Interchanging integral and sum

Question

I have the following dilemma - I wish to prove rigourously the following: $$\int_{0}^{1} \int_{0}^{1} \frac{1}{1-xy} dxdy = \sum_{n\geq1} \frac{1}{n^2} $$ Here the obvious strategy is to use the power series: $$\int_{0}^{1} \int_{0}^{1} \sum_{n\geq0} (xy)^n dx dy$$ And the natural step would be to interchange the sum twice, first yielding us: $$\int_{0}^{1} \sum_{n\geq0} \int_{0}^{1} (xy)^n dx dy$$ $$=\int_{0}^{1} \sum_{n\geq0} \int_{0}^{1} (xy)^n dx dy$$ $$=\int_{0}^{1} \sum_{n\geq0} \frac{y^n}{n+1} dy$$ Repeating this step gives us the wanted result. How can I justify interchanging the two steps without something like the Lebesgue dominated convergence theorem. Showing uniform convergence would be ideal. Thanks!

Answer 1

I would do this in another way:

$$\int_{0}^{1} \int_{0}^{1} \frac{1}{1-xy} dxdy$$

$$= \int_{y=0}^{y=1} \left( \int_{x=0}^{x=1} \frac{1}{1-xy} dx \right) dy$$

So, I'd start with :
$$\int_{x=0}^{x=1} \frac{1}{1-xy} dx$$
$$=-\int_{x=0}^{x=1} \frac{1}{xy - 1} dx$$
$$=-\frac{1}{y} \int_{xy=0}^{xy=y} \frac{1}{xy - 1} d(xy - 1)$$ $$=-\frac{1}{y} \ln{|xy-1|}\bigg\rvert^{xy=y}_{xy=0}$$ (I admit, it's not entirely correct)

And from here on, I continue integrating over $y$.