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Thread: improper integral #5

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

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


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


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

    Spoiler:


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


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


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


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


    Rewrite this as


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


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


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


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


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




    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



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


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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 $\displaystyle \gamma $, namely that $\displaystyle \int^{\infty}_{0} \Big(\frac{1}{1+x} - e^{-x} \Big) \ \frac{dx}{x} = -\psi_{0}(1) = \gamma $.


    Can you also show that $\displaystyle \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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