Convergence to Delta-distribution I was trying to prove that function $\frac{\sin{nx}}{\pi x}$ sutisfies these conditions:
1) $ \forall D \; \exists c : \forall a,b \;\; |a|<D, \; |b|<D \quad \left|\int_a^b  \frac{\sin{nx}}{\pi x} dx\right| \leq c$
2) $ \underset{n \to \infty}{\lim} \int_a^b  \frac{\sin{nx}}{\pi x} dx = \begin{cases} 1 & 0\in(a,b) \\
              0 & 0\notin(a,b)\\
\end{cases} $ 
Then it converges to $\delta$-distribution. I do not know how I can show that the function sutisfies these conditions. Can someone please explain how I should proceed or at least give some hints? 
 A: First, we enforce the substitution $x\to x/n$ to obtain
$$\int_a^b \frac{\sin(nx)}{\pi x} \,dx=\frac1\pi \int_{na}^{nb}  \frac{\sin(x)}{ x}\,dx \tag 1$$

PART $(1):$
Using $\frac{\sin(x)}{x}\le 1$ for $x\in [-\pi ,\pi]$, it is easy to see that for any $na$ and $nb$ we have
$$\left| \frac1\pi \int_{na}^{nb}\frac{\sin(x)}{x}\,dx\right| \le \frac2\pi\int_0^\pi (1)\,dx=2$$

PART $(2)$:
Using $(1)$, we have  
$$\lim_{n\to \infty}\int_a^b \frac{\sin(nx)}{\pi x}\,dx=
\begin{cases}
\frac1\pi\int_{-\infty}^\infty \frac{\sin(x)}{x}\,dx=1 &,a<0<b\\\\
\frac1\pi\int_{\infty}^\infty \frac{\sin(x)}{x}\,dx=0 &,0<a<b\\\\
\frac1\pi\int_{-\infty}^{-\infty} \frac{\sin(x)}{x}\,dx=0 &,a<b<0
\end{cases}$$

NOTE:
To show that $\lim_{n\to \infty}\int_{na}^{nb}\frac{\sin(x)}{x}\,dx=0$ for $0<a<b$ or $a<b<0$, we integrate by parts with $u=1/x$ and $v=-\cos(x)$ to write 
$$\lim_{n\to \infty}\int_{na}^{nb}\frac{\sin(x)}{x}\,dx=\lim_{n\to \infty}\left(\frac{\cos(na)}{na}-\frac{\cos(nb)}{nb}-\int_{na}^{nb}\frac{\cos(x)}{x^2}\,dx\right)=0$$
To show that $\int_0^\infty \frac{\sin(x)}{x}\,dx=\pi/2$, we write
$$\int_0^\infty \frac{\sin(x)}{x}\,dx=\int_0^\infty \sin(x)\int_0^\infty e^{-kx}\,dk\,dx$$
Now, since we have for any $L>0$
$$\int_0^L\int_0^\infty \left|e^{-kx}\sin(x)\right|\,dk\,dx\le L$$
the Fubini-Tonelli Theorem guarantees that 
$$\begin{align}
\int_0^L\int_0^\infty e^{-kx}\sin(x)\,dk\,dx&=\int_0^\infty \int_0^L e^{-kx}\sin(x)\,dx\,dk\\\\
&=\int_0^\infty \frac{1-e^{-kL}(k\sin(L)+\cos(L))}{k^2+1}\,dk\\\\
&=\frac\pi2-\int_0^\infty \frac{e^{-kL}(k\sin(L)+\cos(L))}{k^2+1}\,dk
\end{align}$$
Finally, since $\left|\frac{e^{-kL}(k\sin(L)+\cos(L))}{k^2+1}\right|\le g(k)$ where $g(k)$ is given by
$$g(k)= \begin{cases}e^{-k}&,k\ge 1\\\\\frac{1+\sqrt2}{2}&,k\le 1\end{cases}$$
and $\int_0^\infty g(k)\,dk <\infty$, we can apply the Dominated Convergence Theorem to obtain
$$\begin{align}\lim_{L\to \infty}\int_0^L\int_0^\infty e^{-kx}\sin(x)\,dk\,dx&=\lim_{L\to \infty}\int_0^\infty\int_0^L e^{-kx}\sin(x)\,dk\,dx\\\\
&=\int_0^\infty \lim_{L\to \infty}\left(\frac{1-e^{-kL}(k\sin(L)+\cos(L))}{k^2+1}\right)\,dk\\\\
&= \pi/2
\end{align}$$
