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.

  • $\begingroup$ 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. $\endgroup$ – 010514 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


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

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/

  • $\begingroup$ I have verified it and it is correct. Thank you for taking the time to provide a proof. Also, thank you for the link. $\endgroup$ – 010514 May 10 '17 at 7:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.