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Antiderivative of 1/sqrt(lnx - c)

  1. Jul 28, 2005 #1
    does anybody know the antiderivative of 1/(sqrt(lnx - (a constant))
     
  2. jcsd
  3. Jul 28, 2005 #2
    Try substituting u = lnx.
     
  4. Jul 28, 2005 #3
    i don't think it is possible to solve that problem. could be wrong though
     
  5. Jul 29, 2005 #4

    HallsofIvy

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    Hmmm, the substitution u= ln(x)- C, along the lines Icebreaker suggested, changes the integral to [tex]\int\frac{e^{u+C}du}{\sqrt{u}}[/tex] but I don't see any simple way to do that.

    In fact, the "obvious" substitution [tex]u= \sqrt{ln(x)- C}[/tex] converts the
    integral to [tex]2e^C\int e^{u^2}du[/tex] which has no elementary antiderivative!
     
  6. Jul 29, 2005 #5

    saltydog

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    Your task, should you choose to accept it, is to show how Mathematica comes up with the following:

    [tex]\int \frac{dx}{\sqrt{ln(x)-c}}=-\frac{e^{c}\sqrt{\pi}\left(1-Erf\left[\sqrt{c-ln(x)}\right]\right)\sqrt{-c+ln(x)}}{\sqrt{c-ln(x)}}[/tex]

    I'll try too.
     
    Last edited: Jul 29, 2005
  7. Jul 29, 2005 #6

    lurflurf

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    You can verify the result if you like by showing the derivatives of each side are equal. Seeing how Mathematica comes up with it though is much harder and not very helpful. Mathematica does not do integral like a person would. If Integrate had a ShowSteps option what you would see would likely be ridiculuously long and complicated.
    On the verification front
    the substitution u=i Sqrt(log(x)-C) gives
    [tex]\int \frac{1}{\sqrt{\log(x)-C}}dx=-2i e^c\int e^{-u^2}du[/tex]
    Thus all that is needed to bridge to the mathematica result is to use (under suitable conditions)
    sqrt(C-log(x))=i sqrt(log(x)-C)
     
    Last edited: Jul 29, 2005
  8. Jul 29, 2005 #7

    saltydog

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    Hello Lurflurf. The destination is not the objective but rather the journey. :smile:

    "Equal rights for special functions." :smile:
     
  9. Jul 29, 2005 #8

    lurflurf

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    Of course erf's evil brother erfi could also be used.
     
  10. Jul 29, 2005 #9

    saltydog

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    Thanks Lurflurf. That's very helpful cus' I wasn't getting anywhere in the Real world. Still not happening for me but I'll continue working on it . . . it's a pleasant ride. :smile:
     
  11. Jul 29, 2005 #10

    saltydog

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    I'm making progress with this. Using Lurflurf's substitution, I obtain:

    [tex]\int \frac{dx}{\sqrt{ln(x)-c}}=-2i e^c \int e^{-u^2}du,\quad u(x)=i\sqrt{ln(x)-c}[/tex]

    Since:

    [tex]Erf(t)=\frac{2}{\sqrt{\pi}}\int e^{-t^2}dt[/tex]

    Then:

    [tex]\int e^{-t^2}dt=\frac{\sqrt{\pi}}{2}Erf(t)[/tex]

    Thus:

    [tex]\int \frac{dx}{\sqrt{ln(x)-c}}dx=-i e^c\sqrt{\pi} Erf \left[i\sqrt{ln(x)-c}\right][/tex]

    Now, as long as [itex] c\leq ln(x)[/itex],

    [tex] Erf\left[i\sqrt{ln(x)-c}\right] [/tex]

    will be a pure complex number (prove?) and thus the integral will be a positive real number.

    I don't know how to calculate (numerical or otherwise) the value of erf(ai). Can someone please show me?

    [tex]Erf[ia]=?[/tex]
     
  12. Jul 29, 2005 #11

    saltydog

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    Alright, I got it by considering the Riemann sum:

    [tex]Erf(ia)=\frac{2}{\sqrt{\pi}}\int_0^{ia} e^{-t^2}dt=\frac{2i}{\sqrt{\pi}}\int_0^a e^{t^2}dt[/tex]

    Thus, applying this to the problem above we have:

    [tex]\int \frac{dx}{\sqrt{ln(x)-c}}=-ie^c\sqrt{\pi} Erf \left[ i\sqrt{ln(x)-c}\right][/tex]

    Now:

    [tex]
    \begin{align*}
    Erf \left[i\sqrt{ln(x)-c}\right]&=\frac{2}{\sqrt{\pi}}\int_0^{i\sqrt{ln(x)-c}} e^{-t^2}dt \\
    &=\frac{2i}{\sqrt{\pi}}\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt
    \end{align}
    [/tex]

    Thus we finally have:

    [tex]\int \frac{dx}{\sqrt{ln(x)-c}}=2e^c\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt[/tex]
     
    Last edited: Jul 29, 2005
  13. Jul 29, 2005 #12

    saltydog

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    Well, wait. That's just substituting a tough integral for a more-tough one. So, let's do this:

    Let:

    [tex]Sa[x,c]=\int_0^{\sqrt{ln(x)-c}} e^{t^2}dt;\quad x>0\quad\text{and}\quad ln(x)\geq c[/tex]

    Then:

    [tex]\int \frac{dx}{\sqrt{ln(x)-c}}=2e^cSa(x,c)[/tex]

    See, that's much better. :smile:
     
  14. Jul 29, 2005 #13

    lurflurf

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    Your to quick but I'll say this anyway.
    [tex]erf(i a)=-i erfi(a)=-i\frac{2}{\sqrt{\pi}}\int_0^a e^{z^2} dz[/tex]
    for fun put the integral into mathematica differently and try to get a result in terms of erfi.
    like
    integral=exp(C)sqrt(pi)efi(sqrt(x))
    Also when trying to get the same form as mathematica
    u=i sqrt(log(x)-C)=sqrt(-log(x)+C)
     
    Last edited: Jul 29, 2005
  15. Jul 29, 2005 #14

    saltydog

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    I don't think it's a minus i right? That is:

    [tex]erf[i a]=\frac{2i}{\sqrt{\pi}}\int_0^a e^{z^2} dz[/tex]

    I mean you're better at this than me. :smile:
     
  16. Jul 30, 2005 #15

    lurflurf

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    Drats I was thinking sdrawkcab.
    erf(i a)=i erfi(a)
    I mixed that up with
    erfi(a)=-i erf(i a)
    to produce the error.
     
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