Fourier transform to solve $\int_{- \infty}^{+\infty}\frac{\sin^3(x)}{x^3}\,dx$ I think that I am almost there:
giving $P(x)=1$ for $x \in ]-1/2;1/2[$ and $P(x)=0$ elsewhere.
I managed to find that the convolution product $P*P*P=\frac{3}{4}-x^2$ for $x \in [-1/2;1/2]$
I found that Reverse Fourier transform of $P$ is $p=\frac{1}{2\pi}\times \frac{\sin(x/2)}{x/2}$
knowing that Fourier Transform of 
$p\times p \times p$ is $\frac{1}{4\pi^2}P*P*P$
We then get :
$\int_{- \infty}^{+\infty} p(x)\times p(x) \times p(x)\times e^{ikx}dx =\frac{1}{4\pi^2}P*P*P(k)$
$\int_{- \infty}^{+\infty} \frac{1}{8\pi^3} \frac{\sin^3(x)}{x^3}\times e^{ikx}dx =\frac{1}{4\pi^2}P*P*P(k)$
Then for $k=0$, $\int_{- \infty}^{+\infty} \frac{1}{8\pi^3} \frac{\sin^3(x)}{x^3}dx =\frac{1}{4\pi^2}\frac{3}{4}$
then I find $\int_{- \infty}^{+\infty}  \frac{\sin^3(x)}{x^3}\,dx =\frac{3\pi}{2}$
but this is wrong because $\int_{- \infty}^{+\infty}  \frac{\sin^3(x)}{x^3}\,dx =\frac{3\pi}{4}$
Where is the problem?
Thanks

Your mistake is you forgot $p=\operatorname{sinc}\frac{x}{2}\not\equiv\operatorname{sinc}x$. (In case the notation is unfamiliar, $\operatorname{sinc}x=\frac{\sin x}{x}$, with $\operatorname{sinc}0=1$ for continuity.) We have $\int_{\Bbb R}\operatorname{sinc}^3\frac{x}{2} dx=2\int_{\Bbb R}\operatorname{sinc}^3y dy$ by a trivial substitution.
