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Beta function

  1. Dec 28, 2013 #1
  2. jcsd
  3. Dec 28, 2013 #2

    Ray Vickson

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  4. Dec 28, 2013 #3
    ok, is not necessary write with capital letter the word "DO".

    For to solve the integral [tex] \int^1_0 \frac{dy}{\sqrt{1+y^4}}[/tex] i tried to use the beta function [tex] \beta (x,y) = \int^1_0 t^{x-1}(1-t)^{y-1} dt [/tex]

    but, i don't find values of x and y but for to get of the form [tex] \int^1_0 \frac{dy}{\sqrt{1+y^4}}[/tex]
     
  5. Dec 28, 2013 #4
  6. Dec 28, 2013 #5

    Dick

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    That stackexchange thread is talking about the integral ##\int^1_0 \frac{dy}{\sqrt{1-y^4}}## not ##\int^1_0 \frac{dy}{\sqrt{1+y^4}}##. If that's the one you really want to do then substituting ##t=y^4## is a good first step. Beta functions won't help with ##\int^1_0 \frac{dy}{\sqrt{1+y^4}}##.
     
    Last edited: Dec 28, 2013
  7. Dec 28, 2013 #6
    Yes, I know the hint is for minus and here we have a plus. But it seemed like the approach might be useful.
     
  8. Dec 28, 2013 #7

    Dick

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    It might be, I'm just curious what problem we are actually trying to solve here. It's pretty straightforward with the minus sign.
     
    Last edited: Dec 28, 2013
  9. Dec 28, 2013 #8

    Ray Vickson

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    Capitalization was intended and was needed to get the meaning I wanted to convey.
     
  10. Dec 29, 2013 #9

    lurflurf

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    There are many ways to transform this, for example

    $$\int_0^1 \frac{\mathrm{d}y}{\sqrt{1+y^4}}=\frac{1}{8}\int_0^\infty \frac{u^{1/4}}{\sqrt{1+u}}\frac{\mathrm{d}u}{u}=\frac{3}{2}\int_0^\infty \frac{u^{5/4}}{(1+u)^{5/2}}\frac{\mathrm{d}u}{u}$$
    recall that
    $$\mathrm{B}(m,n)=\int_0^\infty \frac{u^m}{(1+u)^{m+n}}\frac{\mathrm{d}u}{u}=\frac{\Gamma(n)\Gamma(m)}{\Gamma(m+n)}$$
     
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