$2$ equal improper integrals but one integral converges absolutely other is not

Question

I can show by parts
$$\int_{0}^{\infty}\frac{\cos x}{1+x}dx=\int_{0}^{\infty}\frac{\sin x}{(1+x)^2}dx$$.
But How to argue that one of them is converging absolutely and other is not?
Is it possible to argue using integral test ? like $\int_{0}^{\infty}\frac{\sin x}{(1+x)^2}dx$ $\leq \int_{0}^{\infty}\frac{|\sin x|}{(1+x)^2}dx$ which is convergent by integral test as $\sum_{i=0}^{\infty} \frac{1}{(1+x)^2}$ is convergent .
Or there is other way to show .Any Help will be appreciated

Answer 1

The second integrand has absolute value bounded by $1/(1+x)^2$ which has a finite integral. On the other side one has absolute value $$\frac{|\cos x|}{1+x}.$$ The numerator has period $\pi$, so that $$\int_{n\pi}^{(n+1)\pi}\frac{|\cos x|}{1+x}\,dx \ge\frac1{1+(n+1)\pi}\int_{n\pi}^{(n+1)\pi}|\cos x|\,dx =\frac{A}{1+(n+1)\pi}$$ where $$A=\int_0^\pi|\cos x|\,dx>0.$$ So $$\int_0^\infty\frac{|\cos x|}{1+x}\,dx\ge\sum_{n=0}^\infty \frac A{1+(n+1)\pi}$$ which is a divergent series.

Answer 2

An alternative approach: the absolute convergence of $\int_{0}^{+\infty}\frac{\sin(x)}{(x+1)^2}\,dx$ is trivial, but the function $\left|\cos x\right|$ has a positive mean value ($\frac{2}{\pi}$), hence the integral $\int_{0}^{+\infty}\frac{\cos x}{x+1}\,dx$ is not absolutely convergent by Kronecker's lemma.