1
$\begingroup$

I would like to ask the following two:

  • For the integral: \begin{equation} \int_{0}^{t}\left( 1-s^p \right)^{\frac{1-p}{p}}ds \end{equation} I know that it is reduced to the following product involving the hypergeometric function: \begin{equation} \int_{0}^{t}\left( 1-s^p \right)^{\frac{1-p}{p}}ds={}_{2}F_{1}\left(1-\frac{1}{p}, \frac{1}{p}; 1+\frac{1}{p};t^p \right)t \end{equation} I am not yet able to prove that, but my thoughts are to expand $\left( 1-s^p \right)^{\frac{1-p}{p}}$ using the binomial theorem, then use that to approximate the series for the hypergeometric function mentioned.
  • Also, for the following: \begin{equation} x={}_{2}F_{1}\left(1-\frac{1}{p}, \frac{1}{p}; 1+\frac{1}{p};t^p \right)t \end{equation} I would like to ask whether it is possible to solve explicitly for $t$. ${}_{2}F_{1}(a,b;c;z)$ is the Gauss ordinary hypergeometric function and $p>1$. Since I am not familiar with this kind of manipulations, I would like to ask if there is any transformation for this kind of functions, which would help in this situation. Thanks!
Improve this question
$\endgroup$
4
  • $\begingroup$ you might find it helpful to try $t=1$ first, when there is a much simpler answer: $$\int_{0}^{1}\left( 1-s^p \right)^{\frac{1-p}{p}}ds =\frac{\Gamma(1/p)^2}{p \Gamma(2/p)}$$ $\endgroup$
    – Carlo Beenakker
    May 3, 2017 at 19:58
  • $\begingroup$ @CarloBeenakker I see. It is also interesting to see how this is derived, but still, I need to solve for $t$ :/ $\endgroup$
    – Bazinga
    May 3, 2017 at 20:01
  • $\begingroup$ for $p=2$ you have $t=\sin x$; there may be simple expressions for other (small) values of $p$, but a closed form expression for any $p$ is unlikely. $\endgroup$
    – Carlo Beenakker
    May 3, 2017 at 20:19
  • $\begingroup$ @CarloBeenakker Yes you are right, $p=2$ for the integral equals the $arc \sin(x)/x$ as a special case and up until now that is the only one I have discovered too. I was more interested on how to treat relations like the one, on the second part of my question. $\endgroup$
    – Bazinga
    May 3, 2017 at 20:22

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.