I am stuck on some other problems in introductory measure theory on the convergence theorems (monotone convergence theorem and dominated convergence theorem).
The exercise asks to compute the limit as $n\to\infty$ of the following integrals.
$$(1)\quad\int_{(0,\infty)}\frac{\sin x}{x^2}\frac{x^{1/n}}{1+x^{1/n}}\,dx$$
$$(2)\quad \hspace{13pt}\int_{(0,\infty)} \frac{\sin (nx^n)}{nx^{n+\frac12}}\,dx$$
($1$) The monotone convergence theorem clearly doesn't apply since $\sin x$ changes sign on $(0,\infty)$. My hope goes to the dominated convergence theorem.
For all $x\in(0,\infty)$, $\frac{x^{1/n}}{1+x^{1/n}}=\frac{1}{1+\frac{1}{x^{1/n}}}\xrightarrow{n\to\infty} \frac12$, so by positivity it is bounded by $\frac12$ for all $x\in(0,\infty)$ and all $n$.
If I cut the integral into $\int_{0}^1$ and $\int_1^\infty$, then the second part is easy. Indeed, for all $n$ and for all $x\in [1,\infty)$, by the above remark $f_n(x)\xrightarrow{n\to\infty} \frac{\sin x}{2x^2}$; furthermore $|f_n(x)|\leqslant \frac{1}{2x^2}$ which is integrable, so this part follows by the dominated convergence theorem.
However, I am stuck at what to do with $\int_0^1 f_n(x)\,dx$, as I seem to keep the $x$ in the denominator which avoids integrability.
I tried to use the inequality $\left\vert \frac{x^{1/n}}{1+x^{1/n}} \right\vert\leqslant \frac{1}{1+x}$, but for the $\frac{\sin x}{x^2}$ term, the only inequality we can use is $\sin x\leqslant x$ and so we will always keep $x$ in the denominator.
($2$) Here I have the same problem, $|\sin (nx^n)|\leqslant nx^{n}$ for all $x\in (0,\infty)$ and all $n$, but then $\left\vert \frac{\sin (nx^n)}{nx^{n+\frac12}} \right\vert\leqslant \frac{nx^n}{nx^{n+\frac12}}=\frac{1}{x^{1/2}}$, from where we have nowhere to go.
I hope there is some slick trick I don't know about. Any help is welcome.
Best Answer
$$f_n(x)>\frac{c}{x}\frac{x^{1/n}}{2}.$$
Since $\int_0^1x^{1/n-1}\,dx = n,$ $\int_0^1 f_n\to \infty.$ Since you've already shown $\int_1^\infty f_n$ converges, it follows that $\int_0^\infty f_n \to \infty.$
$$|f_n(x)| \le \frac{1}{x^{1/2}}.$$
Also on $(0,1)$ $nx^n\to 0,$ hence $\sin (nx^n)/(nx^n) \to 1.$ It follows by the DCT that
$$\int_0^1 f_n \to \int_0^1 \frac{1}{x^{1/2}}\, dx = 2.$$
Now $\int_1^\infty f_n$ is a different kettle of fish, but it's actually easier than the above. Take a go at it.