Integration from 0 to 0 - why does my calculator say “undefined” in one case, and “0” in another?

I have the following question:

Using my calculator, $\displaystyle\int_0^{0} \frac{1}{x}dx$ is "undefined".

But when I type $\displaystyle\int_0^{0} - \frac{\ln(1-t)}{t} dt$, the result is 0.

What is the difference?

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I think only your calculator manufacturer can answer that. – Peter Phipps May 24 '12 at 19:59
Ok. So what is the right answer then? – characters May 24 '12 at 20:00
The one-point set $\{0\}$ over which you are integrating is zero measurable: so any measurable function should yield $0$ when integrating over it. However, neither of these functions are defined at $0$, so I think the answer should be "undefined" for both integrals. – T. Eskin May 24 '12 at 20:06

It may be there's no "right answer". The hyperbola $\,1/x\,$ isn't defined at zero (it is in fact a discontinuity point of the second kind of the function), so it seems "obvious" that its definite integral from zero to zero makes no sense. OTOH, the same happens with $\,\displaystyle{-\frac{\log(1-t)}{t}}\,$ , but in this case the disc. point is removable, as $$\lim_{t\to 0}-\frac{\log(1-t)}{t}=1$$Perhaps this is what matters to your computer...
That's what I thought. Maybe the calculator tries to compute something like $\lim_{\epsilon \to 0} \int_{0}^{\epsilon} f(x)$ which does not work for $f(x) = 1/x$ but does work for $f(x) = -\log(1-x)/x$. – TMM May 24 '12 at 20:19