Saturday, 5 December 2015

real analysis - Find limrtoinftyintpi0ercos2(theta)dtheta



limrπ0ercos2(θ)dθ




Here is what I did:



Since ercos2(θ) is continuous on [0,π] for any fixed r, we can use MVT:
There is c(0,π) such that



π0ercos2(θ)dθ=ercos2(c)π



Also ex>0xR, so:



0π0ercos2(θ)dθ=ercos2(c)π




Using squeeze theorem, limr0=0=limrercos2(θ), then we have limrπ0ercos2(θ)dθ=0



Is this correct?
Thanks.



Also: I cannot pass the limit inside the integral, right? But why not, exactly?


Answer



For any r0 and for any θ the following inequality holds ercos2θ1. Choose a sequence of positive real numbers (rn) that runs to the infinity. Then
erncos2θ{0 if θπ21 if θ=π2


pointwise as n goes to infinity (but not uniformly). Moreover such a limit does not depend on the sequence we have chosen before. Now apply the dominated Lebesgue’s theorem to any sequence to pass the limit under the integral (indeed, the pointwise convergence is sufficient, you don’t need the uniform convergence). Hence:



limrnπ0erncos2θ=π0limrnerncos2θ=0


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