Compute $\lim\limits_{ n \to \infty}{\int_0^n{\frac{x\sin(1/(nx))}{\sqrt{(x^2+1)}}}}$

Question

$$ \lim_{n \to \infty}\int_{0}^{n}x\,\sin\left(1 \over nx\right)\, {\mathrm{d}x \over \,\sqrt{\, x^{2} + 1\,}\,} $$

• I tried to use the Dominated Convergence Theorem but I get that for $0\leq x \leq 1$, i.e. for $n = 1$ the dominating function is $x/\sqrt{\, x^{2} + 1\,}$ while for $n \geq 2$ I cannot find a dominating function which is integrable i.e. $g \in L^{1}$, as $x/\sqrt{\,x^{2} + 1\,}$ has integral equal to infinity for $x \geq 1$.
• Also, for $n \geq 2$, I could only get that the functions $f_{n}$ are less than one, so I couldn't find a dominating function. Could you help me to solve this problem ?.

Thank you !.

Answer 1

Hint: The change of variables $x = 1/(ny)$ shows the integral equals

$$\frac{1}{n}\int_{1/n^2}^\infty \frac{\sin y}{\sqrt { 1+(ny)^2}y^2}\, dy.$$

Answer 2

Herein, we present a way forward, which is less elegant, but as effective as changing variables.

Note that we can write simply

$$\begin{align} \left|\int_0^n \frac{x\sin(1/(nx))}{\sqrt{x^2+1}}\,dx\right|&\le \frac1n\int_0^n \frac{1}{\sqrt{x^2+1}}\,dx\\\\ &=\frac1n \log\left(n+\sqrt{n^2+1}\right)\\\\ &\to 0\,\,\text{as}\,\,n\to \infty \end{align}$$

And we are done!