@romsek yes that what I was thinking yet the answer the book gets is very baffling indeed.
2Γ
I am sorry to trouble you with my trifles but would it be possible if you could show me what is u and what is dv?
$\displaystyle{\int_0^\infty}~u e^{-u}~du$
let's rewrite this in $x$ to avoid confusion
$\displaystyle{\int_0^\infty}~x e^{-x}~dx$
$u=x,~du=dx$
$dv = e^{-x}~dx,~v = -e^{-x}$
$\left . -x e^{-x}\right |_0^\infty + \displaystyle{\int_0^\infty}~e^{-x}~dx=$
$\displaystyle{\int_0^\infty}~e^{-x}~dx=1$