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Math Help - expressing repeating fractions as integral of 1/x

  1. #1
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    expressing repeating fractions as integral of 1/x

    EDIT: Nevermind about the "fraction" part. What I mean is a number whose last digit repeats infinitely.

    The function f(x)=\frac{1}{x} is not defined at x=0.

    What is the proper way to express numbers with an infinitely repeating last digit as the area under this function?

    Taking a shot at it, if I take as an example the number 8\overline{33333} I would say...

    \lim_{b \to 0 }\; \int_{b}^{1}\frac{1}{x}\; dx=8\overline{33333}

    Is this correct? Is there a less ugly way to do it?

    Thanks
    Last edited by rainer; May 24th 2010 at 11:05 PM.
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  2. #2
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    Quote Originally Posted by rainer View Post
    EDIT: Nevermind about the "fraction" part. What I mean is a number whose last digit repeats infinitely.

    The function f(x)=\frac{1}{x} is not defined at x=0.

    What is the proper way to express numbers with an infinitely repeating last digit as the area under this function?

    Taking a shot at it, if I take as an example the number 8\overline{33333} I would say...

    \lim_{b \to 0 }\; \int_{b}^{1}\frac{1}{x}\; dx=8\overline{33333}

    Is this correct? Is there a less ugly way to do it?

    Thanks
    I have no idea what you are trying to do but \int_{b}^1 \frac{1}{x}dx= \left[ln(x)\right]_b^1= -ln(b) and that has no limit as b goes to 0. It is certainly not 8.\overline{33333}.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    I have no idea what you are trying to do but \int_{b}^1 \frac{1}{x}dx= \left[ln(x)\right]_b^1= -ln(b) and that has no limit as b goes to 0. It is certainly not 8.\overline{33333}.
    Aha I see.

    Ok, so I was asking this question on the basis of a mistaken assumption which I made because I used a calculator to do this integration for me instead of doing it myself like a real man.

    I still don't understand, however, why the calculator gives me 8.33333x10^25...etc. for lower bound values of just slightly greater than 0 and upper bound of 1.

    What is the calculator smoking for it to give me such an output?

    Thanks.
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