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I'm interested in whether or not integrals of the form $I (\alpha , \beta , k) = \displaystyle\int_0^1 \log^k \left(\frac{x}{1-x}\right) x^{\alpha -1} (1-x)^{\beta -1 }dx,$ with $k\in \mathbb{N}, \alpha , \beta \in \mathbb{R}^+,$ exist. In particular, I've been working on the case $k=1$ for some time, and suspect that it works when $\alpha ,\beta > 1$. Despite my best efforts to coerce a lucky identity or sneaky contour out of it, I'm still out of luck in searching for a closed form or tight bounds. I'd be more interested to hear what sort of strategy you might use to answer these questions. That said, if you have an answer, I would appreciate being led in that direction (especially if this somehow turns out to be trivial), and you may of course post it if you find it somehow more convenient.

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up vote 6 down vote accepted

If we expand $\log^k \left( \frac{x}{1-x} \right) = (\log x - \log (1-x))^k $ by the binomial theorem we see that your integral is a linear combination of integrals of the form $$\int^1_0 \log^nx\log^m(1-x) x^{\alpha-1} (1-x)^{\beta-1} dx = \frac{\partial}{\partial^n\alpha \partial^m\beta} \operatorname{B}(\alpha,\beta) $$ where $\operatorname{B}$ is the Beta function. Hence these can be evaluated in terms of the Beta function and it's derivatives.


The Beta function is defined by $$\operatorname{B}(\alpha,\beta) =\int^1_0 x^{\alpha-1} (1-x)^{\beta-1} dx$$

and is well studied. Googling will turn up many results about this function. To get those $\log$ terms into the integral, we differentiate with respect to one of the variables. For example, $$ \frac{\partial}{\partial \alpha} \operatorname{B}(\alpha,\beta) = \frac{\partial}{\partial \alpha} \int^1_0 x^{\alpha-1} (1-x)^{\beta -1} dx = \int^1_0 \frac{\partial}{\partial \alpha} \left( x^{\alpha-1} (1-x)^{\beta-1} \right) dx .$$

Since $\frac{d}{dx} a^x = \log a \cdot a^x$ the effect of the differentiation is to multiply by the base. So in this case, $$\frac{\partial}{\partial \alpha} \operatorname{B}(\alpha,\beta) = \int^1_0 \log x x^{\alpha-1} (1-x)^{\beta -1} dx.$$

If we differentiate with respect to $\alpha$, we bring out a factor of $\log x$, so if we differentiate with respect to $\alpha$ $n$ times, we bring out $\log^n(x).$ Similarly, if we differentiate with respect to $\beta$ $m$ times, we bring out $\log^m(x).$

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Neat - I am not so quick, so if you could explain the last equality I will gladly accept. –  Tim Duff Jun 17 '12 at 9:24
    
@Tim Duff I had edited my post, hopefully it is more informative. Please tell me if you have any more questions. –  Ragib Zaman Jun 17 '12 at 9:59
    
Yes that's more than sufficient - should have recognized the differentiation of the integral there. Thanks! –  Tim Duff Jun 17 '12 at 10:01
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