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Math Help - Integral over Logarithm with Polynomial in the Argument

  1. #1
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    Integral over Logarithm with Polynomial in the Argument

    Hello,
    I would like to solve the following integral
     A(p^2 , m_R^2) = - \int dx \; ln(1 - x(1-x) \frac{p^2}{m_R^2})

    For  p^2 < (2 m_R)^2 and for  p^2 > (2 m_R)^2

    I tried some substitutions without success.

    Could anybody help me solving this?

    Regards,
    Mr. Fogg
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  2. #2
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    Quote Originally Posted by Kai.Grohs View Post
    Hello,
    I would like to solve the following integral
     A(p^2 , m_R^2) = - \int dx \; ln(1 - x(1-x) \frac{p^2}{m_R^2})

    For  p^2 < (2 m_R)^2 and for  p^2 > (2 m_R)^2

    I tried some substitutions without success.

    Could anybody help me solving this?

    Regards,
    Mr. Fogg
    Hi

    Your problem is the same as solving this integral

    I = \int  ln(ax^2+bx+c) \; dx

    In your problem the discriminant b^2-4ac is negative if p^2 < (2 m_R)^2 and positive if p^2 > (2 m_R)^2

    This integral can be solved by parts and then by rearranging a little bit into

    I = \left(x+\frac{b}{2a}\right) ln(ax^2+bx+c) - 2x - \int  \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx

    The last integral depends by the sign of the discriminant

    If negative then you can chose the proper change of variables to get (1+u) at the denominator, and then integrate into arctan

    If positive then you can factor the denominator into 2a(x-x0)(x-x1) and go further in the decomposition of the last integral
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  3. #3
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    Hello,
    thanks for your help, but I don't see an integration by parts in this
     - \int_0^1 dx (1-x \frac{p^2}{m_R^2} + x^2 \frac{p^2}{m_R^2})

    I tried this
     - \int_0^1 dx \; x' (1-x \frac{p^2}{m_R^2} + x^2 \frac{p^2}{m_R^2})
    but it didn't help me. Maybe I made a mistake.

    Could you help me a little more precisely?

    Regards,
    Kai
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  4. #4
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    If you don't mind I will keep this form

    I = \int  ln(ax^2+bx+c) \; dx

    and you will set a = \frac{p^2}{m_R^2} etc ...

    I = \int  ln(ax^2+bx+c) \; dx

    By parts with u'(x) = 1 and v(x) = ln(ax^2+bx+c)

    2ax+b)}{ax^2+bx+c} \; dx " alt="I = x\:ln(ax^2+bx+c) - \int\frac{x\2ax+b)}{ax^2+bx+c} \; dx " />

    I = x\:ln(ax^2+bx+c) - \int\frac{2ax^2+2bx+2c-bx-2c}{ax^2+bx+c} \; dx

    I = x\:ln(ax^2+bx+c) - \int \:2 \; dx + \int\frac{bx+2c}{ax^2+bx+c} \; dx

    I = x\:ln(ax^2+bx+c) - 2\:x + \int\frac{bx+2c}{ax^2+bx+c} \; dx

    And

    \int\frac{bx+2c}{ax^2+bx+c} \; dx = \int\frac{\frac{b}{2a}(2ax+b)-\frac{b^2}{2a}+2c}{ax^2+bx+c} \; dx

    \int\frac{bx+2c}{ax^2+bx+c} \; dx = \frac{b}{2a}\:ln(ax^2+bx+c) -\int \frac{\frac{b^2}{2a}-2c}{ax^2+bx+c} \; dx
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  5. #5
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    Quote Originally Posted by running-gag View Post
    The last integral depends by the sign of the discriminant

    If negative then you can chose the proper change of variables to get (1+u) at the denominator, and then integrate into arctan

    If positive then you can factor the denominator into 2a(x-x0)(x-x1) and go further in the decomposition of the last integral
    Could you tell me the proper change of variables. I don't see it. I am just a physicist, who has not this magic eye for finding the proper change

    Regards,
    Mr. Fogg
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  6. #6
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    Let \Delta = b^2-4ac be the discriminant
    If \Delta < 0 then let u = \frac{2a}{\sqrt{-\Delta}}\:\left(x+\frac{b}{2a}\right) = \frac{2a}{\sqrt{-\Delta}}\:x+\frac{b}{\sqrt{-\Delta}}

    Then

    - \int  \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = - \int  \frac{b^2-4ac}{2a^2\left(x^2+\frac ba\:x+\frac ca\right)} \; dx

    - \int  \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = - \int  \frac{\Delta}{2a^2 \left[\left(x+\frac{b}{2a}\right)^2+\frac ca-\frac{b^2}{4a^2}\right]}  \; dx

    - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = - \int \frac{\Delta}{2a^2\left[\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a^2}\right]} \; dx

    - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = - \int \frac{\Delta}{2a^2\left[-\frac{\Delta}{4a^2}\:u^2-\frac{\Delta}{4a^2}\right]} \; \frac{\sqrt{-\Delta}}{2a}\:du

    - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = \frac{\sqrt{-\Delta}}{a}\:\int \frac{du}{1+u^2}

    - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = \frac{\sqrt{-\Delta}}{a}\:Arctan(u)

    - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = \frac{\sqrt{-\Delta}}{a}\:Arctan \left(\frac{2a}{\sqrt{-\Delta}}\:x+\frac{b}{\sqrt{-\Delta}}\right)
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  7. #7
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    Thanks for your help.

    I went through your calculation, but didn't get the correct solution, which is



    Do you have any idea, how to get this solution?

    Ragards,
    Mr. Fogg
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  8. #8
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    So far I did not check the second case

    But for the first case p^2 < (2m_R^2) I applied the formula I gave you in my previous posts and I could find the good result

    a = \frac{p^2}{m_R^2}

    b = -\frac{p^2}{m_R^2}

    c = 1

    \Delta = \left(\frac{p^2}{m_R^2}\right)^2-4\:\frac{p^2}{m_R^2} = \frac{p^2}{m_R^2}\:\left(\frac{p^2}{m_R^2}-4\right)

    In your problem the discriminant b^2-4ac is negative if p^2 < (2 m_R)^2 and positive if p^2 > (2 m_R)^2

    I = \left(x+\frac{b}{2a}\right) ln(ax^2+bx+c) - 2x - \int  \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx

    And

    - \int \frac{b^2-4ac}{2a(ax^2+bx+c)} \; dx = \frac{\sqrt{-\Delta}}{a}\:Arctan \left(\frac{2a}{\sqrt{-\Delta}}\:x+\frac{b}{\sqrt{-\Delta}}\right)

    Since you are integrating between 0 and 1 you do not have to care about the first term \left(x+\frac{b}{2a}\right) ln(ax^2+bx+c) because for 1, ln(ax+bx+c) gives ln(a+b+c)=ln(1)=0 ; and for 0, ln(ax+bx+c) gives ln(c)=ln(1)=0

    The second term (-2x) gives -2 for 1 and 0 for 0, therefore -2


    For the third term \frac{\sqrt{-\Delta}}{a}\:Arctan \left(\frac{2a}{\sqrt{-\Delta}}\:x+\frac{b}{\sqrt{-\Delta}}\right)

    For 1, Arctan \left(\frac{2a}{\sqrt{-\Delta}}\:x+\frac{b}{\sqrt{-\Delta}}\right) = Arctan \left(\frac{2a+b}{\sqrt{-\Delta}}\right) = Arctan \left(\frac{a}{\sqrt{-\Delta}}\right)

    For 0, Arctan \left(\frac{2a}{\sqrt{-\Delta}}\:x+\frac{b}{\sqrt{-\Delta}}\right) = Arctan \left(\frac{b}{\sqrt{-\Delta}}\right) = Arctan \left(\frac{-a}{\sqrt{-\Delta}}\right)

    But Arctan(-x)=-Arctan(x) therefore for 0 it gives -Arctan \left(\frac{a}{\sqrt{-\Delta}}\right)

    Therefore the third term gives 2\:\frac{\sqrt{-\Delta}}{a}\:Arctan \left(\frac{a}{\sqrt{-\Delta}}\right)

    \frac{\sqrt{-\Delta}}{a} = \frac{\sqrt{4\:\frac{p^2}{m_R^2}-\left(\frac{p^2}{m_R^2}\right)^2}}{\frac{p^2}{m_R^  2}}= \frac{m_R^2}{p^2}\:\sqrt{4\:\frac{p^2}{m_R^2}-\frac{p^4}{m_R^4}}

    \frac{\sqrt{-\Delta}}{a} = \sqrt{\frac{m_R^4}{p^4}\:\left(4\:\frac{p^2}{m_R^2  }-\frac{p^4}{m_R^4}\right)} = \sqrt{4\:\frac{m_R^2}{p^2}-1} = \sqrt{\frac{(2m_R)^2}{p^2}-1}

    At the end the third term gives

    2\:\sqrt{\frac{(2m_R)^2}{p^2}-1}\:\:Arctan \left(\frac{1}{\sqrt{\frac{(2m_R)^2}{p^2}-1}}\right) = 2\:\left(\frac{(2m_R)^2}{p^2}-1\right)^{\frac12}\:\:tan^{-1} \left(\frac{(2m_R)^2}{p^2}-1\right)^{-\frac12}
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