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Integral of a log

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Calculate the following integral:
    [tex]\int_0^a{\left(y-m_1u-c_1\right)\log{\left[u^2+\left(y-m_1u-c_1\right)^2\right]}{\rm d}u}[/tex]
    Where "a" is positive Real, "y, c, m" are Reals.


    2. Relevant equations
    [tex]\int{\log{\left[u^2+c^2\right]}{\rm d}u}=u\log{\left[u^2+c^2\right]}-2u+2c\arctan{(u/c)}[/tex]


    3. The attempt at a solution
    I'm not quite sure how to tackle this problem. With factors inside the log, that would've helped, but it is not the case. A hint would be appreciated.
     
  2. jcsd
  3. Nov 6, 2011 #2
    Need to try and encapsulate things to make it easier. Let y-c=k then you have:

    [tex]2\int(k-mu)\log[u^2-mu+k]du[/tex]

    You can complete the square right? Won't that get one part of it into the form you have? The other then is just a variable substitution say, v=u-c and you can solve:

    [tex]\int v\log(v^2+h^2)dv[/tex]

    right?
     
  4. Nov 6, 2011 #3

    SammyS

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    The integral under "Relevant equations" won't help directly, because there's a u in both terms of the integrand of the given definite integral.
     
  5. Nov 6, 2011 #4
    Shouldn't I get:

    [tex]\int(k-mu)\log[(m^2+1)u^2-2mku+k^2]du[/tex]

    Then, I can find the roots z1 and z2 of the polynomial and express the integral in the form:

    [tex]\int(k-mu)\log[(u-z_1)(u-z_2)]du=\int(k-mu)\log[u-z_1]du+\int(k-mu)\log[u-z_2]du[/tex]

    and then I integrate by parts with f=k-m*u and dg=log[u-z]?
     
  6. Nov 6, 2011 #5
    Ok. Sorry, I mis-read it. That looks good except I think you have to factor out the 1+m^2 before you find the roots and then write (I think):

    [tex]\int (k-mu)\left[\log(1+m^2)+\log(u-a)+\log(u-b)\right]du[/tex]
     
    Last edited: Nov 6, 2011
  7. Nov 6, 2011 #6
    I think for the assumptions on k and m (Reals), the roots can be complex. Isn't there something special to do in that case? I think there is something about the complex plane and singularities, but it's been a while, can't remember very well.
     
  8. Nov 6, 2011 #7
    That brings up an interesting phenomenon in Complex Analysis. Essentially, as long as the antiderivative is differentiable along the entire path of integration, then we need only evaluate the antiderivative at the endpoints of the interval.


    I don't think you'll encounter a problem with this integral though but not sure.
     
    Last edited: Nov 6, 2011
  9. Nov 6, 2011 #8
    Well, I should have a problem at u=z1 or u=z2, since there's a singularity at log(0). However, since I am integrating over the real axis, I guess I only get the singularities if z1 and z2 are reals?
     
  10. Nov 7, 2011 #9
    The integral would remain proper unless the zeros are along the path of integration.
     
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