Integration of the cardinal sine It is said that the integral of the cardinal sine is $1$. How do I integrate the cardinal sine?
$$ \int_{-\infty}^{\infty} \frac{\sin(a)}{a} \, {\rm d} a $$
 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: 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 :)
A: 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 $t \to 0^+$:
$$\int_{0}^{\infty} \frac{\sin x}{x} dx=\frac{\pi}{2}$$
$$\int_{-\infty}^{\infty} \frac{\sin x}{x} dx=\pi$$
A: Regarding the (great) answer of SuperAbound, Aug 9 '14:
I think it is easier to solve integral (2) using Euler's identity and $\alpha = ia-t$ instead of "integrati[on] by parts".
$$\int^\infty_0e^{-xt}\sin{ax}\ dx
=\int^\infty_0e^{-xt}\cdot\frac{e^{iax}-e^{-iax}}{2}\ dx
=\int^\infty_0 e^{\alpha x}-e^{\alpha^* x}\ dx$$
