I was trying to see if there is a simple way to compute the following integral, where $0<a<b<1$, $$ \int_a^b\log(x)\frac{\sqrt{(x-a)(b-x)}}{x(1-x)}dx. $$ Any idea ?

NB : With change of variables and the use of series expansion of $\log(1+x)$, one can reduce the problem to compute for all $k\geq 0$ $$ \int_0^{b-a}y^k\frac{\sqrt{y(b-a-y)}}{(y+a)(1-a-y)}dy, $$ but then I'm stuck ...

  • $\begingroup$ integration by parts may help since $log'(x) = 1/x$. $\endgroup$ – quartz Mar 12 '12 at 13:40
  • $\begingroup$ @Anon : Of course, that's why it is assumed $0<a<x<b<1$. $\endgroup$ – Student Mar 13 '12 at 17:55
  • $\begingroup$ @Quartz : And do you know a primitive of the other part ? $\endgroup$ – Student Mar 13 '12 at 17:55

If we let

$$I = \int_a^b\log(x)\frac{\sqrt{(x-a)(b-x)}}{x(1-x)}dx$$

using the AM GM Inequality

$$ \sqrt{(x-a)(b-x)} \leq \frac{x-a+b-x}{2} = \frac{b-a}{2}$$

and the fact $\hspace{5pt}log(1+x) \leq (1-x) \hspace{5pt}$ for $x>0$,

$$ I \leq \int_a^b \frac{b-a}{2x} dx \leq \frac{1}{2}(b-a)ln\left(\frac{b}{a}\right)$$

I might have to think of lower bound. But this method is only determining the bounds.

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  • $\begingroup$ Thank you for your answer, but I would be very surprised if the upper bound you suggest would be the right answer. Anyway, that's already something. $\endgroup$ – Student Mar 13 '12 at 22:49

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