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Here it is function,

$$\lim_{n\rightarrow \infty}\int\limits_{0}^{\pi}\sqrt{\frac{t}{n}}\sin\sqrt{\frac{n}{t}}\;dt$$

but I read about Dominated convergence theorem and I don't know how to implement it to provie that that limit equals zero?

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Step A. Set $\,\displaystyle f_n(t)=\sqrt{\frac{t}{n}}\sin\sqrt{\frac{n}{t}},$ and observe that $$ \lvert\, f_n(t)\rvert= \left|\sqrt{\frac{t}{n}}\sin\sqrt{\frac{n}{t}}\,\right|\le 1=g(t),$$ for all $t\in[0,\pi]$ and $n\in\mathbb N$, since $$ \lvert \sin x\rvert\le \lvert x\rvert, $$ for all $x$, and clearly $\int_0^\pi g(t)\,dt=\pi<\infty$.

Step B. Next, we have that $$\lim_{n\to\infty}f_n(t)=\lim_{n\to\infty}\frac{\sin\sqrt{\frac{n}{t}}}{\sqrt{\frac{n}{t}}}=0, $$ for all $t\in [0,\pi]$.

Thus, Lebesgue Dominated Convergence Theorem is applicable and provides that $$ \lim_{n\to\infty}\int_0^\pi f_n(t)\,dt=\int_0^\pi\big(\lim_{n\to\infty}f_n(t)\big)\,dt=0. $$

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  • $\begingroup$ @yiorgos-s-smyrlis I think you wronged somewhere since the idea is to prove that equals zero?! $\endgroup$ – JohnvNeumann Nov 2 '14 at 22:21
  • $\begingroup$ There should be a mistake, because $\lim_{x\rightarrow \infty} \frac{sin(x)}{x} = 0$ $\endgroup$ – Marm Nov 2 '14 at 22:21
  • $\begingroup$ You are both right. The answer is indeed zero. $\endgroup$ – Yiorgos S. Smyrlis Nov 2 '14 at 22:23
  • $\begingroup$ @YiorgosS.Smyrlis Perfect! Thanks a lot! $\endgroup$ – JohnvNeumann Nov 2 '14 at 22:24
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Hint:

a) Find an integrable function $g$ on the given domain such that $|\sqrt{\frac{t}{n}}\sin(\sqrt{\frac{t}{n}})|$ $\leq g$ for all $n\in N$ almost everywhere.

b) Exchange limit and integral operations

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No need for heavy machinery here. Note that $$\left|\int\limits_0^\pi\sqrt\frac tn\sin\sqrt\frac nt\;dt\right| \le \int_0^{\pi} \left| \sqrt\frac tn\sin\sqrt\frac nt \right|\,dt$$ Since $|\sin(x)| \leq 1$ for all $x$ this is at most $$\int_0^{\pi} \sqrt\frac tn\,dt$$ $$= {1 \over \sqrt{n}} \int_0^{\pi} \sqrt{t}\,dt$$ You don't even have to do the integral here; take limits as $n$ goes to $\infty$ and get 0.

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This doesn't use the Dominated Convergence Theorem, but suppose we take

$$F(n) = \int\limits_0^\pi \sqrt{\frac{t}{n}}\sin\sqrt{\frac{n}{t}}\,dt$$

By the intermediate value theorem for integrals,

$$F(n) = \pi\sqrt{\frac{\tau}{n}}\sin\sqrt{\frac{\tau}{n}}$$

for some $\tau \in (0, \pi)$.

It follows that $|F(n)| < \frac{\pi^{\frac{3}{2}}}{n^{\frac{1}{2}}}$. (In fact, since $\sin$ is nonnegative over the domain of integration, $0 \le F(n)$, but we don't need this tighter bound). From here, the proof that $F(n) \to 0$ an $n \to +\infty$ follows from a quick application of the squeeze theorem.

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I don't know about the DCT, but what about

$$\left|\int\limits_0^\pi\sqrt\frac tn\sin\sqrt\frac nt\;dt\right|\le\sqrt\frac1 n \int\limits_0^\pi \sqrt t\;dt=\frac2{3\sqrt n}\pi\sqrt\pi\xrightarrow[n\to\infty]{}0\;\;\;?$$

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