Question

Improper integral

Answer 1

I know this integral is ∞, but have can I show it?
\[\int\limits_{1}^{2}\frac{ dx }{ x-1 }\]

Answer 2

Well, you can have it like this:
\[\Large = \lim_{a\rightarrow 1^+}\int\limits_a^2\frac{dx}{x-1}\]

Answer 3

That's basically the definition of an improper integral anyway, and just integrate normally, with the 'added' bonus of evaluating that limit at the end. :D

Answer 4

And ln(a-1) goes to infinity when a goes to 1 from the "+-side"

Answer 5

No, it goes to negative infinity... and that's what you want anyway, since you'll have
\[\Large = \lim_{a\rightarrow 1^+}\int\limits_a^2\frac{dx}{x-1}=\lim_{a\rightarrow1^+}\left.\ln(x-1)\right]_a^2\]

Answer 6

of course.

Answer 7

\[\Large =\lim_{a\rightarrow1^+} \left[\ln(2-1)-\ln(a-1)\right]\]

Answer 8

And the rest is history :)

Answer 9

Thanks..