To find the integral of the Sinc function:
Start with, \begin{equation} I(a)=\int_{-\infty}^{\infty}\frac{\sin\ ax }{x}dx %\hspace{20.0} ; (a>0) \end{equation}
\begin{equation} \Longrightarrow \frac{dI(a)}{da} = \int_{-\infty}^{\infty}\frac{\partial}{\partial a}(\frac{\sin\ ax}{x})dx= \int_{-\infty}^{\infty} \cos\ ax\ dx \end{equation}
\begin{equation} = \\ \textbf{Re}(\int_{-\infty}^{\infty} e^{iax} dx)=2\pi \delta(a) \end{equation}
Take $a>0$; noting that $I(a)$ is an odd function, we get, \begin{equation} \Longrightarrow 2I(a)=\int_{-a}^{a}dI(a^{'})=2\pi \int_{-a}^{a} \delta(a^{'})da^{'}=2\pi \end{equation}
\begin{equation} \Longrightarrow \int_{0}^{\infty}\frac{\sin\ x}{x}dx= \frac{1}{2}I(1)= \frac{\pi}{2} \end{equation} (The $a<0$ case following in a similar vein as well.)
Since the integrand is even, we can reverse the order of integration. If you have learned laplace transforms, review the laplace transform of $\sin{x}$, or you can just integrate by parts. Whichever is easier for you. Next notice that $\arctan(\infty)=\frac{\pi}{2}$ Lastly, use the fact below: $$\int^\infty_0e^{-xt}dt=\frac{1}{x}$$ Therefore: \begin{align} \tag1 &\int^\infty_{0}\frac{\sin{x}}{x}dx\\ \tag2 &=\int^\infty_0\int^\infty_0e^{-xt}\sin{x}dxdt\\ \tag3 &=\int^\infty_0\frac{1}{1+t^2}dt\\ \tag4 &=\frac{\pi}{2} \end{align}
Additionally, the Dirichlet kernel converges to a Dirac delta function: $$\lim_{a\to\infty}\frac{\sin(ax)}{x}=\frac\pi2 \delta(x).$$