Integral in terms of Gamma functions

[ShayanJ]

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

 

[MAGNIBORO]

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

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

 

[ShayanJ]

I'm not quite sure how to get to the Beta function specially with that $\frac 1 {d-1}$ term being there.

 

[MAGNIBORO]

I'm not quite sure how to get to the Beta function specially with that $\frac 1 {d-1}$ term being there.

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 $d<1$

 

[ShayanJ]

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.

 

[MAGNIBORO]

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 don't know about convergence methods of integrals or continuous convergence or anything like that (im a high school 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 don't 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 don't 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

 

[ShayanJ]

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.