Integral yielding part of a harmonic series Why is this true?
$$\int_0^\infty x \frac{M}{c} e^{(\frac{-x}{c})} (1-e^{\frac{-x}{c}})^{M-1} \,dx = c \sum_{k=1}^M \frac{1}{k}.$$
I already tried substituting $u = \frac{-x}{c}$. Thus, $du = \frac{-dx}{c}$ and $-c(du) = dx$. Then, the integral becomes (after cancellation) $\int_0^\infty c u M e^u (1-e^u)^{M-1}\,du$.
I looked at integral-table.com, and this wasn't there, and I tried wolfram integrator and it told me this was a "hypergeometric integral".
Thanks,

if you substitute $t=1-e^{-x/c}$ your integral becomes
$I_M=-c\int_0^1\log(1-t)Mt^{M-1}dt=-c\int_0^1\log(1-t)(t^M)'dt$. Hence $I_M-I_{M-1}=-c\int_0^1\log(1-t)(t^{M-1}(t-1))'dt=c\int_0^1 t^{M-1} dt$ (by per partes) $=c/M$. So you only need to compute $I_1$, which is $c$. 
(there should be a nicer solution)
