Is $(\log x)e^{-x}$ Lebesgue integrable? Could anyone give me any little hints on how to show the following please?
Is $(\log x)e^{-x}$ Lebesgue integrable on $(0, \infty)$ ? 
I cannot see how to do this. I've tried Comparison Test. 
I believe it may require me to use that if $g$ is integrable and $h$ is bounded and measurable, $gh$ is integrable. $e^{-x}$ is bounded and measurable but how is $\log x$ integrable? Comparison again doesn't work and neither does looking at the modulus (since $\log x$ is unbounded at $0$ and $\infty$) 
Thanks for any hints.  

You can show that $f$ is integrable on $(0,\varepsilon)$, $\varepsilon>0$, by comparison with $\frac{1}{\sqrt x}$.  
You can show that $f$ is integrable on $(1,\infty)$ by comparison with $xe^{-x}$, or by noting that $\log(x)e^{-x}=(\log(x)e^{-x/2})e^{-x/2}$ and showing that $\log(x)e^{-x/2}$ is bounded on $(1,\infty)$.  
