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I found an asnwer to my question on the site : Stackexchange.

It is a very nice answer but I still wonder: how do we get between those two lines?

\begin{align} & \int_0^1 \ln x ,\frac{ x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)},dx \[4pt] &= \frac{1}{B(\alpha,\beta)} , \int_0^1 \frac{\partial x^{\alpha-1}(1-x)^{\beta-1}}{\partial \alpha},dx \[4pt] \end{align}

Thanks,

Flora Grappelli
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Sle R.
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1 Answers1

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First, you can take $\frac{1}{B(\alpha , \beta)}$ out of the integral because $B(\alpha , \beta)$ is constant with respect to x.

Then, use the fact that

$\frac{\partial x^{\alpha -1}}{\partial \alpha}=\ln x\cdot x^{\alpha -1}$

to obtain

$\frac{\partial x^{\alpha-1}(1-x)^{\beta-1}}{\partial \alpha} = \ln(x) \, \cdot x^{\alpha-1}(1-x)^{\beta-1} $

Within the integral.

Blueyedaisy
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