Monday, 28 December 2015

calculus - Prove: inti0nftysin(x2),dx converges.




sinx2 does not converge as x, yet its integral from 0 to does.



I'm trying to understand why and would like some help in working towards a formal proof.


Answer



xsin(x2) is integrable on [0,1], so we have to show that limA+A1sin(x2)dx exists. Make the substitution t=x2, then x=t and dx=dt2t. We have A1sin(x2)dx=A21sint2tdt=cosA22A+cos12+12A21costt3/212dt,


and since limA+cosA22A+cos12=cos12 and the integral +1t3/2dt exists (is finite), we conclude that +1sin(x2)dx and so does +0sin(x2)dx.
This integral is computable thanks to the residues theorem.


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