Quote:

Originally Posted by

**Havelock**

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 $\displaystyle \color{red}d_r^3$.

here $\displaystyle d_r^3$ should be $\displaystyle d_r^2.$

Quote:

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)

we have $\displaystyle \frac{1}{1-xy}=\sum_{k=0}^{\infty}x^ky^k.$ put this in (1) and integrate the sum term by term to get (2).

Quote:

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 put $\displaystyle \sigma=0,$ we'll get: $\displaystyle \int^1_0\int^1_0 \frac{x^ry^s}{1-xy}dxdy=\frac{1}{r-s}(\frac{1}{s+1} + \cdots + \frac{1}{r})=\frac{a}{(r-s)b},$ where $\displaystyle b=\text{lcm}\{s+1, \cdots, r \}.$ now since $\displaystyle \{s+1, \cdots, r \} \subseteq \{1, \cdots, r \},$ and $\displaystyle d_r=\text{lcm}\{1,2, \cdots, r \},$

we have $\displaystyle b \mid d_r.$ also since $\displaystyle r-s \leq r,$ we have $\displaystyle r - s \mid d_r.$ thus $\displaystyle (r-s)b \mid d_r^2.$

Quote:

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**.

exactly the same argument as above but this time $\displaystyle \text{lcm}\{(s+1)^2 , \cdots , r^2 \} \mid d_r^2$ and $\displaystyle r-s \mid d_r.$ thus the denominator will divide $\displaystyle d_r^3.$