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Integrating out of the real domain of a function

  1. Nov 27, 2012 #1
    Check this out:

    [tex]\int_{-1}^{0} ln(x) dx[/tex]

    [tex]u=ln(x), dv=dx[/tex]
    [tex]du=\frac{1}{x},v=x[/tex]

    [tex]\int_{-1}^{0} ln(x) dx= x\space ln(x) \|_{-1}^{0} -\int_{-1}^{0} \frac{x}{x} dx[/tex]
    [tex]=x\space ln(x) \|_{-1}^{0}-x \|_{-1}^{0}[/tex]
    [tex]=\lim_{x\to0} x\space ln(x)-(-ln(-1))-(0-(-1))[/tex]
    [tex]\lim_{x\to0} x\space ln(x)=lim_{x\to0} \frac{ln(x)}{1/x}[/tex]
    [tex]=\lim_{x\to0} \frac{1/x}{-1/x^2}[/tex]
    [tex]=\lim_{x\to0} -x=0[/tex]

    [tex]\int_{-1}^{0} ln(x)dx=ln(-1)-1[/tex]
    [tex]e^{i\pi}=-1\to ln(-1)=i\pi[/tex]

    [tex]\int_{-1}^{0}ln(x)\space dx=i\pi-1[/tex]

    Is this legitimate?

    P.S. Why don't my limits look right?
     
    Last edited: Nov 27, 2012
  2. jcsd
  3. Nov 27, 2012 #2
    By the way, this is not a textbook style question, I'm wondering whether or not it makes sense to integrate out of a function's domain.
     
  4. Nov 27, 2012 #3

    Mark44

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    Nothing below is legitimate. For the Fundamental Theorem of calculus to apply, the function (ln(x) here) has to be defined on the interval [-1, 0], and continuous on the interior of this interval. ln(x) is defined only for x > 0.
    Use \lim, not lim.
     
  5. Nov 27, 2012 #4
    So one can never integrate out of a function's domain? What about in complex analysis?
     
    Last edited: Nov 27, 2012
  6. Nov 27, 2012 #5

    haruspex

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    Reinterpreting your integral in the complex plane, we need r >= 0, so over the x interval (-1,0) we will have θ=π:
    [itex]\int_{-1}^0ln(z)dz = \int_{1}^0(ln(r)+i\pi) d(-r) = \left[r ln(r) - r + i\pi r \right]_0^1 = i\pi - 1[/itex]
     
  7. Nov 27, 2012 #6
    How did you go to polar coordinates?

    Also, is integrating the function without reinterpreting it into the complex plane o.k. (i.e. is my way of arriving at the same answer legitimate)?
     
    Last edited: Nov 27, 2012
  8. Nov 27, 2012 #7

    haruspex

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    Not polar co-ordinates (which would be (r,θ)) but the polar form of complex variables (z=re). It's better not to confuse the two.
    No, for the reasons given in other posts.
     
  9. Nov 27, 2012 #8
    OK, but what did you do to transform the integral into that form?
     
  10. Nov 27, 2012 #9

    haruspex

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    First, you need to assume a specific path, P (though for the ln() function it won't matter as long as we don't go around the origin to get there). Assume it's along the negative real axis. So at all points in the path, z is of the form re.
    [itex]\int_Pln(z)dz = \int_P(ln(re^{iπ}) dre^{iπ}= e^{iπ}\int_{r=1}^0(ln(r)+iπ) dr = -\int_{r=1}^0(ln(r)+iπ) dr[/itex]
    [itex] = \int_{r=0}^1(ln(r)+iπ) = \left[(r ln(r)-r+iπr)\right]_0^1 = -1+iπ [/itex]
     
  11. Nov 27, 2012 #10
    Alright. What exactly do you mean by a "path"? And why did the limits of integration change?
     
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