$\displaystyle\int_{0}^{\infty}\dfrac{\cos x}{\ln x}\mathrm{dx}$ determine the convergence or divergence of the integral.
So I tried to distinguish the interval of the integral:
$$\displaystyle\int_{0}^{1}\dfrac{\cos x}{\ln x}\mathrm{dx}+\displaystyle\int_{1}^{\infty}\dfrac{\cos x}{\ln x}\mathrm{dx}$$
And the first integral is proper since $\displaystyle\lim_{x \to 0^{+}} \dfrac{\cos x}{\ln x}=0$. But the other ones I can't evaluate those. Thanks!
Both integrals diverge. Note that $$ \frac{\cos z}{\log z} = \frac{\cos 1}{z-1} + \left(\frac{\cos 1}{2} - \sin 1\right) + O(|z-1|)$$ In particular, $$\lim_{a\to 1^-} \int_0^a \frac{\cos x}{\log x} dx =-\infty$$ and for any $c>1$, $$\lim_{b\to 1^+} \int_b^c \frac{\cos x}{\log x} dx =+\infty$$ so $$\int_0^\infty \frac{\cos x}{\log x} dx$$ has an $\infty-\infty$ singularity at $x=1$. However, it is true that the improper integral $$\int_c^\infty \frac{\cos x}{\log x} dx$$ converges for any $c>1$. You can show this by considering the alternating sum $$\sum_{k=0}^\infty \int_{(k+\frac{1}{2})\pi}^{(k+\frac{3}{2})\pi}\frac{\cos x}{\log x} dx$$
It is interesting to note, that if $\cos x$ were to be replaced by $\cos \frac{\pi x}{2}$, the singularity at $x=1$ would be removed, so the similar-looking integral $$\int_0^\infty \frac{\cos\frac{\pi x}{2}}{\log x} dx$$ is actually convergent.
