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Thread: Improper Integral

  1. #1
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    Improper Integral

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

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

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

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

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

    and $\displaystyle \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 $\displaystyle \int_{0}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}$?

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

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

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

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

    and $\displaystyle \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
    Moo
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    Answer by Wolframalpha :
    Spoiler:
    $\displaystyle \frac{4\Gamma\left(\tfrac 54\right)^2}{\sqrt{\pi}}$
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