Wednesday, 11 June 2014

calculus - Evaluating intinfty0sin(x)sin(2x)sin(3x)ldotssin(nx)sin(n2x)overxn+1,dx



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 a0nk=1|ak|, then 0nk=0sin(akx)xdx=π2nk=1ak.




So it is sufficient to note that if we take a0=n2,ak=k,k=1,,n we have a0=n2n(n+1)2=nk=1ak

hence




0sin(n2x)xnk=1sin(kx)xdx=πn!2.



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