Sorry but I have a problem I am student in first year in economics. I don't have enough knowledge on integration.
I want to compute this integral, for $a>2$, $a>i>0$, $b\in R$ and $d\in R$.
\begin{equation*} \int_{0}^{1} (\ln(p))^{a-i} (\ln(1-p))^{i} \left[ (\frac{p^{2}}{2})- b p+( \frac{b^{2}}{2} + d)\right] dp \end{equation*}
It will be great if you can detail the proof.
Here is a special case $a=3,i=1,b=0,d=0$ $$ \frac{1}{2}\int _{0}^{1}\left( \ln \left( p \right) \right) ^{2}\ln \left( 1-p \right) {p}^{2}{dp} = \frac{\zeta(3)}{3} + \frac{\pi^2}{54} - \frac{131}{216} $$ I doubt they expect first-year economics students to be able to do these.
Using beta function,
\begin{align*} &\int_{0}^{1} \left( \frac{p^{2}}{2} - pb + \frac{b^{2}}{2} + d \right) \log^{a-i}p \log^{i}(1-p) \, dp \\ &= \frac{1}{2} \frac{\partial^{a}\beta}{\partial z^{a-i} \partial w^{i}}(3,1) - b\frac{\partial^{a}\beta}{\partial z^{a-i} \partial w^{i}}(2,1) + \left( \frac{b^{2}}{2} + d \right) \frac{\partial^{a}\beta}{\partial z^{a-i} \partial w^{i}}(1,1). \end{align*}
I don't believe this can be simplified further for general $a$ and $i$ (except for the case you admit the answer that contains complete mess of polygamma functions).
