Showing that $\int_e^\infty \ln x\cdot \cos(e^x)\,dx$ conditionaly converges I look at a proof where they claim such thing: 
$$\int_{e}^{\infty}\ln x\cdot \cos(e^x)\,dx = \left[ t = e^x \right] = \int_{e^e}^{\infty}\frac{\ln(\ln t)}{t}\cos t\,dt $$
Now, they say that the function: 
$$
f(x) = \int_{e^e}^\infty \frac{\ln(\ln t)}{t}\,dt
$$
Diverges using comparison test for $t \geq e^e$ with $1/t$
Now using dirichlet they say that the integral converges. 
Totally they conclude that the integral conditionally converges. 
How did they prove that it doesn't converges for $|\int|$?
They showed that: 
$$
f(x) = \int_{e^e}^{\infty}\frac{\ln(\ln t)}{t}dt
$$
diverges and not the whole expression, the $cost$ is not there, so how they conclude about divergence of the whole integral? 
Thank you. 
 A: Let $I(y)$ be given by 
$$I(y)=\int_e^y \log(x)\cos(e^x)\,dx\tag1$$
After letting $x=\log(t)$ in $(1)$ we find that
$$\begin{align}
\lim_{y\to\infty}I(y)&=\lim_{y\to\infty}\int_{e^e}^{e^y}\frac{\log(\log(t))}{t}\cos(t)\,dt\\\\
&=\lim_{y\to\infty}\left.\left(\frac{\sin(t)\log(\log(t))}{t}\right)\right|_{e^e}^{e^y}-\lim_{y\to\infty}\int_{e^e}^{e^y}\left(\frac{1-\log(t)\log(\log(t))}{t^2\log(t)}\right)\,\sin(t)\,dt\\\\
&=-\frac{\sin(e^e)}{e^e}-\int_{e^e}^{\infty}\left(\frac{1-\log(t)\log(\log(t))}{t^2\log(t)}\right)\,\sin(t)\,dt\tag3
\end{align}$$
We have the estimates
$$\begin{align}
\left|\int_{e^e}^{\infty}\left(\frac{1-\log(t)\log(\log(t))}{t^2\log(t)}\right)\,\sin(t)\,dt\right|&\le \int_{e^e}^\infty \left|\left(\frac{1-\log(t)\log(\log(t))}{t^2\log(t)}\right)\,\sin(t)\right|\,dt\\\\
&\le  \int_{e^e}^{\infty}\frac{\log(t)\log(\log(t))-1}{t^2\log(t)}\,dt\\\\\
&=\frac{1}{e^e}
\end{align}$$
Hence, the original integral converges.
To show that the integral fails to converge absolutely, we note that for $t\ge e^e$, $\log(\log(t))\ge 1$.  Therefore, for $x\ge e^e$, $\left|\frac{\log(\log(t))}{t}\cos(t)\right|\ge \frac{|\cos(t)}{t}$.
Inasmuch as 
$$\int_{e^e}^{\infty}\frac{|\cos(t)|}{t}\,dt=\infty$$
the integral of interest converges conditionally.
A: It's trivial that $\frac{\ln\ln t}{t}$ is decreasing to $0$ in the interval $[e^e,+\infty)$, and
$$J:=\int_{e^e}^x \cos t\,\mathrm{d}t$$
is bounded by 2.
Using Dirichlet test can get that your integral converges.
That $\ln\ln t\geqslant 1$ implies that
$$\int_{e^e}^\infty \frac{\ln\ln t}{t} \left|\cos t\right| \, \mathrm{d}t \geqslant \int_{e^e}^\infty \frac{\left|\cos t\right|}{t} \,\mathrm{d}t=+\infty.$$
Thus your integral converges conditionally.
