\frac12 x\int_0^{\infty} dt \;t^5 \, e^{-x t^3} = \frac1{6 x}$$
Note that this is $O(1/x)$ as is the leading contribution from $t=\pi/2$, so we may add these. The stated result follows.