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Integral of arcsinh(1/x)dx

  1. Nov 26, 2012 #1
    Hello, I am attempting to do a radiolgoical dose over equivalent fields problem. I have the following integral that I am trying to show each step for but am getting tricked up after the first integration. Any help would be greatly appreciated:

    1. The problem statement, all variables and given/known data
    ∫∫((1/√((x^2)+(y^2)))-μ')dxdy (μ' is a constant)also note the limits on X:0→L, and Y:0→W
    Solution: 2L*ln((D+W)/L)+2W*ln((L+D)/W)+μ'LW (where D=sqrt((W^2)+(L^2)))


    2. Relevant equations
    n/a


    3. The attempt at a solution
    I can get through the first integration:
    ∫asinh((L/y)-μ'L)dy
    Not sure what to do with the arcsinh, maybe express as logarithm?
     
  2. jcsd
  3. Nov 27, 2012 #2
    Yes, the inverse hyperbolic sine can be written as a logarithm: http://mathworld.wolfram.com/InverseHyperbolicSine.html . You should be able to write your first integral as ln(2(L + sqrt((L^2 + y^2))) - Lμ', though, which is different from what you got, although I may have read your integral badly. It doesn't get much nicer.
    The term sqrt(x^2 + y^2) may mean an easier integral in polar coordinates. The rectangular region of integration then changes to r in [0, Lsec(t)] and t in [0, arctan(W/L)] plus a second integral with r in [0, Wcsc(t)] and t in [arctan(W/L), pi/2]. Your integral then simplifies to
    [tex]\int_0^{\arctan\left(\frac{W}{L}\right)} \int_0^{L\sec\theta} 1 - \mu'r dr d\theta + \int_{\arctan\left(\frac{W}{L}\right)}^\frac{\pi}{2}\int_0^{W\csc\theta} 1 - \mu'r dr d\theta[/tex]
    This gives the same result with more transparent integrals.
     
  4. Nov 27, 2012 #3

    haruspex

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    Do you know how to convert to polar co-ordinates? The bounds are a bit tricky, but not too bad. For the second integral you'll need to integrate a secant (or two).
     
  5. Nov 27, 2012 #4
    Thank you for your help
     
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