Integration of the cardinal sine

It is said that the integral of the cardinal sine is $1$. How do I integrate the cardinal sine?

$$ \int_{-\infty}^{\infty} \frac{\sin(a)}{a} \, {\rm d} a $$

Use the fact that $$\int^\infty_0e^{-xt}dt=\frac{1}{x}$$ Hence \begin{align} \int^\infty_{-\infty}\frac{\sin{x}}{x}dx \tag1 &=2\int^\infty_{0}\frac{\sin{x}}{x}dx\\ \tag2 &=2\int^\infty_0\int^\infty_0e^{-xt}\sin{x}dxdt\\ \tag3 &=2\int^\infty_0\frac{1}{1+t^2}dt\\ \tag4 &=\pi \end{align} Explanation:
$1)$Integrand is even
$2)$Reverse the order of integration
$3)$Recognise the laplace transform of $\sin{x}$, or integrate by parts.
$4)$ $\arctan(\infty)=\frac{\pi}{2}$

A common method given:

Because the function $f(x)=\frac{\sin x}{x}$, where $f: \mathbb{R} - \{0\} \to \mathbb {R}$ is even we have:

$$\int_{-\infty}^{\infty} \frac{\sin x}{x} dx=2\int_{0}^{\infty} \frac{\sin x}{x} dx$$

Now let:

$$I(t)=\int_{0}^{\infty} \frac{\sin x}{x} e^{-tx} dx$$

Note:

$$\frac{\partial}{\partial t} \frac{\sin x}{x} e^{-tx}=\frac{\sin x}{x} e^{-tx}(-x)$$

So by differentiation under the integral sign we have:

$$I'(t)=-\int_{0}^{\infty} e^{-tx} \sin x dx$$

And through integration by parts twice we have:

$$I'(t)=-\frac{1}{t^2+1}$$

Hence,

$$I(t)=\int -\frac{1}{t^2+1} dt$$

$$I(t)=-\arctan (t) +c$$

But as $t \to \infty$, $I(t) \to 0$ hence:

$$I(t)=\frac{\pi}{2}-\arctan t$$

Let $t \to 0^+$:

$$\int_{0}^{\infty} \frac{\sin x}{x} dx=\frac{\pi}{2}$$

$$\int_{-\infty}^{\infty} \frac{\sin x}{x} dx=\pi$$

If you are using the normalised $\mathrm{sinc}$ function, the area will be $1$ though if not, it is $\pi$. Proofs can be found here and here. Note that the second link still answers your question even though the integrand is squared.

Please consider googling your question before asking :)