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Math Help - Complex Integral

  1. #1
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    Complex Integral

    Explaining your method compute..

    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!

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

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

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

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

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


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

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

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

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

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

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