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Math Help - Problem with a proof

  1. #1
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    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)




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  2. #2
    MHF Contributor

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    Quote Originally Posted by Havelock View Post

    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.



    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).


    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.



    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.
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