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Math Help - Generalized integral

  1. #1
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    Generalized integral

    How do I solve this generalized integral??

    \displaystyle\int_{-\infty}^{\infty}{\frac{1}{1+{x^6^}}} dx

    I appriciate any guidance. Thank you
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  2. #2
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    Have you learned how to solve complex integrals? 'Complexifying' the integral and then using the residue theorem can solve this quite easily, but it won't help if you haven't studied those tools.
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  3. #3
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    Hello, 4Math!

    This is a truly ugly problem, but it can be solved . . .



    \displaystyle \int_{\text{-}\infty}^{\infty}{\frac{dx}{1+x^6}

    The denominator is the sum of cubes: . x^6 + 1 \;=\;(x^2)^3 + (1)^3

    . . which factors: . (x^2 + 1)(x^4 - x^2 + 1)


    The quartic can also be factored . . .

    Add and subtract 3x^2\!:

    . . x^4 - x^2 + 1 + 3x^2 - 3x^2

    . . \;=\;x^4 + 2x^2 + 1 - 3x^2

    . . =\;(x^2 + 1)^2 - (\sqrt{3}\;\!x)^2 .
    (difference of squares)

    . . =\;(x^2+1 - \sqrt{3}\;\!x)(x^2+1 + \sqrt{3}\;\1x)

    . . =\;(x^2 - \sqrt{3}\;\!x + 1)(x^2 + \sqrt{3}\;\!x + 1)


    The denominator is: . x^6+1 \:=\:(x^2 + 1)(x^2 - \sqrt{3}\;\!x + 1)(x^2 + \sqrt{3}\;\!x + 1)


    You know what to do, right?

    Apply partial fraction decomposition . . .

    \displaystyle \frac{1}{(x^2+1)(x^2-\sqrt{3}\;\!x + 1)(x^2 + \sqrt{3}\;\!x + 1)}

    . . . . . . . . . \displaystyle =\;\frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2-\sqrt{3}\;\!x+1} + \frac{Ex + F}{x^2 + \sqrt{3}\;\!x + 1}


    I'll wait in the car . . .
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  4. #4
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    I'll bring the burgers and chips.

    Soroban, I get this picture of square roots, squares and cubes strewn all around inside the boot of your car!
    So educational!
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  5. #5
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     \int_{-\infty}^{\infty} \frac{dx}{1 + x^6}


     = 2   \int_0^{\infty} \frac{dx}{1 + x^6}

    Consider the integral next to  2 :

     I =  \int_0^{\infty} \frac{dx}{1 + x^6}

    Sub.  x \mapsto 1/x ( maybe i have been using this substitution much more times than trigo. sub. since I realized its power and magic ! )


     I =  \int_0^{\infty} \frac{x^4 dx}{1 + x^6}

    Summation gives :

     2I =  \int_0^{\infty} \frac{1+x^4 }{1 + x^6} ~dx

      =  \int_0^{\infty} \frac{(1-x^2 + x^4 )  + x^2}{1 + x^6} ~dx

     =   \int_0^{\infty} \frac{dx}{1 + x^2}   +  \frac{1}{3} \int_0^{\infty} \frac{d(x^3)}{1 + (x^3)^2 }

     = \frac{\pi}{2}  + \frac{1}{3} \frac{\pi}{2} = \frac{2\pi}{3}


    so we find that the answer is  \frac{2\pi}{3}


    If i am right , your generalized integral

     \int_0^{\infty} \frac{dx}{1 + x^n }  should be

     \frac{\pi}{n \sin( \frac{\pi}{n} ) }
    Last edited by simplependulum; September 16th 2010 at 02:04 AM.
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  6. #6
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    Quote Originally Posted by simplependulum View Post
     \int_{-\infty}^{\infty} \frac{dx}{1 + x^6}


     = 2   \int_0^{\infty} \frac{dx}{1 + x^6}

    Consider the integral next to  2 :

     I =  \int_0^{\infty} \frac{dx}{1 + x^6}

    Sub.  x \mapsto 1/x ( maybe i have been using this substitution much more times than trigo. sub. since I realized its power and magic ! )


     I =  \int_0^{\infty} \frac{x^4 dx}{1 + x^6}

    Summation gives :

     2I =  \int_0^{\infty} \frac{1+x^4 }{1 + x^6} ~dx

      =  \int_0^{\infty} \frac{(1-x^2 + x^4 )  + x^2}{1 + x^6} ~dx

     =   \int_0^{\infty} \frac{dx}{1 + x^2}   +  \frac{1}{3} \int_0^{\infty} \frac{d(x^3)}{1 + (x^3)^2 }

     = \frac{\pi}{2}  + \frac{1}{3} \frac{\pi}{2} = \frac{2\pi}{3}


    so we find that the answer is  \frac{2\pi}{3}


    If i am right , your generalized integral

     \int_0^{\infty} \frac{dx}{1 + x^n }  should be

     \frac{\pi}{n \sin( \frac{\pi}{n} ) }

    I took Defunkt's advised (See the attached pdf): Complexifying integral.pdf

    Quote Originally Posted by Defunkt View Post
    Have you learned how to solve complex integrals? 'Complexifying' the integral and then using the residue theorem can solve this quite easily, but it won't help if you haven't studied those tools.
    and complexified the integral. It was not easy. After tedious calculations I finally solved it but didn't know if it was correct. But we both get the same result, so I sure hope its correct. Thank you very much for your response.

    Kind regards
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