Does $\int_0^\infty \frac{\cos x}{1+x}$ absolutly converge?

Question

Does the following indefinite integral converge? $\int_0^\infty \frac{\cos x}{1+x}$

converges, absolutely converges?

i can say that by the Dirichlet test it does converge.

i am trying to prove it diverges absolutely (seems close to $\frac{1}{x}$) unsuccessfully so far

how do i prove /disprove it absolutely converge?

Answer 1

Consider intervals $[2k\pi-\frac{\pi}{4},2k\pi+\frac{\pi}{4}]$ for $k \in \mathbb{N}$, then:

$$\int_{0}^{\infty}\frac{|\cos(x)|}{1+x}dx>\int_{\bigcup_{k \in \mathbb{N}}[2k\pi-\frac{\pi}{4},2k\pi+\frac{\pi}{4}]}\frac{|\cos(x)|}{1+x}dx=\sum_{k=1}^{\infty}\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{|\cos(x)|}{1+x}dx$$

But on each interval $[2k\pi-\frac{\pi}{4},2k\pi+\frac{\pi}{4}]$ we have $\cos(x)>\frac{1}{2}$, so:

$$\sum_{k=1}^{\infty}\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{|\cos(x)|}{1+x}dx>\sum_{k=1}^{\infty}\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{1}{2(1+x)}dx$$

Now we have:

$$\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{1}{2(1+x)}dx>\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{1}{2(1+2k\pi+\frac{\pi}{4})}dx=\frac{\pi}{2}\frac{1}{2(1+2k\pi+\frac{\pi}{4})}$$

So:

$$\sum_{k=1}^{\infty}\int_{2k\pi-\frac{\pi}{4}}^{2k\pi+\frac{\pi}{4}}\frac{1}{2(1+x)}dx>\sum_{k=1}^{\infty}\frac{\pi}{2}\frac{1}{2(1+2k\pi+\frac{\pi}{4})}=\infty$$

Answer 2

In an open neighbourhood of $\pi\mathbb{Z}\cap\mathbb{R}^+$ we have $|\cos x|\geq\frac{1}{2}$, so the integral cannot be absolutely convergent, since $$\int_{\pi\mathbb{Z}+B(0,\epsilon)} \frac{dx}{|1+x|}$$ is not convergent.

Answer 3

Hint. For the convergence, you have, for $M>0$, integrating by parts: $$ \int_0^M \frac{\cos x}{x+1}dx=\left. \frac{\sin x}{x+1}\right|_0^M+\int_0^M \frac{\sin x}{(x+1)^2}dx $$ since $$ \left|\int_0^M \frac{\sin x}{(x+1)^2}dx\right|\leq\int_0^{+\infty} \frac{1}{(x+1)^2}dx<+\infty $$ giving that your initial integral is convergent.

The integral is not absolutely convergent, as showned by Jack d'Aurizio.