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?
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?
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...