Can we solve integrals of the type $\int_{-\infty}^{\infty} (\sin(x)/x)^n \, dx$ using Parseval’s relation?

Question

Can we use Parseval’s relation to evaluate integrals like the following?

EDIT: I found a similar post here.

$$I_n:=\int_{-\infty}^{\infty} \left(\frac{\sin(x)}{x}\right )^n \, dx, \quad n \in \mathbb{N}.$$

I’m not expecting a closed form for all $n$, but perhaps the approach can be extended beyond the $n=2$ and $n=4$ cases I’ve done.

I’ve been able to do

$$I_2=\int_{-\infty}^{\infty} \left(\frac{\sin(x)}{x}\right)^2 \, dx = \pi$$

and

$$ I_4=\int_{-\infty}^{\infty} \left(\frac{\sin(x)}{x}\right)^4 \, dx = \frac{2\pi}{3}$$

Recall that Perseval’s relation states the following:

$$\|\hat{f}(\omega)\|_2 =\|f(x)\|_2,$$

where $\hat{f}(\omega)$ represents the Fourier transform $\mathcal{F}(f(x))$ and the 2-norm of a function $g$ is given by

\begin{equation} \|g\|_2=\sqrt{\int_{-\infty}^{\infty} |g(x)|^2 \,dx} \end{equation}

Using these equations, we can relate the value of an integral to the Fourier transform of its integrand.

$$\int_{-\infty}^{\infty} |f(x)|^2 \,dx = \int_{-\infty}^{\infty} |\hat{f}(\omega)|^2 \,dx.$$

I managed to evaluate $I_2$ using the well known fact that the Fourier transfrom of a step function is the sinc function $\sin(x)/x$.

I’ll sketch how I did $I_4$

Firstly, I was able to show that

$$\mathcal{F} (\Lambda(x)) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\Lambda(x) e^{-i \omega x} \, dx = \sqrt\frac{2}{\pi} \frac{1-\cos(\omega)}{\omega^2}= \frac{1}{\sqrt{2\pi}}\left(\frac{\sin(\omega/2)}{\omega/2}\right)^2,$$

where $\Lambda(x)$ is the tent function defined by $$\Lambda(x) = \begin{cases} 1-|x|, & \text{if } x \in (-1, 0) \text{ or } x \in (0, 1); \\ 0, & \text{otherwise.} \end{cases}$$

From the above, we now have that

$$\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\sin^4(\nu)}{\nu^4} \,d\nu = \int_{-1}^{1}\Lambda^2(x) \, dx, \quad \nu = \frac{\omega}{2}$$

I think I can show that

$$\int_{-1}^{1}\Lambda^k(x)\, dx = \frac{2}{k+1}$$

This clearly implies that $\frac{1}{\pi}I_4 = \frac{2}{3}$, meaning

$$\boxed{I_4=\int_{-\infty}^{\infty} \left(\frac{\sin(x)}{x}\right)^4 \, dx = \frac{2\pi}{3}.}$$

I have no idea if this trick generalises for larger values of $n$. The trick seems to be able to relate the function $\sin^k(x)/x^k$ to the Fourier transform of some easier-to-integrate function, but I’m not sure how to find such a function in general. Is there hope of doing something similar for larger $n$?