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,

Careful, the substitution changes the limits of integration.  It needs to be the integral going over negative real numbers.
You could just expand $(1-e^u)^{M-1}$ in $$\int_0^{-\infty} c u M e^u (1-e^u)^{M-1}\,du$$ to get 
$$cM \int_0^{-\infty} u \sum_{n=0}^{M-1} \binom{M-1}{n} (-1)^n e^{(n+1)u}du.$$  Rearrange the integral and sum to find 
$$cM \sum_{n=0}^{M-1} \binom{M-1}{n} (-1)^n  \int_0^{-\infty}  u e^{(n+1)u}du.$$  The anti-derivative of $xe^{rx}$ is $\frac{x}{r}e^{rx}-\frac{1}{r^2}e^{rx}$. 
Can you solve it from here?
