Integration, computation of integral in $[0,1]$. I am student in first year in economics. I don't have enough knowledge on integration.
I want to compute this integral: 
\begin{equation*}
\int_{0}^{1} (\ln(p))(\ln(1-p)) dp
\end{equation*}
It will be great if you can detail the proof.
 A: By the substitution $p = e^{-u}$,
\begin{align*}
\int_{0}^{1} \log p \log(1-p) \, dp
&= - \int_{0}^{\infty} u e^{-u} \log(1-e^{-u}) \, du.
\end{align*}
Then by the Taylor expansion of the logarithm,
\begin{align*}
\int_{0}^{1} \log p \log(1-p) \, dp
&= \int_{0}^{\infty} u e^{-u} \left\{ \sum_{n=1}^{\infty} \frac{e^{-nu}}{n} \right\} \, du \\
&= \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\infty} u e^{-(n+1)u} \, du \\
&= \sum_{n=1}^{\infty} \frac{1}{n(n+1)^{2}} \\
&= \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} - \frac{1}{(n+1)^{2}} \right) \\
&= 2 - \zeta(2)
 = 2-\frac{\pi^{2}}{6} \\
&\approx 0.355065933151773563527584833354\cdots.
\end{align*}
Or we can just make use of the beta function
$$ \beta(z, w) = \frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)} = \int_{0}^{1} x^{z-1}(1-x)^{w-1} \, dx $$
to obtain
\begin{align*}
\int_{0}^{1} \log p \log(1-p) \, dp
&= \frac{\partial^{2} \beta}{\partial z \partial w}(1, 1) \\
&= \beta(1, 1) \left\{ \left( \psi^{(0)}(1)-\psi ^{(0)}(2) \right)^{2} - \psi^{(1)}(2) \right\} \\
&= 1 \cdot \{ (-1)^{2} - (\zeta(2) - 1) \} \\
&= 2 - \zeta(2).
\end{align*}
