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Thread: Residues

  1. #1
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    Residues

    Evaluate $\displaystyle \int\limits_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}}\ dx$
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Chandru1 View Post
    Evaluate $\displaystyle \int\limits_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}}\ dx$
    Hint:

    Let $\displaystyle a=\frac13 $ and $\displaystyle x=e^t \implies dx = e^tdt $

    Thus $\displaystyle \int_0^\infty \frac{x^a}{x^2+1}\ dx = \int_{-\infty}^\infty \frac{e^{(a+1)t}}{e^{2t}+1}\ dt $
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  3. #3
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    Let $\displaystyle x= t^3 $

    The integral

    $\displaystyle I = 3 \int_0^{\infty} \frac{ t^3 }{1+ t^6} ~dt $

    Sub $\displaystyle t = 1 /s $

    $\displaystyle I = 3 \int_0^{\infty} \frac{s}{1+s^6}~ds$

    Sub $\displaystyle s^2 = u $

    $\displaystyle I = \frac{3}{2} \int_0^{\infty} \frac{du}{1+u^3}$

    $\displaystyle I = \frac{3}{2} \int_0^{\infty} \frac{ 1+ u - u}{1+u^3}~du $

    $\displaystyle I = \frac{3}{2} \left[ \int_0^{\infty} \frac{1+u}{(1+u)(1-u+u^2)}~du - \int_0^{\infty} \frac{u}{1+u^3}~du \right] $

    But by substituting $\displaystyle u= 1/y $ in the integral $\displaystyle \int_0^{\infty} \frac{du}{1+u^3}$ , we have

    $\displaystyle \int_0^{\infty} \frac{du}{1+u^3} = \int_0^{\infty} \frac{u du}{1+u^3}$

    Thus we have

    $\displaystyle \int_0^{\infty} \frac{u du}{1+u^3} = 2I/3 $

    $\displaystyle 2I = \frac{3}{2} \int_0^{\infty} \frac{du}{1-u+u^2} $

    $\displaystyle I = \frac{3}{4} \cdot \frac{4\pi}{3\sqrt{3}} $

    $\displaystyle I = \frac{\pi}{\sqrt{3}} $
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    This is a much better way than my method! It requires no complex analysis.

    My way does generalize this problem though:

    Given $\displaystyle -1<\Re(a)<1 $, $\displaystyle \int_0^\infty \frac{x^a}{x^2+1}\ dx = \frac12\pi\sec\left(\frac{\pi a}{2}\right) $

    p.s. May I ask how you came up with your solution?
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