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By using a mathematical software I was able to deduce that the integral: \begin{equation} \int_{0}^{t}\frac{ds}{\left[\left( 1-s^p \right)^{p-1}\right]^{1/p}} \end{equation} where $p>1$, can be expressed in the following way: \begin{equation} \int_{0}^{t}\frac{ds}{\left[\left( 1-s^p \right)^{p-1}\right]^{1/p}}=_{2}F_{1}\left( 1-\frac{1}{p},\frac{1}{p},1+\frac{1}{p};t^p \right)t \end{equation} where $_{2}F_{1}\left( \alpha,\beta,\gamma;z \right)$ is the ordinary hypergeometric Gauss function: \begin{equation} _{2}F_{1}\left( \alpha,\beta,\gamma;z \right)=\sum_{n=0}^{\infty} \frac{(\alpha)_n(\beta)_n}{(\gamma)_n}\frac{z^n}{n!} \end{equation} with $()_n$ to be the rising Pochhammer symbol. The issue is that I am new to the hypergeometric functions theory and not yet good with this kind of manipulations. Therefore, I am not able to prove analytically that this relation above holds. I tried to go throught the definition of the Incomplete Beta function but it does not always hold that $s>0$.

I would really appreciate some help, at least where to start from. Thank you.

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  • $\begingroup$ I assume that by the Pochhammer symbol, for this context, you mean rising factorial, right? $\endgroup$ – Masacroso May 9 '17 at 10:55
  • $\begingroup$ @Masacroso Correct. $\endgroup$ – Mitscaype May 9 '17 at 10:56
  • $\begingroup$ Take a look at this paper, it have a proof for a integral representation of the hypergeometric function $_2F_1$ $\endgroup$ – Masacroso May 9 '17 at 11:27
  • $\begingroup$ Expand the Integrand into a series using the generalized binary formula, then integrate each term of the series, and compare the result with the definition of the hypergeometric function. $\endgroup$ – Dr. Wolfgang Hintze May 9 '17 at 11:43
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Consider

$$\int^t_0 (1-s^p)^{\frac{1}{p}-1}\,ds$$

Let $s^p = x t^p $ hence $dx = \frac{t}{p} x^{1/p-1}$

$$\frac{t}{p}\int^1_0 x^{1/p-1}(1-t^px)^{\frac{1}{p}-1}\,dx$$

Now use the integral representation

$$\tag{1}\beta(c-b,b) \, _2F_1(a,b;c;z)=\int_0^1 \frac{x^{b-1}(1-x)^{c-b-1}}{(1-xz)^a}\, dx$$

With $b=1/p,c=1+\frac{1}{p},a=1-\frac{1}{p},z = t^p$

$$t\int^1_0 x^{1/p-1}(1-t^px^p)^{\frac{1}{p}-1}\,dx=\frac{t}{p}\beta\left(1,\frac{1}{p}\right) \, _2F_1\left(1-\frac{1}{p},\frac{1}{p};1+\frac{1}{p};t^p\right)$$

Note that

$$\frac{1}{p}\beta\left(1,\frac{1}{p}\right) = \frac{\Gamma(1)\Gamma(1/p)}{p\Gamma(1+1/p)} = 1$$

For the proof (1) http://advancedintegrals.com/2017/01/integral-representation-of-gauss-hypergeometric-function-proof/

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  • $\begingroup$ I have verified it and it is correct. Thank you for taking the time to provide a proof. Also, thank you for the link. $\endgroup$ – Mitscaype May 10 '17 at 7:41

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