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
 A: An elementary closed-form antiderivative is obtainable.  Integration by parts with the choice $u = x$, $du = dx$, $dv = \sqrt{1-e^{-x}} e^{-x} \, dx$, $v = \frac{2}{3}(1-e^{-x})^{3/2}$ yields $$F(x) = \int u \, dv = \frac{2x}{3} (1-e^{-x})^{3/2} - \frac{2}{3} \int (1-e^{-x})^{3/2} \, dx.$$  This suggests a trigonometric substitution of the form $e^{-x} = \cos^2 t$, or $x = -2 \log \cos t$, and $dx = 2 \tan t \, dt$:  $$\begin{align*} \int (1-e^{-x})^{3/2} \, dx &= \int (1-\cos^2 t)^{3/2} \cdot 2 \tan t \, dt \\ &= 2\int \sin^3 t \tan t \, dt \\ &= 2 \int (1-\cos^2 t)^2 \sec t \, dt \\ &= 2 \int \sec t - \cos t - \sin^2 t \cos t \, dt. \end{align*}$$  This latter integral can be easily evaluated term by term using known formulas, which I leave to you.
A: By partial integration we have
$$
\int x \sqrt{1-e^{-x}}e^{-x}dx = \frac{2}{3} x (1-e^{-x})^{3/2} -\frac{2}{3}\int dx (1-e^{-x})^{3/2}
$$
now by letting $y=e^{-x}$, $-\ln y =x$, $dx = -dy / y$ we have
$$
\int dx (1-e^{-x})^{3/2} = \int \frac{-dy}{y}(1-y)^{3/2}
$$
now let $\sqrt{1-y} = \tau$, $y= 1-\tau^2$, $dy = -2\tau d \tau$ so
$$
\int \frac{+2\tau d \tau}{1-\tau^2}\tau^3 = 2\int d\tau \frac{\tau^4-\tau^2+\tau^2-1+1}{1-\tau^2} = -2 \int \tau^2 d\tau - 2\int d\tau +2\int \frac{d\tau}{1-\tau^2}=\\
-\frac{2\tau^3}{3}-2\tau +  \ln \frac{1+\tau}{1-\tau}.
$$
We have to put everything back together in orderly fashion:
$$
\frac{2}{3}x(1-e^{-x})^{3/2}-\frac{2}{3}\left[-\frac{2}{3}(1-e^{-x})^{3/2}-2\sqrt{1-e^{-x}}+\ln \frac{1+\sqrt{1-e^{-x}}}{1-\sqrt{1-e^{-x}}}\right].
$$
Now, as $x\to 0$, the primitive approaches $0$. As $x\to \infty$ we have to handle an indeterminate form:
$$
\lim_{x\to \infty}\left[\ln\frac{1+\sqrt{1-e^{-x}}}{1-\sqrt{1-e^{-x}}}-x(1-e^{-x})^{3/2}\right]=\\
=\lim_{y\to 0}\left[\ln\frac{1+\sqrt{1-y}}{1-\sqrt{1-y}}+\ln y \cdot (1-y)^{3/2}\right]=\\
=\lim_{y\to 0}\left[\ln\frac{1+1-y/2}{1-1+y/2}+\ln y \cdot (1-3y/2)\right] =\lim_{y\to 0}\ln\frac{(2-y/2)y}{y/2} = \ln 4.
$$ 
If we have done everything right our result is:
$$
\frac{2}{3}(-\ln 4+2+\frac{2}{3}) = \frac{2}{9}(8-3\ln 4).
$$
A: $\newcommand{\+}{^{\dagger}}
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Following the ${\large\tt OP}$:
\begin{align}
\int_{0}^{\infty}x\root{1 - \expo{-x}}\expo{-x}\,\dd x&=
2\sum_{k = 1}^{\infty}{1 \over k\pars{3 + 2k}}
=\sum_{k = 0}^{\infty}{1 \over \pars{k + 5/2}\pars{k + 1}}
\\[3mm]&={\Psi\pars{5/2} - \Psi\pars{1} \over 5/2 - 1}
={2 \over 3}\bracks{\Psi\pars{5 \over 2} - \Psi\pars{1}}
\end{align}
where $\ds{\Psi\pars{z}}$ is the Digamma Function ${\bf\mbox{6.3.1}}$.

Also
  $$
\int_{0}^{\infty}x\root{1 - \expo{-x}}\expo{-x}\,\dd x
={2 \over 3}\bracks{\Psi\pars{3 \over 2} + {2 \over 3} - \Psi\pars{1}}
={2 \over 3}\bracks{\Psi\pars{1 \over 2} + {8 \over 3} - \Psi\pars{1}}
$$
  where we used ${\bf\mbox{6.3.2}}$ and ${\bf\mbox{6.3.5}}$. 

However
$$
\Psi\pars{\half}=-\gamma - 2\ln\pars{2}\,,
\qquad\Psi\pars{1} = -\gamma 
$$
$\ds{\gamma}$ is the Euler-Mascheroni Constant ${\bf\mbox{6.1.3}}$.
$$\color{#00f}{\large%
\int_{0}^{\infty}x\root{1 - \expo{-x}}\expo{-x}\,\dd x
={16 \over 9} - {4 \over 3}\,\ln\pars{2}} \approx 0.8536
$$
A: 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 $$\frac{2}{3}(\frac{8}{3}-2\log2)=\frac{4}{3}(\frac{4}{3}-\log2)$$
A: 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:
$$\frac{4}{3}\int_0^1 \frac{1-x^3}{x}\frac{x^2}{1-x^2}\,dx=\frac{4}{3}\int_0^1 \frac{(x^2+x+1)x}{1+x}\,dx=\frac{4}{3}\int_0^1\left(x+\frac{x^3}{1+x}\right)\,dx$$
$$=\frac{4}{3}\int_0^1 \left(x+\frac{x^3+1}{1+x}-\frac{1}{1+x}\right)\,dx=\frac{4}{3}\int_0^1 \left( x^2+1-\frac{1}{x+1}\right)\,dx=\frac{4}{3}\left(\frac{4}{3}-\ln 2\right)$$
$\blacksquare$
A: Your sum doesn't diverge:
$$ \left(\frac{1}{3k} - \frac{2}{3(2k+3)} \right) 
= \frac{(2k+3) - (2k)}{3k(2k+3)}
= \frac{3}{3k(2k+3)}$$
so its convergence behaves like $\frac{1}{2k^2}$, not like $\frac{1}{3k}$.
However, there's an easy approach to working from the point where you introduced the sum: you can simplify the logarithm by using $\log(ab) = \log(a) + \log(b)$.
A: Another approach:
$$
\begin{align}
\int_0^\infty x\sqrt{\vphantom{^1}1-e^{-x}}\,e^{-x}\,\mathrm{d}x
&=-\int_0^1\log(u)\sqrt{1-u}\,\mathrm{d}u\tag{1}\\
&=-\frac23\int_0^1\log(u)\,\mathrm{d}\left(1-\sqrt{1-u}^3\right)\tag{2}\\
&=\frac23\int_0^1\frac{1-\sqrt{1-u}^3}{u}\,\mathrm{d}u\tag{3}\\
&=\frac43\int_0^1\frac{1-v^3}{1-v^2}v\,\mathrm{d}v\tag{4}\\
&=\frac43\int_0^1\left(v^2+1-\frac1{v+1}\right)\,\mathrm{d}v\tag{5}\\
&=\frac43\left(\frac43-\log(2)\right)\tag{6}
\end{align}
$$
Explanation:
$(1)$: substitute $u=e^{-x}$
$(2)$: $\frac23\mathrm{d}\left(1-\sqrt{1-u}^3\right)=\sqrt{1-u}\,\mathrm{d}u$
$(3)$: integrate by parts
$(4)$: substitute $v=\sqrt{1-u}$
$(5)$: polynomial division
$(6)$: integration
