# Thread: expressing repeating fractions as integral of 1/x

1. ## 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

2. Originally Posted by rainer
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 $8.\overline{33333}$.

3. Originally Posted by HallsofIvy
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.