What is the correct integral definition here?

Question

I know $\int_{-\infty}^{\infty}\frac{\sin x}{x}dx=\pi$, but what is the definition of the integral here?

$\int_{-\infty}^{\infty}\left|\frac{\sin x}{x}\right|dx$ is not finite, therefore $\frac{\sin x}{x}$ is not Lebesgue-integrable. But what is the integral definition then?

If we define $\int_{-\infty}^{\infty}\frac{\sin x}{x}dx$ as $\lim_{T\rightarrow\infty}\int_{-T}^{T}\frac{\sin x}{x}dx$, then I know

$$\int_{-\infty}^{\infty}\frac{\sin x}{x}dx\overset{\circ}{=}\lim_{T\rightarrow\infty}\int_{-T}^{T}\frac{\sin x}{x}dx=\pi,$$

but is it true if I define it for example as:

$$\int_{-\infty}^{\infty}\frac{\sin x}{x}dx\overset{\circ}{=}\lim_{T\rightarrow\infty}\int_{-T}^{2T}\frac{\sin x}{x}dx\overset{?}{=}\pi?$$

If the last statement is true, then I would say $\frac{\sin x}{x}$ is integrable in the improper Riemann sense on $\mathbb{R}$, but otherwise I would say it is just convergent in principal value sense. The $\int_{0}^{\infty}\frac{\sin x}{x}dx$ Dirichlet-integral is another question, I know that is convergent in improper Riemann sense and it converges to $\frac{\pi}{2}$.

Answer 1

The integral is only defined in terms of the improper Riemann integral and not the Lebesgue integral. It is defined as

$$\int_{-\infty}^\infty\frac{\sin(x)}{x}dx\overset{\circ}{=}\lim_{T\to\infty}\int_0^T\frac{\sin(x)}{x}dx+\lim_{T\to-\infty}\int_T^0\frac{\sin(x)}{x}dx$$

provided both limits exist. However, since $\frac{\sin(x)}{x}$ is an even function this simply becomes

$$=\lim_{T\to\infty}\int_0^T\frac{\sin(x)}{x}dx+\lim_{T\to\infty}\int_0^T\frac{\sin(x)}{x}dx=2\lim_{T\to\infty}\int_0^T\frac{\sin(x)}{x}dx$$

As you already noted, this limit is $\frac{\pi}{2}$ which implies the original integral has value $\pi$.

Answer 2

$\int_{a(t)}^{b(t)}\frac {\sin x }x dx\to \pi$ whenever $a(t) \to -\infty$ and $b_t \to +\infty$ as $ t \to \infty$. Similar statement holds for the integral from $0$ to $\infty$.

[$F(t)=\int_0^{t} \frac {\sin x} x dx$ defines a continuous function on $[0,\infty)$ and $F(t) \to \frac {\pi } 2$ as $ t \to \infty$].

Answer 3

When we claim that$$\int_{-\infty}^\infty\frac{\sin x}x\,\mathrm dx\tag1$$converges to some number (which turns out to be $\pi$), what this means is that both limits$$\lim_{T\to\infty}\int_a^T\frac{\sin x}x\,\mathrm dx\quad\text{and}\quad\lim_{T\to-\infty}\int_T^a\frac{\sin x}x\,\mathrm dx$$exist, for every real number $a$; then $(1)$ is the sum of those two limts (which happens to be independent of the choice of $a$.

On the other hand, for each $a\in\Bbb R$, neither of the limits$$\lim_{T\to\infty}\int_a^T\left|\frac{\sin x}x\right|\,\mathrm dx\quad\text{and}\quad\lim_{T\to-\infty}\int_T^a\left|\frac{\sin x}x\right|\,\mathrm dx$$exist. One would be enough that one of them did not existe to deduce that the integral $\int_{-\infty}^\infty\left|\frac{\sin x}x\right|\,\mathrm dx$ diverges.