Convergence of improper integral Show that 
$\displaystyle\int_{0}^{\infty}ln(x)e^{-x}dx $
converges.
i used integration by parts but it always diverges. any hints?
 A: Hints: Construct a function $g$ that is integrable on $[0,\infty)$ and satisfies $|\log(x) e^{-x}| \le g(x)$. It may be profitable to define $g$ separately on $[0,1]$ and $[1,\infty)$.
A: Note that for $\int_1^{\infty}\ln x\text{e}^{-x}dx$, the following limit is zero:
$$\lim_{x\to\infty}x^2\ln (x)~\text{e}^{-x}=0$$ so this part converges. Now think of the other part as @Lord noted.
A: We know that the integrals
$$\int_0^1 \log x dx\quad\text{and}\quad \int_1^\infty \frac{dx}{x^2}$$
are convergent.
Now we have
$$\log x e^{-x}\sim_0 \log x\quad\text{and}\quad \log x e^{-x}=_\infty o\left(\frac{1}{x^2}\right)$$
so we can see the convergence of the given integrals.
A: Break up into two parts, say $0$ to $1$ and $1$ to $\infty$. On $(0,1]$, our function has absolute value less than $|\ln x|$. And you can evaluate $\int_0^1 \ln x\,dx$ explicitly. To be formal, you can calculate $\int_c^1 \ln x\,dx$ by parts, and let $c$ approach $0$ from the right.
For the integral from $1$ to $\infty$, note that for large $x$, we have $\ln x\lt e^{x/2}$. We can use for example L'Hospital's Rule to show that $\lim_x\to\infty \frac{\ln x}{e^{x/2}}=0$. In fact, $\ln x\lt e^{x/2}$ for all $x$. So from $1$ to $\infty$, our function is $\lt \frac{e^{x/2}}{e^x}=\frac{1}{e^{x/2}}$.
