expressing repeating fractions as integral of 1/x

Sep 2009
242
6
EDIT: Nevermind about the "fraction" part. What I mean is a number whose last digit repeats infinitely.

The function \(\displaystyle f(x)=\frac{1}{x}\) 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 \(\displaystyle 8\overline{33333}\) I would say...

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

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

Thanks
 
Last edited:

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
EDIT: Nevermind about the "fraction" part. What I mean is a number whose last digit repeats infinitely.

The function \(\displaystyle f(x)=\frac{1}{x}\) 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 \(\displaystyle 8\overline{33333}\) I would say...

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

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

Thanks
I have no idea what you are trying to do but \(\displaystyle \int_{b}^1 \frac{1}{x}dx= \left[ln(x)\right]_b^1= -ln(b)\) and that has no limit as b goes to 0. It is certainly not \(\displaystyle 8.\overline{33333}\).
 
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Sep 2009
242
6
I have no idea what you are trying to do but \(\displaystyle \int_{b}^1 \frac{1}{x}dx= \left[ln(x)\right]_b^1= -ln(b)\) and that has no limit as b goes to 0. It is certainly not \(\displaystyle 8.\overline{33333}\).
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.