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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.

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I corrected one $\log(p)$ to $\log(1-p)$ as the OP indicated. –  GEdgar Nov 4 '13 at 14:44

3 Answers 3

Log[p]^(a - i)*Log[p]^i is just Log[p]^a ?

Assuming[Element[{b, d, a}, Reals] && a > 2, 
 Integrate[Log[p]^a ( p^2/2 - b p + b^2/2 + d), {p, 0, 1}]]

Mathematica graphics

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Thanks for your answer, But it's not $[log(p)]^i$ but $[log(1-p)]^i$, how do you compute your answer, and which software did you used? –  lea Nov 4 '13 at 14:13
@lea Mathematica –  oldrinb Nov 4 '13 at 14:46

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).

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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.

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