Showing convergence of the integral $\int_{1}^{\infty}\frac{\ln(x)\cos(x)}{x^2+1}\,{\rm d}x$

Question

I need to show that the following integral either converges or diverges.

$$\int_{1}^{\infty}\frac{\ln(x)\cos(x)}{x^2+1}\,{\rm d}x$$

I am fairly certain it converges, but am stuck on showing how.

Because $\cos(x)<1$ for all $x$, we know that

$$\frac{\ln(x)\cos(x)}{x^2+1}\leq \frac{\ln(x)}{x^2+1}$$

Help on the next step would be great.

Answer 1

You are just missing one more piece.

$\frac{\ln x}{x^2+1}<\frac{\sqrt{x}}{x^2+1}\sim\frac{1}{x^{3/2}}$

Since $\ln x$ grows slower than any power of $x$, the integral converges.

Answer 2

$$\left|\frac{\ln x \cos x}{x^2+1}\right|\leq\frac{\ln x}{x^2+1}\leq\frac{x^{0.5}}{x^2}=\frac{1}{x^{1.5}}$$ for for all $x\geq 1$ which means: $$0\leq\int_1^{\infty}\left|\frac{\ln x \cos x}{x^2+1}\right|\leq\int_1^{\infty}\frac{1}{x^{1.5}}$$

By the integral test, $\int_1^{\infty}\frac{1}{x^{1.5}}$ converges since $\sum_{n=1}^{\infty}\frac{1}{n^{1.5}}$ converges ($p$-series with $p>1$) and hence $$\int_1^{\infty}\left|\frac{\ln x \cos x}{x^2+1}\right| \;\text{converges which implies}\int_1^{\infty}\frac{\ln x \cos x}{x^2+1} \;\text{converges}$$.