How do I prove that $\int_{-\infty}^x \frac{\sin(ny)}{\pi y}\text{d}y$ tends towards the Heaviside step function?

Question

I want some (ideally short) proof of the statement $$\int_{-\infty}^x \frac{\sin(ny)}{\pi y}\text{d}y\,\,\stackrel{n\rightarrow\infty}{\longrightarrow}\,\,H(x)$$ with which I can later proof that $\frac{\sin(nx)}{\pi x}\stackrel{n\rightarrow\infty}{\longrightarrow}\delta_0$. The proofs I found so far (1 2) always assume prior knowledge of something similar to my question, for instance "obvious" properties of $Si(x)$ function.

Answer 1

Hint $$ \int_{-\infty}^x {\sin ny\over \pi y}dy=\int_{-\infty}^{nx} {\sin ny\over \pi ny}dny =\int_{-\infty}^{nx} {\sin u\over \pi u}du $$ Now, tend $n\to \infty$ for $x>0$ and $x<0$ and substitute the integral bounds with proper values.

Answer 2

If $\lim_{A,B\to \infty}\int_{-A}^B f(y)dy$ converges then $F(x)=\int_{-\infty}^x f(y)dy$ is well-defined and $F(nx)\to F(\infty) H(x)$ boundedly on $\Bbb{R}$ and locally uniformly on $|x|\ge 1$.