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A Integral in terms of Gamma functions

  1. Mar 24, 2017 #1


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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})} ##

  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


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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.

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


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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!

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


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

    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


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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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