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
x↦sin(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=dt2√t. We have ∫A1sin(x2)dx=∫A21sint2√tdt=−cosA22√A+cos12+12∫A21cost⋅t−3/2−12dt,
and since limA→+∞−cosA22√A+cos12=cos12 and the integral ∫+∞1t−3/2dt 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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