I'm having trouble with the integral $\int_{[0,1]\times[0,1]}\frac{1}{(1-xy)^a}dydx$ for the case in which $a\neq{1}$. For the case in which $a=1$ it was not difficult to calculate the integral and I got the convergent series $\sum_{n=1}^\infty\frac{1}{n^2}$.

Now, for the case in which $a>0$ and $a\neq{1}$, first, I calculated the integral $\int_0^1\frac{1}{(1-xy)^a}dy$ and by making a change of variable I got the result $\frac{-1}{x}$($\frac{(1-x)^{1-a}}{1-a}$ - $\frac{1}{1-a}$) which can be simplified to $\frac{1-(1-x)^{1-a}}{x(1-a)}$ (I'm almost sure this is right, but if anyone finds my calculation of this integral with respecto to $y$ is wrong, let me know).

So, then I would have to calculate the integral $\int_o^1\frac{1-(1-x)^{1-a}}{x(1-a)}dx$ and this is where I'm stuck. Maybe I should express one of the expressions in the integrand as a series? Or is there any easier approach to the problem? Any suggestions would be appreciated.

  • $\begingroup$ Did you try a change of variable for $x$ & $y$? $\endgroup$ – user207710 Mar 22 '15 at 23:27
  • $\begingroup$ I would try polar coordinates on the triangle (0,0), (1,0), (1,1). $\endgroup$ – abnry Mar 22 '15 at 23:27
  • $\begingroup$ You should have to integrate something like $\int (1-\tan \theta)^{1-a} d\theta$ if you do polar coordinates. I'd then substitute $u=\tan \theta$ and try to get something like the Beta function. Also, you got to be careful with convergence issues. $\endgroup$ – abnry Mar 22 '15 at 23:29

Given the integral you have reached, $$ \frac{1}{1-a}\int_0^1 \frac{1-(1-x)^{1-a}}{x} \, dx, $$ if you substitute $t=1-x$, you obtain $$ \frac{1}{1-a}\int_0^1 \frac{1-t^{1-a}}{1-t} \, dt = \frac{1}{1-a} H_{1-a}, $$ in fact by the definition of the harmonic numbers for $a<1$. This looks like a cop-out (because it is), but it comes from one of the ways of defining the digamma function.


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