Use Watson's lemma to find the expansion of $$F(\lambda) = \int_{0}^{\infty} e^{- \lambda t} \sin(t)dt,$$ and verfify the answer using Laplace transformation by expanding the answer using Taylor's expansion.
Comparing the above integral with $$F(\lambda) = \int_{0}^{\infty} e^{- \lambda t} f(t)dt.$$
For applying Watson's lemma we need $f(0) \neq 0$, in our case it is $f(0)=0.$ So I integrated by parts before applying Watson's Lemma: $$F(\lambda) = \int_{0}^{\infty} \frac{1}{\lambda}e^{- \lambda t}\ \cos(t) \ dt \sim \frac{1}{\lambda} \sum_{n=0}^{\infty} \frac{f^{n}(0)}{n!} \frac{\Gamma(n+1)}{\lambda^{n+1}} \sim \frac{1}{\lambda} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{\lambda^{n+1}} $$
Using Laplace transform: $$F(\lambda) = \int_{0}^{\infty} e^{- \lambda t} \sin(t)dt = \frac{1}{1+ \lambda^{2}} = \frac{1}{\lambda^{2}} \sum_{n=0}^{\infty} \frac{(-1)^n}{\lambda^{2n}}$$
My Question
I am not able to figure out where I am making mistakes as I am getting different answers with two different methods. Can someone figure out my mistake?
Thanks for the help!