Deceptively cunning integral

FunkyDwarf · Jun 13, 2008

Well either that or im missing something obvious =)
I need to integrate 1/sqrt(u^2)

Now i cant just apply the square root because then i lose all information about the minus values. Ive tried various substitutions, trig ones seem to make it worse and ive run the integral in mathematica and it returns an answer that i agree with and fits with what im trying to do but im not sure how to get it :S

Im sure its something simple that my tired caffeine saturated brain is missing, so be gentle with me :P

Cheers
-G

quasar987 · Jun 13, 2008

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

uman · Jun 13, 2008

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

Edit: quasar beat me to it :-)

tiny-tim · Jun 13, 2008

Hi G!

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

HallsofIvy · Jun 13, 2008

tiny-tim said: ↑

Hi G!

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

That's going to be rather difficult to do!

tiny-tim · Jun 13, 2008

oops!

oah … aah …

maybe not 0

FunkyDwarf · Jun 13, 2008

no i understand its the magnitude its just mathematica gives a result of u Ln(u) /sqrt(u^2) and i dunno how to get that. Its been a while since i did methods of integration, as you can probably tell :P

EDIT: By the way we know that u itself is always negative and im integrating from x0 to zero. Its to show that a particle in a certain potential will take forever to fall in basically so i need either a 1/u term somewhere or log (not Abs(u))

ehj · Jun 14, 2008

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|

FunkyDwarf · Jun 14, 2008

Yes but the only problem being i should get the same answer for u always positive. Basically im applying boundary conditions too soon i think, im supposed to show that it is always negative from the integral. im not sure what im allowed to assume, have a tute with lecturer tomorrow will ask then.

Thanks for your help guys

Count Iblis · Jun 14, 2008

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.

HallsofIvy · Jun 15, 2008

FunkyDwarf said: ↑

no i understand its the magnitude its just mathematica gives a result of u Ln(u) /sqrt(u^2) and i dunno how to get that. Its been a while since i did methods of integration, as you can probably tell :P

EDIT: By the way we know that u itself is always negative and im integrating from x0 to zero. Its to show that a particle in a certain potential will take forever to fall in basically so i need either a 1/u term somewhere or log (not Abs(u))

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.

Count Iblis · Jun 15, 2008

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

FunkyDwarf · Jun 15, 2008

Yeh no i could see it wasnt simplified, just thought maybe there was a catch i was missing. Ok guys seems i was reading too much into it :P thanks !