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A Integral in terms of Gamma functions

  1. Mar 24, 2017 #1

    ShayanJ

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    Does anyone know how I can prove the following equation?

    ##\displaystyle \frac 1 {d-1}-\int_0^1 \frac{dy}{y^d} \left( \frac 1 {\sqrt{1-y^{2d}} }-1\right)=-\frac{\sqrt \pi \ \Gamma(\frac{1-d}{2d})}{2d \ \Gamma(\frac 1 {2d})} ##

    Thanks
     
  2. jcsd
  3. Mar 24, 2017 #2
    the right side look like ##-\frac{\beta (\frac{1}{2},\frac{1-d}{2d})}{2d}##

    and the beta function is $$\beta (x,y)=2\int_{0}^{\frac{\pi}{2}}cos^{2x-1}(\theta )\, sin^{2u-1}(\theta ) \, d\theta $$

    the part of ##\frac{1}{y^{d}\sqrt{1-y^{2d}}}## could be transfor of the trigonometric identity
     
  4. Mar 24, 2017 #3

    ShayanJ

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    I'm not quite sure how to get to the Beta function specially with that ## \frac 1 {d-1} ## term being there.
     
  5. Mar 25, 2017 #4
    Try to separate the integral in 2 , and solve the part of ##y^{-d}## then the ## \frac 1 {d-1} ## Are canceled.
    With the integral you have
    $$\int_0^1 \frac{dy}{y^d \sqrt{1-y^{2d}}}$$

    you can do a symple trigonometric substitution.

    note:
    when you solve the integral
    $$\int_0^1 \frac{dy}{y^{d }}$$
    you has to restrict ##d<1##
     
  6. Mar 25, 2017 #5

    ShayanJ

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    Your solution is reasonable but I'm trying to go through the calculations of a paper and in the context of that paper, this solution doesn't work. Actually the result is the same, but it shouldn't be, and that's confusing!
    I should calculate the integral ## \displaystyle \int_0^1 \frac{dy}{y^d\sqrt{1-y^{2d}}} ##. d is the number of spatial dimensions and so clearly it can't be smaller than 1 and so the integrand has a singularity at 0. So we need to regularize the integral by introducing a cut-off like below:

    ## \displaystyle \int_{\varepsilon}^1 \frac{dy}{y^d \sqrt{1-y^{2d}}}= \int_{\varepsilon}^1 \frac{dy}{y^d}+\int_0^1 \frac{dy}{y^d} \left( \frac 1{\sqrt{1-y^{2d}}}-1 \right) ##

    The second integral doesn't use the cut-off because its not divergent(I don't know how to figure that out but the author says so!). Then the first integral gives two terms, one is the ## \frac 1 {d-1} ## in the OP, and the other is a term that depends on the cut-off. But if we calculate the second integral by separating its two terms, we're just undoing all the things we've done before. So there should be a different way of doing it. But the confusing thing is that the result is the same as you say, so I'm confused!

    EDIT:
    Page 33 of this paper and section 7.1.1 of this paper.
     
    Last edited: Mar 25, 2017
  7. Mar 25, 2017 #6
    Let's start with

    $$-\int_0^1 \frac{dy}{y^d \sqrt{1-y^{2d}}} =-\frac{\sqrt \pi \ \Gamma(\frac{1-d}{2d})}{2d \ \Gamma(\frac 1 {2d})}$$

    this is true for d<1 , but you need a d >1 .
    so we do a cut-off and we get
    $$\displaystyle \int_{\varepsilon}^1 \frac{dy}{y^d \sqrt{1-y^{2d}}}= \int_{\varepsilon}^1 \frac{dy}{y^d}+\int_\varepsilon^1 \frac{dy}{y^d} \left( \frac 1{\sqrt{1-y^{2d}}}-1 \right)$$

    then we apply a limit
    $$\lim_{\varepsilon\rightarrow 0\, }\displaystyle \int_{\varepsilon}^1 \frac{dy}{y^d \sqrt{1-y^{2d}}}= \lim_{\varepsilon\rightarrow 0\, } \int_{\varepsilon}^1 \frac{dy}{y^d}+ \lim_{\varepsilon\rightarrow 0\, }\int_\varepsilon^1 \frac{dy}{y^d} \left( \frac 1{\sqrt{1-y^{2d}}}-1 \right)$$

    i dont know about convergence methods of integrals or continuous convergence or anything like that (im a highschool student)
    so suppose that the second integral its fine and converges, we get
    $$\int_{0}^1 \frac{dy}{y^d \sqrt{1-y^{2d}}}= \lim_{\varepsilon\rightarrow 0\, } \int_{\varepsilon}^1 \frac{dy}{y^d}+ \int_0^1 \frac{dy}{y^d} \left( \frac 1{\sqrt{1-y^{2d}}}-1 \right)$$

    solving:

    $$\int_{0}^1 \frac{dy}{y^d \sqrt{1-y^{2d}}}= \lim_{\varepsilon\rightarrow 0\, } \frac{1}{\varepsilon^{d-1} (d-1) }-\frac{1}{d-1}+ \int_0^1 \frac{dy}{y^d} \left( \frac 1{\sqrt{1-y^{2d}}}-1 \right)$$

    then if d>1 and we apply the limit we get a divergent result, i dont understand how it works.

    i guess that adding and substract the ##y^{-d}## term is just a "trick" that make the result
    $$
    \displaystyle \frac 1 {d-1}-\int_0^1 \frac{dy}{y^d} \left( \frac 1 {\sqrt{1-y^{2d}} }-1\right)=-\frac{\sqrt \pi \ \Gamma(\frac{1-d}{2d})}{2d \ \Gamma(\frac 1 {2d})}$$

    valid for d>1 , i dont know about make things rigorous so I'm sorry that I can not help you

    edit: When you find out how it works, I'd appreciate it if you could tell me =D
     
  8. Mar 27, 2017 #7

    ShayanJ

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    I got it!
    Wolframalpha gives an expression in terms of the hypergeometric function for the integral ## \int \frac {dy}{y^d}\left( \frac 1 {\sqrt{1-y^{2*d}}}-1\right) ##. Then using the formula(16) in this page, everything falls into place.
     
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