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Math Help - Improper Integral

  1. #1
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    Improper Integral

    Is there any method to evaluate  \int_{0}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}?

    If there is , please just say yes and hide the solution

    and i think this integral should converge to a value :
     \int_{0}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}

     =  \int_{0}^{1} \frac{dx}{ \sqrt{ 1 + x^4 }} + \int_{1}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}

    but  \int_{0}^{1} \frac{dx}{ \sqrt{ 1 + x^4 }} = \int_{1}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}

    and  \int_{1}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}  <  \int_{1}^{\infty} \frac{dx}{x^2} which converges to 1

    Thus , the integral should be smaller than 2
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by simplependulum View Post
    Is there any method to evaluate  \int_{0}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}?

    If there is , please just say yes and hide the solution

    and i think this integral should converge to a value :
     \int_{0}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}

     =  \int_{0}^{1} \frac{dx}{ \sqrt{ 1 + x^4 }} + \int_{1}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}

    but  \int_{0}^{1} \frac{dx}{ \sqrt{ 1 + x^4 }} = \int_{1}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}

    and  \int_{1}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}  <  \int_{1}^{\infty} \frac{dx}{x^2} which converges to 1

    Thus , the integral should be smaller than 2
    yeah there is
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  3. #3
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    Answer by Wolframalpha :
    Spoiler:
    \frac{4\Gamma\left(\tfrac 54\right)^2}{\sqrt{\pi}}
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