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Thread: Complex Integral

  1. #1
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    Complex Integral

    Explaining your method compute..

    $\displaystyle J=\int_{-\infty }^{\infty }\frac{x^{2}dx}{1+x^{6}}$

    Is it right to say that we use a contour consisting of the real interval [-R,R].

    and to write the integral again in terms of f(z)?
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  2. #2
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    Hello, zizou1089!

    $\displaystyle J=\int_{-\infty }^{\infty }\frac{x^2\,dx}{1+x^6}$

    We have: .$\displaystyle \int\frac{x^2\,dx}{1 + (x^3)^2} $

    Let: .$\displaystyle u \:=\:x^3 \quad\Rightarrow\quad du \:=\:3x^2\,dx \quad\Rightarrow\quad x^2\,du \:=\:\tfrac{1}{3}du$

    Substitute: .$\displaystyle \int\frac{\frac{1}{3}\,du}{1+u^2} \;=\;\tfrac{1}{3}\arctan u + C$

    Back-substitute: .$\displaystyle \tfrac{1}{3}\arctan(x^3) + C$


    Evaluate: .$\displaystyle \tfrac{1}{3}\arctan(x^3)\,\bigg]^{\infty}_{\text{-}\infty}$

    . . . . . . $\displaystyle =\;\lim_{b\to\infty} \tfrac{1}{3}\arctan(x^3)\,\bigg]^b_{\text{-}b} $

    . . . . . . $\displaystyle = \;\frac{1}{3}\cdot\lim_{b\to\infty}\bigg[\arctan(b) - \arctan(\text{-}b)\bigg]$

    . . . . . . $\displaystyle =\;\frac{1}{3}\bigg[\frac{\pi}{2} - \left(\text{-}\frac{\pi}{2}\right)\bigg]$

    . . . . . . $\displaystyle =\;\frac{\pi}{3}$

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