TITLE: Integration of the cardinal sine
QUESTION [27 upvotes]: It is said that the integral of the cardinal sine is $1$. How do I integrate the cardinal sine?
$$ \int_{-\infty}^{\infty} \frac{\sin(a)}{a} \, {\rm d} a $$

REPLY [14 votes]: 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 $t \to 0^+$:
$$\int_{0}^{\infty} \frac{\sin x}{x} dx=\frac{\pi}{2}$$
$$\int_{-\infty}^{\infty} \frac{\sin x}{x} dx=\pi$$