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Math Help - A integral problem hard to solve

  1. #1
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    Smile A integral problem hard to solve

    I am stuck for weeks...

    integral from 0 to infinity 1/(1+x^(2n)) dx

    someone told me tips... but I still can't solve

    tips:x^n=tan

    Could anyone teach me how to solve it step by step?
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  2. #2
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    Quote Originally Posted by mymaydayya View Post
    I am stuck for weeks...

    integral from 0 to infinity 1/(1+x^(2n)) dx

    someone told me tips... but I still can't solve

    tips:x^n=tan

    Could anyone teach me how to solve it step by step?
    let x=(\tan t)^{\frac{1}{n}}. then dx=\frac{1}{n}(1 + \tan^2 t)(\tan t)^{\frac{1}{n} - 1}dt. therefore:

    \int_0^{\infty} \frac{dx}{1+x^{2n}}=\frac{1}{n}\int_0^{\frac{\pi}{  2}}(\tan t)^{\frac{1}{n} - 1} dt=\frac{1}{n} \int_0^{\frac{\pi}{2}} (\sin t)^{\frac{1}{n}-1}(\cos t)^{1 - \frac{1}{n}} dt

    =\frac{1}{2n}B \left(\frac{1}{2n}, 1- \frac{1}{2n} \right)=\frac{1}{2n} \Gamma \left(\frac{1}{2n} \right) \Gamma \left(1 - \frac{1}{2n} \right) = \frac{\pi}{2n \sin \left(\frac{\pi}{2n} \right)}.

    the last identity is true by Euler's reflection formula.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    let x=(\tan t)^{\frac{1}{n}}. then dx=\frac{1}{n}(1 + \tan^2 t)(\tan t)^{\frac{1}{n} - 1}dt. therefore:

    \int_0^{\infty} \frac{dx}{1+x^{2n}}=\frac{1}{n}\int_0^{\frac{\pi}{  2}}(\tan t)^{\frac{1}{n} - 1} dt=\frac{1}{n} \int_0^{\frac{\pi}{2}} (\sin t)^{\frac{1}{n}-1}(\cos t)^{1 - \frac{1}{n}} dt

    =\frac{1}{2n}B \left(\frac{1}{2n}, 1- \frac{1}{2n} \right)=\frac{1}{2n} \Gamma \left(\frac{1}{2n} \right) \Gamma \left(1 - \frac{1}{2n} \right) = \frac{\pi}{2n \sin \left(\frac{\pi}{2n} \right)}.

    the last identity is true by Euler's reflection formula.

    sorry I don't know
    "B(1/2n,1-1/2n)" and the next one
    stand for
    could you give me link to wiki about it?
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  4. #4
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    Quote Originally Posted by mymaydayya View Post
    sorry I don't know
    "B(1/2n,1-1/2n)" and the next one
    stand for
    could you give me link to wiki about it?
    they are called "gamma" and "beta" functions. google them!
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