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Recently I've come across two integrals that seemed hard to check for me. Here they are: $$\int_0^\infty \frac{x \sin \ln x}{x^2 + \cos x} \, \mathrm{d}x$$ And another: $$\int_1^\infty \frac{\sin \ln x}{(\ln x)^3} \, \mathrm{d}x$$

I've already figured out how to prove the first one's convergence upon $\int_0^e$, but have no proper ideas to let me proceed to infinity as applying equivalence is restricted by sign changes of sinus. But I'm convinced they both diverge :) Any help?

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Look at the asymptotics as $x\rightarrow \infty$ on the first integral

\begin{equation} \frac{x\sin(\ln(x))}{x^2+\cos(x)} \sim \frac{x\sin(\ln(x))}{x^2} = \frac{\sin(\ln(x))}{x} \end{equation}

If we change variables

\begin{equation} \int_{N}^{\infty} \frac{\sin(\ln(x))}{x} dx = \int_{\ln(N)}^{\infty} \sin(u) du \end{equation}

For the second integral, use a change of variables

\begin{equation} \int_{N}^{\infty} \frac{\sin(\ln(x))}{(\ln(x))^3} dx = \int_{\ln(N)}^{\infty} \frac{e^{u}}{u^3}\sin(u)du \end{equation}

It seems that both diverge, but this is all imprecise. The details would need to be filled in. If you restrict to intervals of the form $[e^{\pi M},e^{\pi M+\pi}]$, then the sign in the numerator with be either positive or negative and you can make the necessary comparisons.

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  • $\begingroup$ your reply is much appreciated :) $\endgroup$ – GrapeGreen May 25 '16 at 20:06

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