<s>[INST] Trying to evaluate integral$\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx$ I am trying to integrating
$$
\int_0^\infty  x \sqrt{1-e^{-x}}\,e^{-x}dx\equiv I
$$
but cannot get the answer, I would like a proof not a numerical answer.  My attempt at proof: 
$$
y=\sqrt{1-e^{-x}}\\
y(0)=0, \ y(\infty)=1\\
y^2=1-e^{-x}\\
2ydy=e^{-x}dx\\
e^{-x}=1-y^2\\
x=\ln\frac{1}{1-y^2} \rightarrow\\
I=2\int_0^1 y^2\ln\frac{1}{1-y^2} dy=\\
-2\int_0^1y^2\ln(1-y^2)\,dy=\\2\int_0^1y^2\sum_{k=1}^\infty \frac{y^{2k}}{k}=\\
2\sum_{k=1}^\infty \frac{1}{k}\int_0^1 y^{2(k+1)}dy=\\
2\sum_{k=1}^\infty \frac{1}{k(3+2k)}=\\
2\sum_{k=1}^\infty \left(\frac{1}{3k}-\frac{2}{3(2k+3)}    \right)
$$
but this diverges because $\sum_k\frac{1}{k}\to\infty$?  Mistakes I made...
Please help if can on doing the sum or integral. Thank you, Grazie
 [/INST] Using the well-Known Series formula for the Digammafunction http://en.m.wikipedia.org/wiki/Digamma_function we have $$2\sum_{k=1}^{\infty}\frac{1}{n(2n+3)}=\frac{2}{3}\sum_{n=0}^{\infty}\frac{\frac{5}{2}-1}{(n+1)(n+\frac{5}{2})}=\frac{2}{3}(\gamma +\psi(\frac{5}{2}))$$
Recall the recurrence relation formula for Digamma, i.e $$\psi(x+1)=\psi(x)+\frac{1}{x}$$ hence it reduces to $$\frac{2}{3}(\gamma +\frac{8}{3}+\psi(\frac{1}{2}))$$ It is pretty simple to calculate that $\psi(\frac{1}{2})=-\gamma -2\log2$ So your sum finally evaluates to $$\frac{2}{3}(\frac{8}{3}-2\log2)=\frac{4}{3}(\frac{4}{3}-\log2)$$
</s>