In another question here on the site I was looking for an integral bound and it turned out one has to know how to deal with oscillatory integrals of the form $$\int_0^{1} \sin^2(\pi x) f(x) dx,$$ where $f$ is a smooth function. There should be some trick involving Fourier-analysis and the Riemann-Lebesgue Lemma, but I've never heard of this and a Google search didn't help much.
Can anyone show me what the general method is in solving such integrals?
Thanks!
$$\int_{0}^{1}\sin^2(\pi x)\,f(x)\,dx = \frac{1}{2}\int_{0}^{1}f(x)\,dx-\frac{1}{2}\int_{0}^{1}\cos(2\pi x)f(x)\,dx $$ and if $f$ is a $C^1$ function we have $$ \frac{1}{2}\int_{0}^{1}\cos(2\pi x)\,f(x)\,dx = -\frac{1}{4\pi}\int_{0}^{1}\sin(2\pi x)\,f'(x)\,dx $$ by integration by parts. If $f(x)$ behaves like $\frac{1}{x+k}$, this process greatly improves the convergence and allows to estimate $\int_{0}^{1}\sin^2(\pi x)\,f(x)\,dx$ with an arbitrary precision.
For short: separate the mean value from the mean-zero part and apply integration by parts to the mean-zero part.