# Dominated convergence theorem (computing limit)

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?

• Just as a side remark: $g(x) = \frac{1}{x^2}$ is not a good choice for the dominating function as $g$ is not integrable: $$\int g(x) \, dx \geq \int_0^1 g(x) \, dx = \infty.$$ – saz Nov 21 '15 at 6:31
• Sorry, I forgot that g needs to be integrable. Can you suggest any other? – user262860 Nov 21 '15 at 6:33
• For example $g(x) := 1$ for $|x| \leq 1$ and $g(x) := \frac{1}{x^2}$, $|x| \geq 1$ does the job. To see that this is a dominating function (for $|x| \leq 1$) use the elementary inequality $|\sin(x)| \leq |x|$ (which follows e.g. directly from the mean value theorem). And are you sure that you have to compute the limit for the integrand $\frac{\sin(x^n)}{x^2}$ and not $\frac{(\sin (x))^n}{x^2}$....? – saz Nov 21 '15 at 6:41
• Ok. after that should I compute $\displaystyle\lim_{n \rightarrow \infty} \frac{\sin (x^n)}{x^2} \ = f$ and then compute the integral of $f$ which will give me my answer? – user262860 Nov 21 '15 at 6:46
• It is not so clear if the function converges pointwisely at all to any function when $x>1$/ – user99914 Nov 21 '15 at 7:13

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 $[0,1]$.