Evaluating limn→∞1n∫0π2sin2nxsin2xf(x)dx for continuous f



Answered question


Evaluating limn1n0π2sin2nxsin2xf(x)dx for continuous f

Answer & Explanation

Emilie Booker

Emilie Booker

Beginner2022-01-30Added 14 answers

The limit is π2f(0) Note that |0π/2sin2(nx)nsin2xf(x)dxπ2f(0)| (*) |0δsin2(nx)nsin2x[f(x)f(0)]dx|+|δπ/2sin2(nx)nsin2x[f(x)f(0)],dx| Since f is continuous on [0,π/2] and, hence, bounded we have |f(x)|M and |f(x)f(0)|2M and an estimate for the second integral on the RHS of (*) is |δπ/2sin2(nx)nsin2x[f(x)f(0)]dx|2Mnδπ/21sin2xdxn0 Since the second integral converges to 0 for any δ, it remains to prove that the first integral on the RHS of (*) can be made arbitrarily close to 0 with a suitable choice for δ. For any ϵ>0 we can chose δ such that |f(x)f(0)|<ϵ for 0xδ Using sinx>2x/π we get |0δsin2(nx)nsin2x[f(x)f(0)]dx|π2ϵ4n0δsin2(nx)x2,dx =π2ϵ40nδsin2uu2du π2ϵ40sin2uu2du=π3ϵ8 Since, ϵ can be arbitrarily close to 0 we are done.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?