Conditional and absolute convergence of an integral If you know that $$\int\limits_x^{+\infty} e^{-t^2}\mathrm dt=e^{-x^2}\biggl(\frac{1}{2x}+o\biggl(\frac{1}{x^2}\biggl)\biggl)$$
examine conditional and absolute convergence of this integral: $$\int\limits_1^{+\infty}\biggl(e^{x^2}\cos{\biggl(x+\frac {1}{x}\biggl)}\int\limits_x^{+\infty}e^{-t^2}\mathrm dt\biggl)\mathrm dx.$$
$\cos{\biggl(x+\frac{1}{x}\biggl)}$ we can write as $\cos{\biggl(x+\frac{1}{x}\biggl)}=\cos x \cos \frac{1}{x}-\sin x \sin \frac {1}{x}$ and by inserting the above formula we get that our integral is: 
$$\int\limits_1^{+\infty}\biggl(e^{x^2}\cos{\biggl(x+\frac {1}{x}\biggl)}\int\limits_x^{+\infty}e^{-t^2}\mathrm dt\biggl)\mathrm dx=\frac {\cos x}{2x}\cos\frac{1}{x}-\frac {\sin \frac {1}{x}}{2x}\sin x+o \biggl(\frac {1}{x^2}\biggl), x \to +\infty$$
I think that $\int\limits_1^{+\infty}\frac {\cos x}{2x} \mathrm dx$ converges by Dirichlet, but I don't know how to show that $\cos x$ has limited partial sums. Also, if $\cos {\frac{1}{x}}$ is monotone and limited how I can show that? So, I believe that $\int\limits_1^{+\infty}\frac {\cos x}{2x}\cos \frac {1}{x} \mathrm dx$ converges by Abel. 
Absolute convergence of $\int\limits_1^{+\infty}\frac {\cos x}{2x}\cos \frac {1}{x}$:
If someone know why this is true:
$$|\frac {\cos x}{2x}\cos \frac {1}{x}| \sim \frac {|\cos x|}{2x}$$
and $\int\limits_1^{+\infty}\frac {|\cos x|}{2x} \mathrm dx$ diverges. So the convergence of $\int\limits_1^{+\infty}\frac {\cos x}{2x}\cos \frac {1}{x} \mathrm dx$ is conditional.
And this one, I don't understand:
$$|\frac {\sin \frac{1}{x}}{2x}\sin x| \le \frac {|\sin {\frac{1}{x}}|}{2x} \sim \frac{1}{2x^2}$$
so $\int\limits_1^{+\infty} \frac {\sin \frac{1}{x}}{2x}\sin x \mathrm dx$ absolute converges by the comparative criterion. $\int\limits_1^{+\infty} o \biggl(\frac {1}{x^2}\biggl) \mathrm dx$ also converges by the comparative criterion. 
So, the conclusion is  $$\int\limits_1^{+\infty}\biggl(e^{x^2}\cos{\biggl(x+\frac {1}{x}\biggl)}\int\limits_x^{+\infty}e^{-t^2}\mathrm dt\biggl)\mathrm dx.$$
converges conditional.
 A: Honestly, you did most of the hard work here. So let's conclude your analysis.
First, about $\int_1^\infty \frac{\cos x}{2x} dx$, you have
$$
\int_1^\infty \frac{\cos x}{2x} dx = \int_1^{\frac\pi2} \frac{\cos x}{2x} dx+\sum_{n=0}^\infty \int_{n\pi+\frac\pi2}^{(n+1)\pi+\frac\pi2} \frac{\cos x}{2x} dx,
$$
where $\int_{n\pi+\frac\pi2}^{(n+1)\pi+\frac\pi2} \frac{\cos x}{2x} dx$ goes to $0$ when $n\to\infty$, and is positive if $n$ is odd and negative otherwise. Hence, using  the alternating series test, you deduce that the sum (and hence the integral) is convergent. By your analysis, you deduce that the integral $\int\limits_1^{+\infty}\biggl(e^{x^2}\cos{\biggl(x+\frac {1}{x}\biggl)}\int\limits_x^{+\infty}e^{-t^2}\mathrm dt\biggl)\mathrm dx$ is conditionally convergent.
Second, $\cos u\to1$ when $u\to 0$ and $\frac1x\to 0$ when $x\to\infty$, hence $\cos\frac1x\to 1$ when $x\to \infty$, so that $|\frac {\cos x}{2x}\cos \frac {1}{x}| \sim \frac {|\cos x|}{2x}$. To show that $\int\limits_1^{+\infty}\frac {|\cos x|}{2x} \mathrm dx$ diverges, use the same decomposition as above to obtain
$$
\int_1^\infty \frac{|\cos x|}{2x} dx \geq \sum_{n=0}^\infty  \int_{n\pi+\frac\pi2}^{(n+1)\pi+\frac\pi2}\frac{|\cos x|}{2((n+1)\pi+\frac\pi2)}dx=\sum_{n=0}^\infty  \frac{1}{(n+1)\pi+\frac\pi2}=+\infty.
$$
Finally, $\sin u\sim u$ when $u\to 0$, and $\frac{1}{x}\to 0$ when $x\to \infty$, so that $\sin \frac1x\sim \frac1x$ when $x\to \infty$. Per your analysis, we deduce that the integral $\int\limits_1^{+\infty}\biggl(e^{x^2}\cos{\biggl(x+\frac {1}{x}\biggl)}\int\limits_x^{+\infty}e^{-t^2}\mathrm dt\biggl)\mathrm dx$ is not absolutely convergent.
