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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

jacky
  • 5,194

2 Answers2

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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 $a>0$.∎

Masacroso
  • 30,417
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By plugging the integral in Wolfram alpha we have

$$ \int \dfrac{\sin^2(5x)}{x} dx = \frac{1}{2}\left(\log\left(5x\right)-\operatorname{Ci}(10x)\right) $$ Where $Ci(x)$ is the cosine integral that vanishes at infinity (see here )

So, by dropping the constant terms to the numerator we have

$$ \frac{1}{2} \lim_{u \to \infty} \dfrac{\log(5\pi u)- \operatorname{Ci}(10 \pi u)}{\log(u^4-1)-2\log(u)}=\frac{1}{2} \lim_{u \to \infty} \dfrac{\log( u)}{4\log(u)-2\log(u)}=\frac{1}{4} $$

Marco
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