If $\phi \in C^1_c(\mathbb R)$ then $ \lim_n \int_\mathbb R \frac{\sin(nx)}{x}\phi(x)\,dx = \pi\phi(0)$. 
Let $\phi \in C^1_c(\mathbb R)$. Prove that $$
\lim_{n \to +\infty} \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx = \pi\phi(0).
$$

Unfortunately, I didn't manage to give a complete proof. First of all, I fixed $\varepsilon>0$. Then there exists a $\delta >0$ s.t. 
$$
\vert x \vert < \delta \Rightarrow \vert \phi(x)-\phi(0) \vert < \frac{\varepsilon}{\pi}.
$$
Now, I would use the well-known fact that 
$$
\int_\mathbb R \frac{\sin x}{x} \, dx = \pi.
$$
On the other hand, by substitution rule, we have also
$$
\int_\mathbb R \frac{\sin(nx)}{x} \, dx = \int_\mathbb R \frac{\sin x}{x} \, dx = \pi.
$$
Indeed, I would like to estimate the quantity
$$
\begin{split}
& \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \pi \phi(0) \right\vert = \\ 
& = \left\vert \int_\mathbb R \frac{\sin(nx)}{x}\phi(x) \, dx - \phi(0)\int_\mathbb R \frac{\sin{(nx)}}{x}dx \right\vert \le \\ 
& \le \int_\mathbb R \left\vert \frac{\sin(nx)}{x}\right\vert \cdot \left\vert  \phi(x)-\phi(0) \right\vert dx
\end{split}
$$
but the problem is that $x \mapsto \frac{\sin(nx)}{x}$ is not absolutely integrable over $\mathbb R$. Another big problem is that I don't see how to use the hypothesis $\phi$ has compact support.
I think that I should use dominated convergence theorem, but I've never done exercises about this theorem. Would you please help me? Thank you very much indeed.
 A: A quick proof of this can be given using the Riemann-Lebesgue lemma, which is covered in Rudin and a number of other texts. Write your limit as $$\lim_{n \rightarrow \infty} \int_\mathbb R \sin(nx)\frac{\phi(x) - \phi(0)\chi_{[-1,1]}(x)}{x}\, dx 
+ \lim_{n \rightarrow \infty} \int_\mathbb R \sin(nx)\frac{\phi(0)\chi_{[-1,1]}(x)}{x}\, dx $$
Here $\chi_{[-1,1]}(x)$ denotes the characteristic function of $[-1,1]$. Since $\phi(x) \in C_c^1({\mathbb R})$, the function ${\displaystyle \frac{\phi(x) - \phi(0)\chi_{[-1,1]}(x)}{x}}$ is a bounded function with compact support; the only place you have to worry about is $x = 0$ and you can use the mean value theorem for example to show it's bounded near $x = 0$. Since the function is bounded function with compact support it is in $L^1$, which is enough to apply the Riemann-Lebesgue lemma and say the first term goes to zero. As for the second term, after changing variables to $y = nx$ we may rewrite it as 
$$\lim_{n \rightarrow \infty} \int_\mathbb R \sin(y)\frac{\phi(0)\chi_{[-n,n]}(y)}{y}\, dy $$
$$= \phi(0)\lim_{n \rightarrow \infty} \int_{-n}^n \frac{\sin(y)}{y}\, dy $$
$$= \phi(0)\int_\mathbb R \frac{\sin(y)}{y}\, dy $$
$$= \pi \phi(0)$$
So this will be the overall limit. 
A: Assume that $\phi(x)$ is supported in $|x|< L$. Since $\phi$ is differentiable, $\frac{\phi(x)-\phi(0)}{x}$ is bounded and therefore integrable on $|x|<L$.
$$
\begin{align}
\int_{-\infty}^\infty\frac{\sin(nx)}{x}\phi(x)\,\mathrm{d}x
&=\pi\,\phi(0)+\int_{-\infty}^\infty\sin(nx)\frac{\phi(x)-\phi(0)}{x}\,\mathrm{d}x\\
&=\pi\,\phi(0)+\color{#C00000}{\int_{-L}^L\sin(nx)\frac{\phi(x)-\phi(0)}{x}\,\mathrm{d}x}\\
&-2\,\phi(0)\color{#00A000}{\int_{nL}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x}\\
\end{align}
$$
As $n\to\infty$, the red integral vanishes by the Riemann-Lebesgue Lemma and the green integral vanishes because The Dirichlet Integral converges. This leaves us with
$$
\lim_{n\to\infty}\int_{-\infty}^\infty\frac{\sin(nx)}{x}\phi(x)\,\mathrm{d}x=\pi\,\phi(0)
$$
A: We can assume WLOG that $\phi\in C^3_c(\Bbb R)$, since this subset is dense for the supremum norm and $\frac{\sin x}x$ is integrable over compact subsets. We have 
$$\phi(x)-\phi(0)=x\phi'(0)+x\int_0^1(1-s)\phi''(sx),$$
hence, if the support of $\phi$ is contained in $[-R,R]$
\begin{align}
\small \int_{-\infty}^{+\infty}\frac{\sin(nx)}x\phi(x)dx&=\small\phi(0)\int_{-R}^R\frac{\sin(nx)}xdx+\phi'(0)\int_{-R}^R\sin(nx)dx+\int_{—R}^R\sin(nx)\int_0^1(1-s)\phi''(sx)dsdx\\
&=\small\phi(0)\int_{-nR}^{nR}\frac{\sin t}tdt-\left[\frac{\cos(nx)}n\int_0^1(1-s)\phi''(sx)ds\right]_{-R}^R+\int_{—R}^R\frac{\cos(nx)}n\int_0^1
s(1-s)\phi'''(sx)dsdx
\end{align}
We have 
$$\lim_{n\to+\infty}\int_{-nR}^{nR}\frac{\sin t}tdt=\int_{-\infty}^{+\infty}\frac{\sin t}tdt;$$
$$\left|\left[\frac{\cos(nx)}n\int_0^1(1-s)\phi''(sx)ds\right]_{-R}^R\right|\leq \frac 2n\sup_{t\in \Bbb R}|\phi''(t)|,$$
and 
$$\left|\int_{—R}^R\frac{\cos(nx)}n\int_0^1
s(1-s)\phi'''(sx)dsdx\right|\leq \frac{2R}n\sup_{t\in \Bbb R}|\phi'''(t)|$$
A: Note that
$$
\small
\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\phi(x)dx-\pi\phi(0)=
\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\phi(x)dx-\phi(0)\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}dx=
\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx
$$
Denote
$$
\small
I_{m}(n):=\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx\qquad
I(n):=\int\limits_{\mathbb{R}}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx
$$
We claim that $I_m(n)$ converges to $I(n)$ uniformly by $n\in\mathbb{N}$ when $m\to\infty$. Indeed, since $\phi$ is compactly supported
$$
\small
\lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|I_m(n)-I(n)\right|=
\lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx\right|=
$$
$$
\small
\lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(nx)}{x}(-\phi(0))dx\right|=
\lim\limits_{m\to\infty}\sup\limits_{n\in\mathbb{N}}\left|\phi(0)\int\limits_{\mathbb{R}\setminus[-\pi mn,\pi mn]}\frac{\sin(y)}{y}dy\right|=
$$
$$
\small
|\phi(0)|\lim\limits_{m\to\infty}\left|\;\int\limits_{\mathbb{R}\setminus[-\pi m,\pi m]}\frac{\sin(y)}{y}dy\right|=0
$$
Since convergence is uniform by $n\in\mathbb{N}$ we can say
$$
\small
\lim\limits_{n\to\infty} I(n)=
\lim\limits_{n\to\infty} \lim\limits_{m\to\infty} I_m(n)=
\lim\limits_{m\to\infty}\lim\limits_{n\to\infty} I_m(n)=
\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx
$$
Since $\varphi\in C_c^1(\mathbb{R})$, then the function $x^{-1}(\varphi(x)-\varphi(0))$ is in $L^1([-\pi m,\pi m])$ for all $m\in\mathbb{N}$. Then by Riemann–Lebesgue lemma
$$
\small
\lim\limits_{n\to\infty}\int\limits_{[-\pi m,\pi m]}\frac{\sin(nx)}{x}\left(\phi(x)-\phi(0)\right)dx=0
$$
so $\small\lim\limits_{n\to\infty}I(n)=0$. This is exactly what we wanted to prove.
