Dominated convergence theorem (computing limit)

Question

I need to compute $\displaystyle\lim_{n \rightarrow \infty} \int \frac{\sin (x^n)}{x^2} \, dx$ using Dominated Convergence theorem. I have taken the function $g$ such that $|f_n| \leq g$ , where $f_n =\dfrac{\sin (x^n)}{x^2} $ to be $\dfrac{1}{x^2}$. I am not sure how to proceed forward. Do I need to find another function $f$ s.t. $f_n \rightarrow f $ almost everywhere. If not then what should be my approach?

Answer 1

I think that you must do some changes before using DCT.

a) Change of variable $u=x^n$. Your integral is now $$I_n=\int_0^{+\infty}\frac{\sin(x^n)}{x^2}dx=\frac{1}{n}\int_0^{+\infty}\frac{\sin(u)}{u^{1+1/n}} du$$

b) Integration by parts, using $(1-\cos(u))^{\prime}=\sin(u)$ :

$$I_n=\frac{n+1}{n^2}\int_0^{+\infty}\frac{1-\cos(u))}{u^{2+1/n}}du$$

And now you have to use the DCT on $\displaystyle \int_0^{+\infty}\frac{1-\cos(u))}{u^{2+1/n}}du$. Note that $\displaystyle |1-\cos(u)|=2\sin(u/2)^2\leq \frac{u^2}{2}\leq u^2$, and as $n\geq 2$, we have $\displaystyle \frac{1}{u^{1/n}}\leq \frac{1}{\sqrt{u}}$ on $[0,1]$.