2
$\begingroup$

I am looking for the general term of the following sequence defined by $$ u_n = \int_0^1 F\left(-n,n+\frac{3}{2}; 2, r^2\right) \, \mathrm{d} r \, , \quad(n \ge 0) \, , $$ wherein $F$ is the hypergeometric function.

Using computer software algebra e.g. Maple, it is possible to know the terms for specified values of $n$ such as $u_0 = 1 $, $u_1 = 7/12$, $u_2=43/120$ etc.

It would be great if someone here could provide with an idea that helps in determining the expression of $u_n$.

Thanks

R

$\endgroup$
3
  • 1
    $\begingroup$ put $r^2=t$. Then apply the Euler Transformation with $z=1, c=1/2, d=3/2$ to get a higher order hypergeometric function $_3F_2$... en.wikipedia.org/wiki/Generalized_hypergeometric_function $\endgroup$
    – tired
    Sep 28, 2016 at 10:15
  • 1
    $\begingroup$ Furthermore Dixon's and Saalschütz's formulas will reduce the resulting equation into Gammavalues for a large class of parameters...just play around a litte bit and see what you get $\endgroup$
    – tired
    Sep 28, 2016 at 10:22
  • 1
    $\begingroup$ @tired thanks indeed. By setting $t=r^2$ Maple gives directly the desired result as ${}_3F_2 \left( \frac{1}{2}, -n, n+\frac{3}{2}; \frac{3}{2}, 2;1 \right)$. $\endgroup$
    – preuss
    Sep 28, 2016 at 11:38

1 Answer 1

Reset to default
1
$\begingroup$

This is not an answer but just the result from a CAS.

$$u_n=\int_0^1\, _2F_1\left(-n,n+\frac{3}{2};2;r^2\right)\,dr=\frac{\pi \, _3\tilde{F}_2\left(-n,\frac{1}{2}-n,\frac{3}{2};2,\frac{3}{2}-n;1\right)}{4 \Gamma \left(n+\frac{3}{2}\right)}$$ where appears the regularized generalized hypergeometric function.

$n$ being an integer, the above can simplify to $$u_n=\frac{(-1)^{n+1}}{(2 n-1) (2 n+1)} \, _3F_2\left(\frac{3}{2},\frac{1}{2}-n,-n;2,\frac{3}{2}-n;1\right)$$ where appears the generalized hypergeometric function.

$\endgroup$
2
  • $\begingroup$ Thanks for your quick answer. What CAS have you used? Maple doesn't succeed to evaluate it.. $\endgroup$
    – preuss
    Sep 28, 2016 at 9:22
  • 1
    $\begingroup$ @RobertoFermi. I still use a very old version of Maxima which we adapted for some specific needs. $\endgroup$
    – Claude Leibovici
    Sep 28, 2016 at 9:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .