So firstly Watson's lemma states that for $\phi(t)=t^\lambda g(t)$, where $g(t)$ has Taylor series $g(t)=\sum^{\infty}_{n=0}\frac{t^n}{n!}\frac{d^ng}{dt^n}(0)$ about $t=0$, with $g(0)\neq 0$, and $\lambda$ is a real constant such that $\lambda > -1$, then as $x \rightarrow \infty$,
$$f(x) = \int^{\infty}_{0}t^\lambda g(t)e^{-xt}dt \sim \sum^{\infty}_{n=0}\frac{\Gamma(\lambda+n+1)}{n!x^{\lambda+n+1}}\frac{d^ng}{dt^n}(0).$$
Now I am trying to use Watson's lemma to show the relation below holds:
$$\int^{\infty}_{0}\{1+\sin(t^2)\}e^{-xt}dt \sim \frac{1}{x}+\frac{2}{x^3}-\frac{120}{x^7}+\mathcal{O}(\frac{1}{x^{11}}).$$
Unfortunately I am struggling a fair amount..
I can see that $1+\sin(t^2)= 1+t^2-\frac{t^6}{6}+\frac{t^{10}}{120}+...$ which my professor has advised is the right starting point.
Now I will take $\lambda = 0$ and $g(t)=1+\sin(t^2)$, which works as then $g(0) \neq 0$.
So then we find that $g(t)= 1 + \sum^{\infty}_{n=0}\frac{(-1)^n(t^2)^{2n+1}}{(2n+1)!}$.
Now this bit I am not so sure about but I think $\frac{d^ng}{dt^n}(0)=1 + \sum^{\infty}_{n=0}\frac{(-1)^n}{(2n+1)!}. $
So now what is left is for me to apply the formula:
$$f(x) \sim \sum^{\infty}_{n=0}\frac{\Gamma(4n+3)}{(2n+1)!} \centerdot1+ \frac{(-1)^n}{x^{2n+1}}.$$
At this point however I can see I am definitely wrong somewhere when comparing to the solutions I am looking for...however I'm really not sure where. Any help would really be appreciated, I've spent too much time on this question already but my curiosity will drive me crazy!
You need to note two things:
$\Gamma(n+1)/n!=1$
for $g^{(n)}$, the Taylor series of $g$ is $$1+\sum_{n=0}^\infty \frac{(-1)^n(4n+2)!}{(2n+1)!}\,\frac{t^{4n+2}}{(4n+2)!},$$ and now you can read $g^{(4n+2)}(0)$ from the series. That is, $g(0)=1$, $g^{(n)}(0)=0$ when $n$ is not of the form $4m+2$, and $$ g^{(4n+2)}(0)=\frac{(-1)^n(4n+2)!}{(2n+1)!}. $$ In particular, $$ g^{(6)}(0)=-\frac{6!}{3!}=-120. $$