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Math Help - An Integral #2

  1. #1
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    Red face An Integral #2

    Here is the integral :
    \int_0^1{\sqrt[4]{1-x^4}}dx


    thanks very much.....
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  2. #2
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    Quote Originally Posted by Xingyuan View Post
    Here is the integral :
    \int_0^1{\sqrt[4]{1-x^4}}dx


    thanks very much.....
    let x=\sqrt{\sin t}. then \int_0^1 \sqrt[4]{1-x^4} \ dx=\frac{1}{2} \int_0^{\frac{\pi}{2}}(\sin t)^{-\frac{1}{2}} (\cos t)^{\frac{3}{2}} \ dt

    =\frac{1}{4}B\left(\frac{1}{4}, \frac{5}{4}\right)=\frac{\Gamma(\frac{1}{4})\Gamma  (\frac{5}{4})}{4\Gamma(\frac{3}{2})}=\frac{1}{8 \sqrt{\pi}}\left(\Gamma\left(\frac{1}{4} \right)\right)^2. \ \ \ \square
    Last edited by NonCommAlg; May 27th 2008 at 01:03 PM.
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  3. #3
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    another approach, probably easier, is to put x^4=t. then dx=\frac{1}{4}t^{-\frac{3}{4}}dt. thus:

    \int_0^1 \sqrt[4]{1-x^4} \ dx = \frac{1}{4}\int_0^1 t^{-\frac{3}{4}}(1-t)^{\frac{1}{4}} \ dt=\frac{1}{4}B \left(\frac{1}{4},\frac{5}{4} \right) = ...

    the rest is the same as what i did in my previous post.
    .................................................. .................................................. ..

    for those viewers who might not be familiar with Beta function:

    by definition: B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1} \ dt, \ \ \text{Re}(x) > 0, \ \text{Re}(y) > 0.

    then it can be shown that: B(x,y)=2\int_0^{\frac{\pi}{2}} (\sin t)^{2x-1}(\cos t)^{2y-1} \ dt=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}.

    see here for more details.
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