3

If $n=1$ we can compute $\int _\mathbb{R} \frac{sin^{2}t}{t^{2}}dt$ by using Parseval's formula since $\widehat{1_{[-1,1]}}(x)=2\frac{\sin x}{x}$. We obtain $\int _\mathbb{R} \frac{sin^{2}t}{t^{2}}dt=\pi$. Is there a similar way in order to compute $\int _\mathbb{R}\frac{sin^{2n}t}{t^{2n}}dt$ ?

It seems to be difficult to compute $\int 1_{[-1,1]}\ast \ldots \ast 1_{[-1,1]}(x)^2dx$.

Patissot
  • 2,495

1 Answers1

1

Since the integrand is even, your integral is $2\int_0^\infty\dots dt$. For this MathWorld (38) gives a closed form in terms of Eulerian numbers $$\int_0^\infty \left(\frac{\sin x}{x}\right)^{2n} \mathrm{d} x = \frac{\pi}{2(2n-1)!} \genfrac{<}{>}{0}{}{2n-1}{n-1}$$

gammatester
  • 18,827