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Question regarding certain standard integrals.

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    OK, this is something that stumped me.

    2. Relevant equations

    [itex]\int_{}^{}{\frac{dZ}{ A^{2}-Z^{2}}} = - \int_{}^{}{\frac{dZ}{ Z^{2}-A^{2} }}[/itex]

    Right?

    [itex]\int_{}^{}{\frac{dZ}{ A^{2}-Z^{2} }}=\; \frac{1}{2A}\ln \left\{ \frac{A+Z}{A-Z} \right\}+\mbox{C}[/itex]

    [itex]\int_{}^{}{\frac{dZ}{ Z^{2}-A^{2} }}=\; \frac{1}{2A}\ln \left\{ \frac{Z-A}{Z+A} \right\}+\mbox{C}[/itex]

    3. The attempt at a solution

    But I don't see how,

    [itex]\frac{1}{2A}\ln \left\{ \frac{A+Z}{A-Z} \right\}+\mbox{C} = - \frac{1}{2A}\ln \left\{ \frac{Z-A}{Z+A} \right\}+\mbox{C}[/itex]

    It should be,

    [itex]\frac{1}{2A}\ln \left\{ \frac{A+Z}{A-Z} \right\}+\mbox{C} = - \frac{1}{2A}\ln \left\{ \frac{A-Z}{A+Z} \right\}+\mbox{C}[/itex]

    No?

    I derived both (OK one of them; the other I just read from the textbook), the integrations "check out", but I don't understand why!? Why the term inside "ln" is "multiplied by -1"?
     
  2. jcsd
  3. Feb 2, 2012 #2
    Well, the integral actually comes out as the logarithm of the absolute value of something. So regardless of whether it is Z-A or A-Z, the two values will give the same answer; i.e. it doesn't matter which way around they are. (Remember there is no ln(-1).)
     
  4. Feb 2, 2012 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Are you forgetting the absolute value?
    [tex]\int \frac{1}{z}dz= ln|z|+ C[/tex]
    NOT ln z+ C unless you are sure z is positive.
     
  5. Feb 2, 2012 #4
    Oh ... that would explain it.

    My textbook didn't put the absolute value brackets**. -_-

    Thanks.

    **Ya, it's not the best book, quite a few obvious typos, but it's easy enough to read. Should really get a more rigorous and accurate "second opinion" ...
     
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