Mlejd5

## Answered question

2022-01-29

Evaluating $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}{\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{2}nx}{{\mathrm{sin}}^{2}x}f\left(x\right)dx$ for continuous f

### Answer & Explanation

Emilie Booker

Beginner2022-01-30Added 14 answers

The limit is $\frac{\pi }{2}f\left(0\right)$ Note that $|{\int }_{0}^{\pi /2}\frac{{\mathrm{sin}}^{2}\left(nx\right)}{n{\mathrm{sin}}^{2}x}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx-\frac{\pi }{2}f\left(0\right)|$ (*) $⩽|{\int }_{0}^{\delta }\frac{{\mathrm{sin}}^{2}\left(nx\right)}{n{\mathrm{sin}}^{2}x}\left[f\left(x\right)-f\left(0\right)\right]\phantom{\rule{0.167em}{0ex}}dx|+|{\int }_{\delta }^{\pi /2}\frac{{\mathrm{sin}}^{2}\left(nx\right)}{n{\mathrm{sin}}^{2}x}\left[f\left(x\right)-f\left(0\right)\right],dx|$ Since f is continuous on $\left[0,\pi /2\right]$ and, hence, bounded we have $|f\left(x\right)|⩽M$ and $|f\left(x\right)-f\left(0\right)|⩽2M$ and an estimate for the second integral on the RHS of (*) is $|{\int }_{\delta }^{\pi /2}\frac{{\mathrm{sin}}^{2}\left(nx\right)}{n{\mathrm{sin}}^{2}x}\left[f\left(x\right)-f\left(0\right)\right]\phantom{\rule{0.167em}{0ex}}dx|⩽\frac{2M}{n}{\int }_{\delta }^{\pi /2}\frac{1}{{\mathrm{sin}}^{2}x}\phantom{\rule{0.167em}{0ex}}dx{\to }_{n\to \mathrm{\infty }}0$ Since the second integral converges to 0 for any $\delta$, it remains to prove that the first integral on the RHS of (*) can be made arbitrarily close to 0 with a suitable choice for $\delta$. For any $ϵ>0$ we can chose $\delta$ such that $|f\left(x\right)-f\left(0\right)|<ϵ$ for $0⩽x⩽\delta$ Using $\mathrm{sin}x>2x/\pi$ we get $|{\int }_{0}^{\delta }\frac{{\mathrm{sin}}^{2}\left(nx\right)}{n{\mathrm{sin}}^{2}x}\left[f\left(x\right)-f\left(0\right)\right]\phantom{\rule{0.167em}{0ex}}dx|⩽\frac{{\pi }^{2}ϵ}{4n}{\int }_{0}^{\delta }\frac{{\mathrm{sin}}^{2}\left(nx\right)}{{x}^{2}},dx$ $=\frac{{\pi }^{2}ϵ}{4}{\int }_{0}^{n\delta }\frac{{\mathrm{sin}}^{2}u}{{u}^{2}}du$ $⩽\frac{{\pi }^{2}ϵ}{4}{\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}u}{{u}^{2}}\phantom{\rule{0.167em}{0ex}}du=\frac{{\pi }^{3}ϵ}{8}$ 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?