The following integral is divergent:
$$\int _1^{\infty }\:\frac{\ln\left(x\right)}{x+1}\:\sin\left(2x\right)dx$$
I am struggling to understand why, because it passes Dirichlet test as $\sin(2x)$ is bounded and $\frac{\ln\left(x\right)}{x+1}$ is positive in the range, and monotonous down to zero.
I guess I have here some wrong assumptions - I'd like to understand what they are.
Thanks,
Dirichlet's test claims that for two continuous functions $f,g\in[a,\infty]$ where $f,g\geq 0$, if a certain $M$ exists such that $\left|\int_a^bf(x)dx\right|\leq M$ for every $a\leq b$, and $g(x)$ is monotonically decreasing, and $\lim_{X\to\infty}g(x)=0$, then $\int_a^\infty fg$ is convergent.
So let's check this here, with $f(x)=\sin 2x$ and $g(x)=\frac{\log x}{1+x}$. The function $g(x)$ decreases as soon as $x\geq e$, and we have $\lim_{x\to\infty}g(x)=0$. Moreover, for all $b\geq 1$ $$\left|\int_ 1^b\sin 2x\,dx\right|=\frac{1}{2}|\cos 2-\cos 2b|\leq 1$$ It follows from Dirichlet's test that the integral $\int_1^{\infty}\frac{\log x}{1+x}\sin 2x\,dx$ converges.
So it seems your textbook has it wrong.