Limit of integral (involving sine integral)

Question

Consider $\phi \in C_c'(\mathbb{R})$. I want to prove that \begin{equation} \lim_{n\rightarrow \infty} \int_{\mathbb{R}}\frac{\sin(nx)}{x}\phi(x)dx=\pi \phi(0) \; . \end{equation} One could observe that for $M\in \mathbb{R}$ sufficiently large, \begin{equation} \int_{\mathbb{R}}\frac{\sin(nx)}{x}\phi(x)dx=\int_{-nM}^{nM}\frac{\sin(y)}{y}\phi\left(\frac{y}{n}\right)dx \end{equation} but now how could I manage the limit for $n\rightarrow\infty$ for both the integrand and the domain?

Answer 1

Since $\int_{\mathbb{R}}\frac{\sin(x)}{x}dx=\pi$, it follows that $$\int_{\mathbb{R}}\frac{\sin(nx)}{nx}d(nx)=\int_{\mathbb{R}}\frac{\sin(nx)}{x}dx=\pi.$$ So you are left to prove that $$\lim_{n\rightarrow \infty} \int_{\mathbb{R}}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right) dx=\lim_{n\rightarrow \infty} \int_{a}^b\sin(nx)\cdot \frac{\phi(x)-\phi(0)}{x} dx=0$$ where $[a,b]$ contains the support of $\phi$. Now by noting that $$F(x):=\frac{\phi(x)-\phi(0)}{x}\sim \phi'(0)$$ you have that $F$ is integrable on $[a,b]$. Finally use the Riemann-Lebesgue Lemma and the proof is complete.