improper integral question - $\int_{1}^{\infty}\!e^{-x}\ln x\,dx$ I ran into this integral question:
does this integral converge:
$$\int_{1}^{\infty}\!e^{-x}\ln x\,dx$$
?
Thank you very much in advance,
Yaron
 A: Hints: Use integration by part by letting $u=\ln(x)$ and then use the exponential integral $E_1(x)$
$$ E_{1}(x) = \int_{x}^{\infty}\frac{e^{-t}}{t}dt. $$
The final answer should be 
$$E_1(1).$$
Added: For convergence test, you can use the fact that $\ln(x)<x$ which implies
$$ \int_{1}^{\infty} \ln(x) e^{-x}dx < \int_{1}^{\infty} x e^{-x}dx < \infty. $$
A: Using integration by parts, we let:
$$u = \ln x; du = \frac{1}{x}$$
$$dv = e^{-x}dx; v = -e^{-x}$$
Then the integral becomes:
$$\int_{1}^{\infty}e^{-x}\ln x dx = -e^{-x}\ln x|^{\infty}_{1} + \int_{1}^{\infty}\frac{e^{-x}}{x}dx$$
The integral $$\int_{1}^{\infty}\frac{e^{-x}}{x}dx$$
is special, and is denoted the exponential integral (Ei$(-1)$).
Via wolfram, we have that $$\lim_{A\rightarrow -\infty} \text{Ei}(A) = 0$$
Furthermore, have $$\text{Ei}(-1) = .2193$$
to four decimal places. 
Hence, our integral is: 
$$\int_{1}^{\infty}e^{-x}\ln x dx = -e^{-x}\ln x|^{\infty}_{1} + \int_{1}^{\infty}\frac{e^{-x}}{x}dx = (0 - 0) + (\lim_{A\rightarrow -\infty} \text{Ei}(A) - \text{Ei}(-1)) = -.2193$$
A: Note that for large enough $x$, we have $\ln x\lt e^{x/2}$. So for large enough $x$, our integrand is $\lt e^{-x/2}$. And it is well known that $\int_1^\infty e^{-x/2}\,dx$ converges: indeed we can easily find an explicit expression for it. So by Comparison, our integral also converges. 
Remark: The point is that $\ln x$ grows very feebly in comparison to $e^{x}$, so the $e^{-x}$ term crushes poor $\ln x$. 
If we want to be formal about the relationship between $e^{x/2}$ and $\ln x$, consider $\dfrac{\ln x}{e^{x/2}}$. By L'Hospital's Rule, or in other ways, we can show that this $\to 0$ as $x\to \infty$.
