A "tricky" integral: $\int_0^{\infty} t e^{-nct} (1-e^{-ct})^m dt$ In an article in the current (May 2013) issue of the College Mathematics Journal,
they say that the following integral is
"tricky to evaluate":
$\int_0^{\infty} t e^{-nct} (1-e^{-ct})^m dt$
where $n$ and $m$ and n are non-negative integers
and $c$ is a positive real.
So, what is the value of this integral?
It would be nice if the derivation was
as simple and understandable as possible.
 A: Actually we also have
$$ \int_{0}^{\infty} t e^{-nct} (1-e^{-ct})^{m} \, dt = \frac{1}{c^2}\frac{(n-1)!m!}{(n+m)!}\sum_{k=0}^{m}\frac{1}{n+k}.$$
This is easily proved by the integration by parts by introducing
$$ J(n, m) = c^{2} \frac{(n+m)!}{(n-1)!m!} \int_{0}^{\infty} t e^{-cnt} (1 - e^{-ct})^{m} \, dt$$ 
and noticing that $J(n, m) = \frac{1}{n} + J(n+1,m-1)$ with $J(n, 0) = \frac{1}{n}$.
Also we can invoke the digamma function $\psi_0$ to evaluate it directly. Let $x = e^{-ct}$ so that
\begin{align*}
\int_{0}^{\infty} t e^{-nct} (1-e^{-ct})^{m} \, dt
&= - \frac{1}{c^2} \int_{0}^{1} x^{n-1} (1-x)^{m} \log x \, dx \\
&= - \frac{1}{c^2} \frac{\partial}{\partial n} \int_{0}^{1} x^{n-1} (1-x)^{m} \, dx \\
&= - \frac{1}{c^2} \frac{\partial}{\partial n} \frac{\Gamma(n)\Gamma(m+1)}{\Gamma(n+m+1)} \\
&= \frac{1}{c^2} \frac{\Gamma(n)\Gamma(m+1)}{\Gamma(n+m+1)} [ \psi_{0}(n+m+1) - \psi_{0}(n) ] \\
&= \frac{1}{c^2} \frac{(n-1)!m!}{(n+m)!} [ H_{n+m} - H_{n-1} ],
\end{align*}
which is the desired result.
A: Here is a start. Using the binomial theorem makes the evaluation of the integral easy 
$$ \int_0^{\infty} t e^{-nct} (1-e^{-ct})^m dt= \sum_{k=0}^{m} {m\choose k}(-1)^k \int_{0}^{\infty} t e^{-nct}e^{-ckt} \,dt$$ 
$$ =\sum_{k=0}^{m} {m\choose k}(-1)^k \int_{0}^{\infty} t e^{-c(n+k)t} \,dt \dots\,. $$
Now, just use integration by parts or the Laplace transform technique to evaluate the above integral.
A: Expand by the binomial theorem, and go from there. Note that $\Gamma(z) = (z - 1)! = \int_0^\infty t^{z - 1} e^{-t} dt$, so $\Gamma(2) = 1$ will show up all over the place... we will need $\int_0^\infty t e^{-\alpha t} dt = 1/\alpha^2$.
$$
\begin{align*}
\int_0^\infty t e^{-n c t} (1 - e^{- c t})^m d t
  &= \sum_{0 \le k \le m} (-1)^k \binom{m}{k} \int_0^\infty t e^{-(n + k) c t} d t\\
  &= c^{-2} \sum_{0 \le k \le m} (-1)^k \binom{m}{k} (n + k)^{-2}
\end{align*}
$$
Sadly, maxima' Zeilberger package says this isn't Gosper summable :-(
A: I make the substitution $u = e^{- c t}$, $t=-(1/c) \log{u}$ and get
$$-\frac{1}{c^2} \int_0^1 du \, \log{u} \, u^{n-1} (1-u)^m$$
Expand the binomial and use the fact that
$$\int_0^1 du\, u^p \, \log{u} = -\frac{1}{(p+1)^2}$$
The result I get is
$$\frac{1}{c^2} \sum_{k=0}^m (-1)^k \binom{m}{k} \frac{1}{(n+k)^2}$$
A: Let $I(c,n,m)$ be the integral
$$\int_0^{\infty} t e^{-nct}(1-e^{-ct})^m dt$$
It is clear $I(c,n,m) = \frac{1}{c^2} I(1,n,m)$. 
We can simplify the integral by differentiate under the integral sign:
$$
I(1,n,m) = \int_0^{\infty} t e^{-nt}(1-e^{-t})^m dt
= -\frac{\partial}{\partial n} \int_0^{\infty}  e^{-nt}(1-e^{-t})^m dt\\
= \frac{\partial}{\partial n} \int_0^{\infty}  e^{-(n-1)t}(1-e^{-t})^m d e^{-t}
$$
By a change of variable to $y = e^{-t}$, the last integral is in the form of a beta function with value:
$$-\frac{\partial}{\partial n} \int_0^1  y^{n-1}(1-y)^m d y
 = -\frac{\partial}{\partial n} \frac{\Gamma(n)\Gamma(m+1)}{\Gamma(n+m+1)}
 = -\frac{\partial}{\partial n}\frac{m!}{\prod_{k=0}^m ( n+k )}\\
 = m! \sum_{l=0}^m \frac{1}{(n+l)^2} \left( \prod_{k=0,\ne l}^{m} \frac{1}{n+k} \right)
 = \frac{m!}{\prod_{k=0}^m ( n+k )} \sum_{l=0}^m \frac{1}{(n+l)}
$$
We get
$$I(c,n,m) = \frac{m! (n-1)!}{c^2 (n+m)!}\sum_{l=0}^m \frac{1}{(n+l)} =\frac{1}{nc^2\binom{n+m}{m}}\sum_{l=0}^m \frac{1}{(n+l)}\tag{*}$$
Please note that despite the different looking, the last expression in $(*)$ gives the
same value as the expression derived in other answer:
$$\frac{1}{c^2} \sum_{k=0}^m (-1)^k \binom{m}{k} \frac{1}{(n+k)^2}$$
