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Math Help - improper integral #5

  1. #1
    Super Member Random Variable's Avatar
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    improper integral #5

    Evaluate  \int^{\infty}_{0} \frac{e^{-x^{p}} - e^{-x^{q}}}{x} \ dx \ \ p,q>0 .


    Then argue that  \int^{\infty}_{0} \frac{e^{-x^{p}} - e^{-x^{q}}}{x} \ dx = \int^{\infty}_{0} \frac{\cos (x^{p}) - \cos(x^{q})}{x} \ dx .
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: improper integral #5

    Quote Originally Posted by Random Variable View Post
    Evaluate  \int^{\infty}_{0} \frac{e^{-x^{p}} - e^{-x^{q}}}{x} \ dx \ \ p,q>0 .


    Then argue that  \int^{\infty}_{0} \frac{e^{-x^{p}} - e^{-x^{q}}}{x} \ dx = \int^{\infty}_{0} \frac{\cos (x^{p}) - \cos(x^{q})}{x} \ dx .

    Spoiler:


    Let I_{p,q} denote our integral. Then, evidently


    \displaystyle I(p,q)=-\int_0^{\infty}\int_q^p e^{-x^y}x^{y-1}\log(x)\;dy\,dx


    Switch the order of integration (Fubini's theorem is satisfied) to get


    \displaystyle I(p,q)=-\int_0^{\infty}\int_q^p x e^{-x^y}x^y\log(x)\;dx\,dy


    Rewrite this as


    \displaystyle I(p,q)=-\int_q^p\frac{1}{y^2}\int_0^{\infty}ye^{-x^y}x^{y-1}\log\left(x^y\right)\; dx\,dy


    Make the obvious substitution ( z=x^y) on the inside to turn it into


    \displaystyle I(p,q)=\left(-\int_q^p\frac{dy}{y^2}\right)\left(\int_0^{\infty}  e^{-x}\log(x)\text{ }dx\right)


    Thus, the first integral being easy, and the second being the definition of the Euler-Mascheroni constant we find that


    \displaystyle I(p,q)=\frac{q-p}{pq}\gamma




    To prove the second assertion merely write the exponentials in complex notation and separate the real and imaginary parts. The theorem then becomes immediate from the common theorem that



    \displaystyle \int_0^{\infty}\frac{\sin(ax)-\sin(bx)}{x}\;dx=0\quad a\ne b


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  3. #3
    Super Member Random Variable's Avatar
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    Re: improper integral #5

    I didn't even think there was a way to write it as a double integral that would make evaluating it easier.


    I used another integral representation of  \gamma , namely that  \int^{\infty}_{0} \Big(\frac{1}{1+x} - e^{-x} \Big) \ \frac{dx}{x} = -\psi_{0}(1) = \gamma .


    Can you also show that  \int^{\infty}_{0} \frac{ e^{-x^{p}} - e^{-x^{q}}}{x} \ dx = \int_{0}^{\infty} \frac{\cos(x^{p}) - e^{-x^{q}}}{x} \ dx ?
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  4. #4
    Super Member Random Variable's Avatar
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    Re: improper integral #5

    Drexel28

    Could you explain your reasoning for the second part of the question in more detail? I feel really stupid because I don't understand where the integral of the form you posted even comes into play.
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