Integral of $re^{-r/a}$ 
How can I integrate $r\,e^{-r/a}$ from $0$ to $\infty$? 

I integrated by parts, and then was left with something in an indeterminate form. Is there an alternative to solving this without using L'Hospital's Rule?
EDIT: I flipped the exponent.

Making the change of variables $r=\frac{1}{t}$ gives rise to the integral

$$ \int _{0}^{\infty }\!{\frac {{{\rm e}^{-at}}}{{t}^{3}}}{dt}. $$

So, can you see if it diverges or converges? Note that, the integrand near zero behaves like 

$$ \frac{1}{t^3}. $$ 

Added: For the new integral, using the change of variables $t=r/a$ and the gamma function

$$ \Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,{\rm d}t,  $$

yields
$$ a^2\int _{0}^{\infty }\!t{{\rm e}^{-t}}{dt} = a^2 \Gamma(2)=a^2 1!=a^2. $$
Note: You can use integration by parts with $u=r$ to evaluate he integral without using the gamma function.
