Integral of 1/ln(x)

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


    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

    Last edited: Dec 26, 2006
  2. jcsd
  3. Make a substitution u=log(m/mo). I assumed that mo in the denominator was supposed to be mo.
  4. 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. 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. I guess that was wrong.
    Last edited: Dec 27, 2006
  7. How do you get that? I'm pretty sure that the first integral is a fairly straightforward computation once you make a substitution.
  8. [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. 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. MWM

    MWM 2

    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. 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. MWM

    MWM 2

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

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

    surely the integral would be


  14. How do you get to that? Try using the substitution u=log(m/m0) the integral will turn out very nicely.
  15. Gib Z

    Gib Z 3,347
    Homework Helper

    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. 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. Gib Z

    Gib Z 3,347
    Homework Helper

    O yes sorry, forgot about the Constant, my bad. Ty for correcting me.
  18. [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. Gib Z

    Gib Z 3,347
    Homework Helper

    Yes but then you have another integral to solve via more substitution. Nothing wrong with your method, just longer.
  20. 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. Gib Z

    Gib Z 3,347
    Homework Helper

    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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