Deceptively cunning integral

Well, sqrt(u^2)=|u|.

 

isn't this just the same as [tex]|\frac{1}{u}|[/tex]?

Edit: quasar beat me to it :-)

 

Hi G!

Integrate from -∞ to 0, and integrate separately from 0 to +∞, and add!

 

That's going to be rather difficult to do!

 

oops!

oah … aah …

maybe not 0 ​

 

Well IF u is always negative, as you mentioned, then 1/sqrt(u^2) = 1/|u| = 1/(-u) cant you just integrate that? and get -ln|u|

 

You can also add an infitessimal imaginary term to the denominator:

[tex]\frac{1}{|x|} \longrightarrow \frac{1}{|x + i\epsilon|}=\frac{1}{\sqrt{x^2 + \epsilon^{2}}}[/tex]

This is easy to integrate. You can let the [tex]\epsilon[/tex] tend to zero at the end of the calculations.

 

I don't know why Mathematica is avoiding |u|. Again, sqrt(u^2)= |u| and u/|u|= 1 if u is positive, -1 if u is negative (and is undefined at u= 0). What mathematica is telling you is that the integral is ln(u) if u is positive and -ln(|u|)= -ln(-u) if u is negative.

That's true because your integral reduces to [itex]\int du/u= ln(|u|)= ln(u)[itex] if u is positive and [itex]-\int du/u= -ln(|u|)= -ln(-u)[/itex] if u is negative.

("+ C" of course)

If you are integrating from x0 to 0, the (improper) integral does not exist.

 

Sometimes you need to tell Mathematica to "Simplify", to "FullSimplify, to "FunctionExpand", to "PowerExpand" etc. etc.