calculate a limit of an integral please how we calculate 
$$
\lim_{n \to +\infty} \displaystyle\int_{-a}^a \dfrac{\sin(nx)}{x} dx?
$$
with $a \in \mathbb{R}$.
I have no idea to calculate 
$$
\displaystyle\int_{-a}^a \dfrac{\sin(nx)}{x} dx
$$
 A: Assuming $a\in\mathbb{R}^+$ and letting $x=az$ we have
$$ \int_{-a}^{a}\frac{\sin(nx)}{x}\,dx = 2\int_{0}^{1}\sin(nax)\frac{dx}{x}=\underbrace{2\left[\frac{1-\cos(nax)}{na x}\right]_{0}^{1}}_{O\left(\frac{1}{n}\right)}+\frac{2}{na}\int_{0}^{1}\frac{1-\cos(nax)}{x^2}\,dx $$
by integration by parts. The problem boils down to estimating
$$ 2\int_{0}^{na}\frac{1-\cos x}{x^2}\,dx $$
whose limit as $n\to +\infty$, by the dominated convergence theorem, is given by
$$ 2\int_{0}^{+\infty}\frac{1-\cos x}{x^2}\,dx = \color{red}{\pi}.$$
The main trick here is to transform $\frac{\sin x}{x}$ into $\frac{1-\cos x}{x^2}\in L^1(\mathbb{R})$. By removing the sign constraint on $a$ we get that the wanted limit, in the general case, equals $\color{red}{\pi\,\text{Sign}(a)}$.
A: Assume $a>0$. Let $t=nx$ so that we get:
$$\int^{na}_{-na} \frac{\sin(t) } {t} \, dt$$ 
When $n\to\infty$ we get:
$$\int^\infty_{-\infty} \frac{\sin(t) } {t} \, dt$$ which its value is well known (look for Dirichlet's integral) and that leads to:
$$\lim_{n\to\infty} \int^a_{-a} \frac{\sin(nx) } {x}\, dx=\int^\infty_{-\infty} \frac{\sin(t) } {t} \, dt=\pi$$
The case $a<0$ can be done similarly. 
