eep said:
A thought... suppose we are trying to do the integral of 1/|x| from -1 to 1. We make the substitution u^2 = x, therefore...
<br />
2udu = dx<br />
<br />
\int_{-1}^{1} \frac{1}{|x|} dx = \int_{\sqrt{-1}}^{\sqrt{1}} \frac{2u}{|u^2|} du<br />
<br />
= \int_{\sqrt{-1}}^{\sqrt{1}} \frac{2}{u}<br />
<br />
= 2 (\ln{|1|} - \ln{|1|})<br />
<br />
= 0<br />
<br />
I can't immediately see what's wrong with this but I'm sure this can't be right as there is obviously only positive area under the graph of 1/|x|.
In your first post, you asked
how to do an integral like this and, in his first response, Hurkyl told you how. Perhaps, "what's wrong with this" is that you are simply ignoring what he said! In general, if there is a singularity in the middle of an interval, you
cannot just ignore it and evaluate the anti-derivative at the two endpoints.
Since 1/|x| is not defined at 0, integrate from -1 to some negative \alpha, integrate from some positive \beta and add. Then take the limit, if it exists, as \alpha and \beta go to 0. That is:
\int_{-1}^\alpha \frac{dx}{|x|}+ \int_\beta^1 \frac{dx}{|x|}
Since |x|= -x for x< 0 and |x|= x for x> 0, this is just
-\int_{-1}^\alpha \frac{dx}{x}+ \int_\beta^1 \frac{dx}{x}
= -ln|x|\left|_{-1}^\alpha+ ln|x|left|_\beta^1
= -ln|\alpha|+ ln|-1|+ ln|\beta|- ln|1|= ln|\beta|- ln|\alpha|
Finally, take the limit as \alpha and \beta go to 0
independently. In this case, that limit does not exist so the integral does not exist..
