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Prove that $\displaystyle\int_0^1\frac{n\cos x}{1+x^2n^{\frac32}}dx\rightarrow0$ as $n\rightarrow\infty$.

$f_n(x)=\frac{n\cos x}{1+x^2n\sqrt{n}}$ tends to zero function pointwise. It just converges uniformly on every compact interval $[\epsilon,1]$, but$f_n(0)=n$ tends to $+\infty$ and we can not use replacement of limit and integral with this reasoning.

So, the other useful method to handle limit of integrals, is Lebesgue Dominated Convergence Theorem. But I can't still find the proper dominating nonnegative and measureble function over $[0,1]$ for the sequence $\{f_n\}$.

First it seems the term $\cos x$ doesn't have impact on the result. Then for each $p<1$, there's natural $m$ that $f_m(x)$ is not dominated by $g(x)=\frac{1}{x^p}$ on the whole interval. (I observed it via plotting).

I think in the process of finding a good inequality, we must generate $n$ in denominator and use $n>1$.

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    $\begingroup$ The $\cos x$ does appear to have an impact on the result. If we remove it, then the value of the integral becomes $n^{\frac{1}{4}} \tan^{-1} (n^{\frac{3}{4}}) \geq \frac{\pi}{4} n^{\frac{1}{4}} \to \infty$. $\endgroup$ – Shalop Apr 6 '15 at 21:32
  • $\begingroup$ yes, you're actually right $\endgroup$ – Fardad Pouran Apr 6 '15 at 21:36
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    $\begingroup$ On the other hand, $\cos x \geq \cos 1 >0$ for all $x \in [0,1]$, so $\int_0^1 \frac{n \cos x}{1+n^{1.5}x^2} dx \geq \cos 1 \int _0^1 \frac{n}{1+n^{1.5}x^2}dx =\cos 1 \cdot n^{\frac{1}{4}} \tan ^{-1}(n^{\frac{3}{4}})\geq \frac{\pi}{4} \cos 1 \cdot n^{\frac{1}{4}} \to \infty.$ Maybe I made a mistake? $\endgroup$ – Shalop Apr 6 '15 at 21:42
  • $\begingroup$ If we interchange $n$ and $x$ in the denominator, the result comes out. Because of AM-GM, you can find the dominating function $\displaystyle g(x)=\frac{1}{2x^{\frac34}}$ . :) $\endgroup$ – Fardad Pouran Apr 6 '15 at 21:54
  • $\begingroup$ @Shalop, In the result of the final integral you've written above, the term $\tan^{-1}(n^{\frac34})-\frac{\pi}{4}$ must be written rather that just $\tan^{-1}(n^{\frac34})$. But still $n^{\frac14}(\tan^{-1}(n^{\frac34})-\frac{\pi}{4})$ goes to infinity $\endgroup$ – Fardad Pouran Apr 6 '15 at 22:05
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Make the substitution: $x = \left(\dfrac{t}{n^{3/2}}\right)^{1/2}$

Then, $\displaystyle \int_0^1\frac{\cos x}{1+x^2n^{\frac{3}{2}}}\,dx = \frac{n^{-3/4}}{2}\int_0^{n^{3/2}}\frac{t^{-1/2}}{1+t}\cos \left(\dfrac{t}{n^{3/2}}\right)^{1/2}\,dt$

Consider, the sequence of functions $f_n: [0,\infty] \to \mathbb{R}$:

$\displaystyle f_n(t) = \frac{t^{-1/2}}{1+t}\cos \left(\dfrac{t}{n^{3/2}}\right)^{1/2}\chi_{n}(t)$, where, $\displaystyle \chi_n(t) = \begin{cases}1,\textrm{ when } t \in [0,n^{3/2}] \\ 0, \textrm{ otherwise } \end{cases}$

Then, $\displaystyle \lim\limits_{n \to \infty} f_n(t) = \frac{t^{-1/2}}{1+t}\cos 0 = \frac{t^{-1/2}}{1+t}$ and $f_n$ is bounded by an integrable function:

$\displaystyle |f_n(t)| \le \frac{t^{-1/2}}{1+t}$

Then, it follows from the Lebesgue Convergence Theorem:

$\displaystyle \begin{align} \lim\limits_{n \to \infty} \int_0^1\frac{n^{3/4}\cos x}{1+x^2n^{\frac{3}{2}}}\,dx &= \lim\limits_{n \to \infty} \frac{1}{2}\int_0^{\infty}\frac{t^{-1/2}}{1+t}\cos \left(\dfrac{t}{n^{3/2}}\right)^{1/2}\chi_n(t)\,dt \\ &= \lim\limits_{n \to \infty}\frac{1}{2}\int_0^{\infty} \frac{t^{-1/2}}{1+t}\,dt = \frac{\pi}{2}\end{align}$

Hence, $\displaystyle \int_0^1\frac{n\cos x}{1+x^2n^{\frac{3}{2}}}\,dx \sim \frac{\pi}{2}n^{1/4}$ (diverges).

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