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.


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 $\log^m(x).$

  • $\begingroup$ Neat - I am not so quick, so if you could explain the last equality I will gladly accept. $\endgroup$ – user17794 Jun 17 '12 at 9:24
  • $\begingroup$ @Tim Duff I had edited my post, hopefully it is more informative. Please tell me if you have any more questions. $\endgroup$ – Ragib Zaman Jun 17 '12 at 9:59
  • $\begingroup$ Yes that's more than sufficient - should have recognized the differentiation of the integral there. Thanks! $\endgroup$ – user17794 Jun 17 '12 at 10:01

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