# Math Help - evaulate the integral...

1. ## evaulate the integral...

the integral from 0 to infinity of x(e^x) dx

2. Originally Posted by tiga killa
the integral from 0 to infinity of x(e^x) dx
It diverges.

$\int_0^{\infty} xe^x\,dx = xe^x-e^x\big|_0^{\infty} = \infty+1 = \infty$

3. Originally Posted by redsoxfan325
It diverges.

$\int_0^{\infty} xe^x\,dx = xe^x-e^x\big|_0^{\infty} = \infty+1 = \infty$
I guess the "correct" question is $\int_0^{\infty} xe^{-x}\,dx$

4. Originally Posted by tiga killa
the integral from 0 to infinity of x(e^{-x}) dx
Use integration by parts:

$\int_0^{\infty}xe^{-x}\,dx = -xe^{-x}-e^{-x}\big|_0^{\infty} = \lim_{a\to\infty}-ae^{-a}-e^{-a} + 1$

$-ae^{-a} = -e^{\ln a}e^{-a} = -e^{\ln(a)-a} = -e^{\ln(a)-\ln(e^a)} = -e^{\ln(\frac{a}{e^a})}$

$\lim_{a\to\infty}\frac{a}{e^a} = \lim_{a\to\infty}\frac{1}{e^a} = 0$

So $\lim_{a\to\infty}-e^{\ln(\frac{a}{e^a})} = -e^{\ln(0)} = -e^{-\infty} = 0$