Evaluation of $\displaystyle \lim_{u\rightarrow \infty}\frac{\int^{\pi u}_{1}\frac{\sin^2(5x)}{x}dx}{\ln(u^2+u^{-2})}$
What I Try: Using newton leibniz formula
$\displaystyle \lim_{u\rightarrow \infty}\frac{\bigg(\int^{\pi u}_{1}\frac{\sin^2(5x)}{x}dx\bigg)'}{\bigg(\ln(u^2+u^{-2})\bigg)'}$
$\displaystyle \lim_{u\rightarrow \infty}\frac{\frac{\sin^2(5\pi u)}{\pi u}\cdot\pi -0\cdot 0 }{\frac{1}{u^2+u^{-2}}\cdot (2u-2u^{-3})}$
$\displaystyle \lim_{u\rightarrow \infty}\frac{\sin^2(5\pi u)}{5\pi u}\cdot \frac{(u^2+u^{-2})}{(2u-2u^{-3})}=0$
But answer is $\displaystyle \frac{1}{4}$.
Please have a look on that problem , Thanks
Best Answer
The last limit is not zero, it just doesn't exists, note that $$ \begin{align*} \lim_{u\to \infty }\frac{\sin ^2(5\pi u)}{5\pi u}\cdot \frac{u^2+u^{-2}}{2u-2u^{-3}}&=\frac1{10\pi}\lim_{u\to \infty }\sin ^2(5\pi u)\lim_{u\to \infty }\frac{u^2+u^{-2}}{u^2-u^{-2}}\\&=\frac1{10\pi}\lim_{u\to \infty }\sin ^2(5\pi u) \end{align*} $$
And as the sine function is periodic, is clear that the last limit doesn't exists. However if we assume that $u\in \mathbb{N}$ and the original limit represents the limit of a sequence then, applying Stolz-Cesáro's theorem to the original limit expression, it can be shown that the limit of this sequence is indeed $1/4$.
Suppose that the given limit is a functional limit, then noticing that $\ln(u^2+u^{-2})\sim_{\infty }2\ln u$ and that $\sin ^2(5x)=\frac{1-\cos (10x)}{2}$ then for any chosen $a>0$ we find that $$ \lim_{u\to \infty }\frac1{\ln (u^2+u^{-2})}\int_{a}^{\pi u}\frac{\sin ^2(5x)}{x}\,d x=\lim_{u\to \infty }\frac1{4\ln u}\int_{10a}^{10\pi u}\frac{1-\cos t}{t}\,d t $$
using the change of variable $10x=t$. However the last limit is easy to see that is $1/4$ as $\left|\int_{a}^{\infty }\frac{\cos t}{t}\,d x\right|<\infty$ for any chosen $a>0$.∎