How can we calculate
∫∞0sin(x)sin(2x)sin(3x)…sin(nx)sin(n2x)xn+1dx?
I believe that we can use the Dirichlet integral
∫∞0sin(x)xdx=π2
But how do we split the integrand?
Answer
We have (theorem 2, part (ii), page 6) that:
If a0,…,an are real and a0≥∑nk=1|ak|, then ∫∞0n∏k=0sin(akx)xdx=π2n∏k=1ak.
So it is sufficient to note that if we take a0=n2,ak=k,k=1,…,n we have a0=n2≥n(n+1)2=n∑k=1ak
hence
∫∞0sin(n2x)xn∏k=1sin(kx)xdx=πn!2.
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