Physics Help Forum Please help in vector potential involve line integral with log.
 User Name Remember Me? Password

 Advanced Electricity and Magnetism Advanced Electricity and Magnetism Physics Help Forum

 Apr 29th 2011, 02:42 PM #11 Member   Join Date: Nov 2010 Posts: 40 Let me work out this and see whether I am correct: $\int_{-b}^{-a}\frac {dx}{|x|} = sgn(-a)ln|-a|-sgn(-b)ln|-b| = -ln|-a|-(-ln|-b|) = ln\left | \frac {-b}{-a}\right | = ln\left | \frac {b}{a}\right |$ And $\int_{-b}^{a}\frac {dx}{|x|} = sgn(a)ln|a|-sgn(-b)ln|-b| = ln|a|-(-ln|-b|) = ln|a(-b)|= ln(ab)$ Am I correct? Thanks Alan Last edited by yungman; Apr 29th 2011 at 02:55 PM.
Apr 29th 2011, 04:40 PM   #12

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,465
 Originally Posted by yungman Let me work out this and see whether I am correct: $\int_{-b}^{-a}\frac {dx}{|x|} = sgn(-a)ln|-a|-sgn(-b)ln|-b| = -ln|-a|-(-ln|-b|) = ln\left | \frac {-b}{-a}\right | = ln\left | \frac {b}{a}\right |$ And $\int_{-b}^{a}\frac {dx}{|x|} = sgn(a)ln|a|-sgn(-b)ln|-b| = ln|a|-(-ln|-b|) = ln|a(-b)|= ln(ab)$ Am I correct? Thanks Alan
Looks good to me.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Apr 29th 2011, 05:05 PM #13 Member   Join Date: Nov 2010 Posts: 40 Thanks so much. You are of big help Dan. Alan

 Tags integral, involve, line, log, potential, vector

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post yungman Advanced Electricity and Magnetism 4 May 5th 2011 11:33 AM yungman Advanced Electricity and Magnetism 10 May 4th 2011 01:35 AM greencheeseca General Physics 0 Mar 11th 2010 05:49 PM rem General Physics 2 Feb 24th 2009 10:48 AM ariol Kinematics and Dynamics 1 Jan 8th 2009 11:08 PM