the integral from 0 to infinity of x(e^x) dx
Use integration by parts:
$\displaystyle \int_0^{\infty}xe^{-x}\,dx = -xe^{-x}-e^{-x}\big|_0^{\infty} = \lim_{a\to\infty}-ae^{-a}-e^{-a} + 1$
$\displaystyle -ae^{-a} = -e^{\ln a}e^{-a} = -e^{\ln(a)-a} = -e^{\ln(a)-\ln(e^a)} = -e^{\ln(\frac{a}{e^a})}$
$\displaystyle \lim_{a\to\infty}\frac{a}{e^a} = \lim_{a\to\infty}\frac{1}{e^a} = 0$
So $\displaystyle \lim_{a\to\infty}-e^{\ln(\frac{a}{e^a})} = -e^{\ln(0)} = -e^{-\infty} = 0$