# Applying Watson's lemma $\int^{\infty}_{0}\{1+\sin(t^2)\}e^{-xt}dt$

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 have conflated the order of the derivative $n$ with the dummy summation index $n$. – Mark Viola Apr 21 '18 at 2:36

• $\Gamma(n+1)/n!=1$
• 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, $$g^{(6)}(0)=-\frac{6!}{3!}=-120.$$
• Perhaps this is a silly question but how did you arrive at that form for the Taylor series? Also for me i would cancel out the $(4n+2)!$ terms and continue from there. However clearly they are essential! My approach to finding the taylor series here was to first consider the series for $\sin(x)$ and then replace $t$ by $t^2$. Ofcourse this gives me the simplified version but how would I know I needed to introduce the $(4n+2)!$? – Evan Apr 21 '18 at 11:10
• What you did is perfect. But to read the derivatives you need to write the series as $g(t)=\sum_n a_n\,\frac{t^n}{n!}$; that's when $g^{(n)}(0)=a_n$. You need the factorial to be the same as the power of $t$. – Martin Argerami Apr 21 '18 at 15:18
• I feel like I am very close.. my final conclusion was that it should look like $\sum^{\infty}_{n=0} \frac{(-1)^n(4n+2)!}{x^{4n+3}(2n+1)!}$. So this does give me the correct values, but it does not give me the initial $\frac{1}{x}$.. is there a reason for this or is my solution incorrect? – Evan Apr 21 '18 at 16:02
• I see in the given formula $x$ should look like $x^{\lambda+n+1}$, since $n$ is the dummy variable it should correspond with the power of $t$ so I turn this in to $4n+2$. But then $x^{4n+3} \neq x$ for any value of $n \in [0,\infty)$.. – Evan Apr 21 '18 at 16:05
• I just told you. From your Watson's Lemma you have that the first term is $$\frac{(-1)^0\Gamma(0+1)}{0!\,x^{0+1}}\,g(0)=\frac1x.$$ – Martin Argerami Apr 21 '18 at 16:22