Evaluating $\int_{0}^{\infty}{\sin(x)\sin(2x)\sin(3x)\ldots \sin(nx)\sin(n^{2}x) \over x^{n + 1}}\,dx $ How can we calculate
$$
\int_{0}^{\infty}{\sin\left(x\right)\sin\left(2x\right)\sin\left(3x\right)\ldots
\sin\left(nx\right)\sin\left(n^{2}x\right) \over x^{n + 1}}\,\mathrm{d}x ?
$$

I believe that we can use the Dirichlet integral

$$
\int_{0}^{\infty}{\sin\left(x\right) \over x}\,\mathrm{d}x =
{\pi \over 2}
$$

But how do we split the integrand?
We have (theorem $2$, part $(ii)$, page 6) that:


  If $a_{0},\dots,a_{n}
 $ are real and $a_{0}\geq\sum_{k=1}^{n}\left|a_{k}\right|$, then $$\int_{0}^{\infty}\prod_{k=0}^{n}\frac{\sin\left(a_{k}x\right)}{x}dx=\frac{\pi}{2}\prod_{k=1}^{n}a_{k}.$$ 


So it is sufficient to note that if we take $a_{0}=n^{2},\, a_{k}=k,\, k=1,\dots,n
 $ we have $$a_{0}=n^{2}\geq\frac{n\left(n+1\right)}{2}=\sum_{k=1}^{n}a_{k}
 $$ hence 


  $$\int_{0}^{\infty}\frac{\sin\left(n^{2}x\right)}{x}\prod_{k=1}^{n}\frac{\sin\left(kx\right)}{x}dx=\frac{\pi n!}{2}.$$
If you know the answer can you verify with my answer 

$$\frac12[\pi(n^2 n!)-\frac{\pi}{2^n(n+1)!}((1+2+..+n+n^2)-2)^2]$$

If you think it is not correct, then verify it by reducing the terms