# Problem with a proof

• December 17th 2008, 02:12 PM
Havelock
Problem with a proof
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 $r,s$ be non negative intgers. If $r>s$, then

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

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

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

The proof is given as follows:

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

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

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

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

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

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

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

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

and summation (2) becomes

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

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)

• December 17th 2008, 06:23 PM
NonCommAlg
Quote:

Originally Posted by Havelock

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

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

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

Quote:

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

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

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

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

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

Quote:

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

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

If we put $\sigma = 0$ then assertion a) follows immediately.

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

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

Quote:

If we differentiate with respect to $\sigma$ and put $\sigma = 0$, then integral (1) changes into

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

and summation (2) becomes

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

and assertion (b) now follows straight away.

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