Calculate $\lim_{n\rightarrow \infty}\int_{[0,1]}\frac{n\cos(nx)}{1+n^2 x^{\frac{3}{2}}}$

Question

I have tried several methods but even I can not calculate.

$$\lim_{n\rightarrow \infty}\int_{[0,1]}\frac{n\cos(nx)}{1+n^2 x^{\frac{3}{2}}}\,dx$$ If anyone can help, it was part of a test and still I can not solve.

Answer 1

Note that

$$\begin{align*} \frac{n \cos(nx)}{1+n^2 x^{3/2}} &\leq \frac{n}{n^2x^{3/2}} \\ &= x^{-3/2} \frac{1}{n} &&(1) \\ \end{align*}$$

Since $2ab \leq a^2+b^2$, it follows (with $a =1$ and $b=nx^{3/4}$)

$$ \begin{align*} \frac{n \cos(nx)}{1+n^2 x^{3/2}} &\leq \frac{n}{2nx^{3/4}} \\ & = \frac{1}{2} x^{-3/4} &&(2) \end{align*} $$ You can use this and the dominated convergence theorem to get that the limit is 0.

(1) gives you the (almost everywhere) pointwise convergence to 0.

(2) shows that the sequence is dominated by an integrable function.

Answer 2

Consider, $$f(x)=\frac{n}{1+n^2x^{3/2}}.$$

Then, $f$ is continuous in $[0,1]$ and so Riemann integrable in $[0,1]$.

So, $$\lim_{n\to \infty}\int_0^1f(x)\cos(nx)\,dx=0$$

Answer 3

We have $$|f_n (x)|=\left|\frac{n\cos (nx) }{1+n^2 \sqrt{x^3} }\right|\leq \frac{n }{1+n^2 \sqrt{x^3} }\leq \frac{n}{2\sqrt{1\cdot n^2 \sqrt{x^3}}} =2^{-1} x^{-\frac{3}{4}}=\varphi (x)$$ and $$\int_0^1\varphi (x) dx =\int_0^1 2^{-1} x^{-\frac{3}{4}} dx =2<\infty$$ therefore the sequence $(f_n ) $ is dominated by integrable function $\varphi .$

Moreover $$f(t)=\lim_{n\to\infty} f_n (t)=\begin{cases} 0 \mbox{ for } 0<t\leq 1\\ \infty \mbox{ for } t=0\end{cases}$$ thus by Lebesgue Dominated Convergence theorem $$\lim_{n\to\infty}\int_0^1 f_n (t)dt =\int_0^1 f(t) dt =0.$$