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Math Help - integral with two parameters

  1. #1
    Super Member Random Variable's Avatar
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    integral with two parameters

     \int_{0}^{1} \frac{\sin(p \ln x) \cos(q \ln x)}{\ln x} \ dx where  p, q >0

    To me this integral is just begging to be written as a triple integral. But I can't think of a way of doing it that will make evaluating the integral any easier.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: integral with two parameters

    Quote Originally Posted by Random Variable View Post
     \int_{0}^{1} \frac{\sin(p \ln x) \cos(q \ln x)}{\ln x} \ dx where  p, q >0

    To me this integral is just begging to be written as a triple integral. But I can't think of a way of doing it that will make evaluating the integral any easier.
    Setting \ln x = t the integral becomes...

    I= \int_{0}^{\infty} \frac{\sin p t \ \cos q t }{t}\ e^{-t}\ dt = \mathcal{L} \{\frac{\sin p t \ \cos q t }{t}\}_{s=1} (1)

    Now You have to remember that...

    \sin p t\ \cos q t = \frac{1}{2}\ \{\sin (p-q) t + \sin (p+q) t\} (2)

    ... and that...

     \mathcal{L} \{\frac{\sin a t}{t}\} = \int_{s}^{\infty} \frac{a}{a^{2}+ u^{2}}\ du  =  \tan^{-1} \frac{a}{s} (3)

    ... and, combining (1), (2) and (3) You obtain...

    I= \frac{1}{2}\ \{ \tan^{-1} (p-q) + \tan^{-1} (p+q) \} (4)

    Kind regards

    \chi \sigma
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  3. #3
    Super Member Random Variable's Avatar
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    Re: integral with two parameters

    I'm pretty sure you meant  t = - \ln x .

    Interestingly, Maple returns the wrong answers for different values of the parameters while Wolfram Alpha returns the correct values. Like for p = 3 and q =2, Maple returns a negative value of  - \frac{1}{2} \arctan \Big( \frac{3}{2} \Big) which is clearly absurd when you look at the graph.
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  4. #4
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    Re: integral with two parameters

    Here's another way. Define

    I(p,q) = \int_0^1 \frac{\sin p \ln x \cos q \ln x}{\ln x} dx

    noting that I(0,q) = 0. Then make the substitution u = \ln x (as suggested) so

    I = \int_{- \infty}^0 \frac{\sin p u \cos q u \;e^u}{u} du.

    Then

    I_p = \int_{- \infty}^0 \cos p u \cos q u e^u du.

    Now let \cos pu \cos qu = \frac{1}{2} \cos (p-q)u + \frac{1}{2}\cos (p+q)u.

    Integration by part twice gives

    I_p = \frac{1}{2} \frac{1}{1 + (p-q)^2 } + \frac{1}{2} \frac{1}{1 + (p+q)^2}

    Integrating gives

    I = \frac{1}{2} \tan^{-1} (p-q) + \frac{1}{2} \tan^{-1} (p+q) + F(q)

    Since I(0,q) = 0 then F(q) = 0 thus giving what chisigma gave.
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