Does the integral from $1$ to infinity of $\frac{\ln(x)}{x}$ converge or diverge? $$
\int_1^\infty \frac{\ln x} x \, dx
$$
The integral is from $1$ to infinity. Solve by taking the limit as $t$ approaches infinity. Not sure how to set up from here.
 A: When $x>e$ then $\ln x>1,$ so we have:
$$
\int_e^t \frac{\ln x} x\, dx > \int_e^t \frac 1 x \, dx \to\infty\text{ as } t\to\infty.
$$
The part from $1$ to $e$ does not alter this bottom line.
A: We can explicitly calculate the integral using $u$-substitution.
$$\int_1^\infty \frac{\ln(x)} x \,dx = \lim_{t \to \infty} \int \limits_1^t \frac{\ln(x)} x \,dx.$$
Now, for each $t$, if we let $u = \ln(x)$, then $du = \frac{1}{x} \,dx$ and when $x = 1$, $u = 0$ while when $x = t$, $u = \ln(t)$, giving: 
$$ \int_1^t \underbrace{\ln(x)}_u \underbrace{\frac{1}{x} \,dx}_{du} = \int_0^{\ln(t)} u \,du = \left.\frac{u^2}{2} \right|_0^{\ln(t)} = \frac{(\ln(t))^2} 2$$
Now we take $t$ to $\infty$.  Since $\lim \limits_{t \to \infty} \ln(t) = \infty$, then $\lim \limits_{t \to \infty} \frac{(\ln(t))^2} 2 = \infty$, implying $\int\limits_1^\infty \frac{\ln(x)} x \,dx = \infty$, as desired.
A: At first, $x \mapsto \frac{\ln(x)}{x}$ is continuous and locally integrable at $[1,+\infty)$.
Let $X>1$
if you put $u=\ln x$ then
$$F(X)=\int_1^X\frac{\ln(x)}{x}~\mathrm dx=\int_0^{\ln(X)}u~\mathrm du=\frac{1}{2}(\ln(X))^2$$
$$\implies \lim_{X\to+\infty}F(X)=+\infty$$
thus the integral is divergent.
A: $\int \frac{\log(x)}{x} dx= \frac{\log(x)^2}{2}$.
$\int\limits_{1}^x \frac{\log(t)}{t}dt=\frac{\log(x)^2-1}{2}$ which diverges to infinity when $x$ goes to infinity.
