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

You can continue with your method too. Use partial fraction decomposition this way:
$$\frac{4}{3}\sum_{k=1}^{\infty} \left(\frac{1}{2k}-\frac{1}{2k+3}\right)=\frac{4}{3}\int_0^1 \sum_{k=1}^{\infty} \left(x^{2k-1}-x^{2k+2}\right)\,dx=\frac{4}{3}\int_0^1 \,dx\left(\frac{1}{x}-x^2\right)\sum_{k=1}^{\infty} x^{2k}$$
where I used the following two:
$$\int_0^1 x^{2k-1}\,dx=\frac{1}{2k}$$
$$\int_0^1 x^{2k+2}\,dx=\frac{1}{2k+3}$$
Since $\displaystyle \sum_{k=1}^{\infty} x^{2k}=\frac{x^2}{1-x^2}$, you get:
$$\frac{4}{3}\int_0^1 \frac{1-x^3}{x}\frac{x^2}{1-x^2}\,dx=\frac{4}{3}\int_0^1 \frac{(x^2+x+1)x}{1+x}\,dx=\frac{4}{3}\int_0^1\left(x+\frac{x^3}{1+x}\right)\,dx$$
$$=\frac{4}{3}\int_0^1 \left(x+\frac{x^3+1}{1+x}-\frac{1}{1+x}\right)\,dx=\frac{4}{3}\int_0^1 \left( x^2+1-\frac{1}{x+1}\right)\,dx=\frac{4}{3}\left(\frac{4}{3}-\ln 2\right)$$
$\blacksquare$
