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Integral of 1/ln(x)

  1. Dec 26, 2006 #1
    I have a formula to solve, which I am having trouble solving, it is a differential equation fo sorts

    [tex]\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}[/tex]

    could someone help me solve the indefinite integral of such an equation!!

    the m multiply of the right handside is suppose to be dm and likewise with the t on the left hand side sorry for any confusion!!!
    Thanks for any help

    N
     
    Last edited: Dec 26, 2006
  2. jcsd
  3. Dec 26, 2006 #2
    Make a substitution u=log(m/mo). I assumed that mo in the denominator was supposed to be mo.
     
  4. Dec 26, 2006 #3
    How can this work? Its an indefinite integral and the answer's definate. One's with respect to m and the other t. I'm sure it doesn't change the u substitution, but shouldn't everything be noted as rigorously as possible?
     
  5. Dec 26, 2006 #4
    I think everything is as already as rigorous as possible, at least for all that the original poster has given, however, I have no idea what you mean by "Its an indefinite integral and the answer's definate."
     
  6. Dec 27, 2006 #5
    I guess that was wrong.
     
    Last edited: Dec 27, 2006
  7. Dec 27, 2006 #6
    How do you get that? I'm pretty sure that the first integral is a fairly straightforward computation once you make a substitution.
     
  8. Dec 27, 2006 #7
    [tex]\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}[/tex]

    Let m=e^x then we get [tex]\int{\frac{dx}{x-log m(0)}[/tex]
     
  9. Dec 27, 2006 #8
    You're missing a factor of e-x which then gives you something you can't integrate, but if you make the substitution u=log(m/m0) then you end up with the integral of 1/u with respect to u.
     
  10. Dec 27, 2006 #9

    MWM

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    We may just notice that 1/m is the 1st derivative of (log(m/m0)) and that, then, the first integral contains an expression dividing its derivative: integrate[D(log(m/m0))]/log(m/m0)]dm. But this is just log|log(m/m0)|+c; isn't it?
     
  11. Dec 27, 2006 #10
    Yes that's it, and this is the same solution you arrive at using the substitution I mentioned, and is pretty much the sam process.
     
  12. Dec 27, 2006 #11

    MWM

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    Ah, you're right. Ok... as if I hadn't said anything :-)
     
  13. Dec 27, 2006 #12
    Ok thanks guys for your replies,

    the thing is I don't get where the m has gone!!

    surely the integral would be

    [tex]\int{\frac{1}{m_{0}ue^{u}}}du[/tex]

    ??
     
  14. Dec 27, 2006 #13
    How do you get to that? Try using the substitution u=log(m/m0) the integral will turn out very nicely.
     
  15. Dec 28, 2006 #14

    Gib Z

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    Let [itex]u=\log(m/mo)[/itex]
    The integral:[tex]\int{\frac{1}{m\log(m/mo)}{dm}[/tex] becomes [tex]\int{\frac{1}{m\cdot u}{dm}[/tex]. Whats the derivative of log x? 1/x of course.
    [tex]u=\log (m/mo)[/tex]
    [tex]\frac {du}{dm} =\frac{1}{m}[/tex]
    Substituting that in, our integral becomes [tex]\int \frac{1}{u} \frac {du}{dm} dm[/tex].
    Dm's cancel out, thats why your m goes away!!
    And obviously [tex]\int \frac{1}{u} du = \log u [/tex].

    Now that looks the same form as the right hand side of your first post. Therefore [itex]t=\log (m/mo)[/itex]. Hopefully that was what you are looking for.
     
  16. Dec 29, 2006 #15

    ssd

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    You are right but I want to add,
    ln(m/mo)=t+c, c is the constant of integration
    Simplifying, ln(m)=t+k, k= c+ ln(mo).
     
  17. Dec 30, 2006 #16

    Gib Z

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    O yes sorry, forgot about the Constant, my bad. Ty for correcting me.
     
  18. Dec 30, 2006 #17
    [tex]\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}[/tex]

    What was wrong with substiting X=Log m?

    [tex]\int{\frac{dx}{x-log m(0)}[/tex]

    Remember, if e^x=m then, e^xdx =dm; so ms in the top and bottem cancel out.
     
    Last edited: Dec 30, 2006
  19. Dec 30, 2006 #18

    Gib Z

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    Yes but then you have another integral to solve via more substitution. Nothing wrong with your method, just longer.
     
  20. Dec 31, 2006 #19

    ssd

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    My natural instinct was to substitute u= ln(m/mo) , for the probable shortest way, I saw later 'Gib Z' also did the same substitution.
     
  21. Dec 31, 2006 #20

    Gib Z

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    Theres a special order to choose your u. Its very helpful: LIATE. Logs, Inverse Trig, Algebraic, Trig, and Exponentials. Go down the list and pick the first one as your u. In that case its obvious this time its the Log.
     
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