Why do we have $\psi^{(m)} (z)=(-1)^{m+1}\int_{0}^{\infty}\frac{t^me^{-zt}}{1-e^{-t}}dt$? Why is the following representation true?
$$\psi^{(m)} (z)=(-1)^{m+1}\int_{0}^{\infty}\frac{t^me^{-zt}}{1-e^{-t}}dt,$$  
where $\psi^{(m)} (z)$ denotes the Polygamma function.
 A: The question really is: why do we have
$$\psi^{(1)}(z)=\int_0^{\infty}\frac{te^{-zt}dt}{1-e^{-t}},\tag{1}$$
since the statement for $m>1$ is obtained by straightforward differentiation with respect to $z$. (Also note that we cannot set $m=0$ since the integral will become divergent.)
The formula (1) is an easy consequence of the infinite product representation for the gamma function (see formula (16) here):
$$\Gamma(z)=\left[ze^{\gamma z}\prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)e^{-z/k}\right]^{-1}.$$
Taking the logarithm at both sides of this relation and differentiating twice w.r.t. $z$, we find
$$\psi^{(1)}(z)=\frac{d^2}{dz^2}\ln\Gamma(z)=\frac{1}{z^2}+\sum_{k=1}^{\infty}\frac{1}{(z+k)^2}=\sum_{k=0}^{\infty}\frac{1}{(z+k)^2}.\tag{2}$$
To show that (1) and (2) are equivalent, it suffices to expand $\displaystyle \frac{1}{1-e^{-t}}$ in (1) into geometric series and observe that
$$\int_{0}^{\infty}te^{-(z+k)t}dt=\frac{1}{(z+k)^2}.$$
A: Start by the following 
$$\int^{\infty}_0 \frac{e^{-t}}{t}-\frac{e^{-(x+1)t}}{1-e^{-t}}\, dt$$
Making the substitution $e^{-t}=y$
$$\int^{1}_0 \frac{1}{-\log(y)}-\frac{y^{t}}{1-y}\, dy$$
By a little trick we have 
$$\int^{1}_0 \frac{1}{-\log(y)}-\frac{1}{1-y}\, dy +\int^1_0 \frac{1-y^{t}}{1-y}\, dy$$
The left hand is another form of the Euler-Mascheroni constant.
$$-\gamma +\int^1_0 \frac{1-y^{t}}{1-y}\, dy=\psi(t+1)$$
Hence we have 
$$\psi(x+1)=\int^{\infty}_0 \frac{e^{-t}}{t}-\frac{e^{-(x+1)t}}{1-e^{-t}}\, dt$$
or
$$\psi(x)=\int^{\infty}_0 \frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}}\, dt$$
The first derivative shows
$$\psi'(x)=\int^{\infty}_0 \frac{te^{-xt}}{1-e^{-t}}\, dt$$ 
And the result follows by continuous differentiation with respect to $x$.
Look at http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant#Integrals
