I am considering the integral $\int_a^{\infty}f(t)\cos(\omega t)dt$ and I want to find the asymptotic expansion using the Riemann-Lebesgue Lemma where as $\omega\rightarrow \infty$, $a,\omega$ real $f(t)$ is $C^\infty$ and $f^{(s)}(t)\rightarrow 0$ as $t\rightarrow \infty$
I am litte bit confused here because I know that $\int_a^{b}f(t)e^{ikt}dt\sim\sum_{n=0}^{N}\frac{(-1)^n}{(ik)^{n+1}}(f^{(n)}(b)e^{ikb}-f^{(n)}(a)e^{ika})$ where we assume that $f(t)$ has $N+1$ contin. derivatives and $f^{(N+2)}$ is piecewise contin on $[a,b]$.
I also know that $e^{i\omega t}=\cos(\omega t)+i\sin(\omega t)$ but can I connect these things?
Intuitively, the integral of $\cos$ over one period is zero. In the case of large $ω$, $\cos(ωt)$ has the very small period $\frac{2\pi}ω$. Since $f$ is smooth, it can be considered almost constant over such a period, suggesting that the integral is zero.
To rule out the obvious, and make the intuitive statement exact, use the already discussed partial integration method, $$ \int_a^\infty f(t)\cos(ωt)dt=\left[f(t)\tfrac{\sin(ωt)}{ω}\right]_a^\infty-\int_a^\infty f'(t)\tfrac{\sin(ωt)}{ω}dt $$ which allows to write $$ \left|\int_a^\infty f(t)\cos(ωt)dt\right|\le\frac1ω\left(|f(a)|+\int_a^\infty |f'(t)|\right)dt $$ which allows to conclude the convergence to $0$ if $f'\in L^1$, or by repeating the partial integration, if any $f^{(n)}\in L^1$.