It is well-known that the integral of sinc equals $\pi$:
$$ \int_{-\infty}^\infty \frac{\sin x}{x}\ dx = \lim_{A \to \infty} \int_{-A}^A \frac{\sin x}{x}\ dx = \pi. $$
Here is a way to get this, but I think this argument is a nonsense:
The above integral equals the value at $0$ of the Fourier transform of sinc function:
$$ \sqrt{2\pi}\cdot \mathcal F_x \left[ \frac{\sin x}{x} \right](0), $$
where the normalization convention is
$$ \mathcal F[f](t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) e^{-itx}\ dx. $$
But we know that the sinc function is an even function and
$$ \mathcal F \left[ \mathbf 1_{[-1,\ 1]} \right](t) = \sqrt{\frac{2}{\pi}} \cdot \frac{\sin t}{t}, $$
having the inverse transform
$$ \mathcal F_x \left[ \frac{\sin x}{x} \right] = \sqrt{\frac{\pi}{2}} \cdot \mathbf 1_{[-1,\ 1]}. \ \ \ \ \ \ \ \ ……[*] $$
Hence we have
$$ \int_{-\infty}^\infty \frac{\sin x}{x}\ dx = \sqrt{2\pi} \cdot \mathcal F_x \left[ \frac{\sin x}{x} \right] (0) = \pi.\ \ \ \ \ \ \ \ ……[**] $$
So what's wrong? The above argument uses the inverse transform. But the Fourier inversion theorem applies only if both $f$ and $\hat f$ are of $L^1$, which is not the case we are dealing. So we have to rely on the Plancherel theorem, which says that the Fourier transform gives a Hilbert space isomorphism of $L^2(\mathbb R)$.
Then we do have the equation $[*]$, but the equality in $[*]$ is not a usual one; it only means the two funcitons are equal almost everywhere. So the specific value $\mathcal F [\text{sinc}](0)$ is not uniquely-deterined.
Question: Yet many people uses the above argument to compute the sinc integral. Can this argument be justified by another way?
Best Answer
There is a way to avoid the problem about the well-definedness of $\mathcal F_x [\sin(x)/x](0)$. We do not use Fourier inversion, but Plancherel theorem.
Let $c$ be a positive number less than $1$. Define the function $f_n$ by $$ f_n(t) := \int_{-n}^n \frac{\sin x}{x} \cdot e^{-itx}\ dx. $$ Then we prove that $f_n$ converges uniformly on $[-c,\ c]$ to some constant function $K$, resembling Mark Viola's proof in the link: https://math.stackexchange.com/a/4219782/833020 (Of course we may not use the fact that $ \lim_{n\to\infty} \int_{-n}^n \sin(x)/x \ dx = \pi$, but it's okay; It sufficies for us right now to prove that the limit is just a constant.) So we conclude that $f_n$ converges to $K$ in $L^2([-c,\ c])$.
By Plancherel theorem we know that $f_n$ converges to $\pi\cdot\mathbf 1_{[-1,\ 1]}$ in $L^2(\mathbb R)$, hence to the constant function $\pi$ in $L^2([-c,\ c])$. This means that two constant functions $K$ and $\pi$ coincicdes almost everywhere on $[-c,\ c]$. Hence $K=\pi$.