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Let $p<0$. Is it true (and if it is, then how to prove) that the improper integral: $$ \int_1^{+\infty}\frac{\sin(e^x)dx}{\ln^p(1+x)} $$ is conditionally divergent?

As far as I know, the integral: $$ \int_1^{+\infty}f(x)dx $$ can be conditionally convergent even if $f(x)$ is continuous but not bounded on $[1,+\infty)$. Therefore, I can't use that $f(x)$ in the integral above is not bounded to prove its divergence. Are there another ways to show it?

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Suggestion: If what you are asking is whether $\int^\infty_1 \Big|\frac{\sin(e^x)}{\log^p(1+x)}\Big|\,dx$

the change of variable $u=e^x$ gives you $$ \int^\infty_e \frac{|\sin u|}{u\log^p(1+\log u)}\,du$$ The function $\log^p(\log u)$ increases very slowly in comparison to $u$ and (this needs a little justification), the integrand should be similar to $\frac{|\sin u|}{u}$.


Here is a sketch of a solution. Details are left to the OP.

Since $p<0$,

\begin{align} \int^\infty_e \frac{|\sin u|\log^{-p}(1+\log u)}{u}\,du&\geq \log^{-p}(2) \sum^\infty_{n=1}\int^{(n+1)\pi}_{n\pi}\frac{|\sin u|}{u}\,du\\ &\geq \log^{-p}(2)\sum^\infty_{n=1}\frac{1}{\pi(n+1)}\int^{(n+1)\pi}_{n\pi}|\sin u|\,du\\ &=\frac{ \log^{-p}(2)}{\pi}\Big(\int^\pi_0|\sin u|\,du\Big)\sum^\infty_{n=2}\frac{1}{n} \end{align}


As for the existence of the limit $\lim_{a\rightarrow\infty}\int^\infty_e \frac{\sin u\log^{-p}(1+\log u)}{u}\,du$, a similar method can break the integral as an alternating series for which you may use standard results.

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