I'm having alot of trouble with the proof below. The parts I don't understand will be bolded, but I have written the whole proof so you can also follow the argument.

Let $\displaystyle r,s$ be non negative intgers. If $\displaystyle r>s$, then

a) $\displaystyle \int^1_0\int^1_0 \frac{x^ry^s}{1-xy}dxdy$ is a rational number whose denominator is a divisor of $\displaystyle d_r^3$

b) $\displaystyle \int^1_0\int^1_0 -\frac{\log(xy)}{1-xy}x^ry^sdxdy$ is a rational number whose denominator is a divisor of $\displaystyle d_r^3$

Where $\displaystyle d_r$ is the lowest common multiple of $\displaystyle 1,2, \ldots r$.

The proof is given as follows:

Let $\displaystyle \sigma$ be any non-negative number. Consider the integral

$\displaystyle \int^1_0\int^1_0 \frac{x^{r+\sigma}y^{s+\sigma}}{1-xy}dxdy$ (1)

**Develop** $\displaystyle (1-xy)^{-1}$ **into a geometrical series and perform the double integration. Then we obtain**

$\displaystyle \sum^{\infty}_{k=0} \frac{1}{(k+r+\sigma + 1)(k + s + \sigma +1)}$ (2)

Assume that $\displaystyle r>s$. Then we can write this sum as

$\displaystyle \sum^{\infty}_{k=0} \frac{1}{r-s}\{\frac{1}{k+s+\sigma +1}-\frac{1}{k+r+\sigma+1}\} = \frac{1}{r-s}\{\frac{1}{s+\sigma +1}+ \cdots + \frac{1}{r+\sigma}\} $

**If we put **$\displaystyle \sigma = 0$

** then assertion a) follows immediately.** If we differentiate with respect to $\displaystyle \sigma$ and put $\displaystyle \sigma = 0$, then integral (1) changes into

$\displaystyle \int^1_0\int^1_0 \frac{\log(xy)}{1-xy}x^ry^sdxdy$

and summation (2) becomes

$\displaystyle \frac{-1}{r-s}\{\frac{1}{(s+1)^2} + \ldots + \frac{1}{r^2}\}$

**and assertion (b) now follows straight away**.

Thank you thank you thank you to anyone who can explain those bolded sentences, I will be forever indebted to you!

Havelock (formerly HTale)