= Solution
<Watson lemma> states that if
$$
f(t)\sim\sum_{n=0}^{\infty}a_nt^{\lambda_n-1}
\quad(t\downarrow0),
\qquad
0<\lambda_0<\lambda_1<\cdots,
$$
and the <Laplace integral> has suitable growth control away from zero, then
$$
\int_0^b e^{-xt}f(t)\,dt
\sim\sum_{n=0}^{\infty}
a_n\Gamma(\lambda_n)x^{-\lambda_n}
\quad(x\to+\infty).
$$
The <Taylor series> of the amplitude is
$$
\sin(t^2)
=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}t^{4k+2}.
$$
Termwise application of Watson lemma therefore gives the <asymptotic expansion>
$$
\boxed{
I(x)\sim
\sum_{k=0}^{\infty}
\frac{(-1)^k\Gamma(4k+3)}{(2k+1)!x^{4k+3}}
=\frac2{x^3}-\frac{120}{x^7}
+\frac{30240}{x^{11}}-\cdots.}
$$
In particular, the leading asymptotic approximation is $I(x)\sim2x^{-3}$.
Back to article page
