4
$\begingroup$

I am trying to compute the indefinite integral $$ \int_0^u {}_2F_1\left(\frac{1}{4},\frac{5}{4},2,1-v^2\right)^2 dv $$ for $0<u<1$. Using Clausen's formula for the square of the hypergeometric function, this can be written as $$ \int_0^u {}_3F_2\left(\frac{1}{2},\frac{3}{2},\frac{5}{2};2,3;1-v^2\right)dv = \frac{1}{2}\int_{1-u^2}^{1} \frac{1}{\sqrt{1-x}}{}_3F_2\left(\frac{1}{2},\frac{3}{2},\frac{5}{2};2,3;x\right)dx, $$ where I have performed a change of variables $x=1-v^2$. From Brudnikov & Brychkov, Volume 3, I know the following indefinite integral for generalized hypergeometric functions ${}_pF_q$: $$ \int x^{\alpha-1}{}_pF_q\left((a_p);(b_q);x\right)=\frac{x^\alpha}{\alpha}\,{}_{p+1}F_{q+1}\left((a_p),\alpha;(b_q),\alpha+1;x\right). $$ One ansatz is to expand the factor of $1/\sqrt{1-x}$ in a Taylor series, to integrate each summand and then to sum the series. Doing this, one obtains the expression $$ \left.\frac{1}{2}\sum_{n=0}^\infty\frac{\left(\frac{1}{2}\right)_{n}}{(n+1)!}x^{n+1}{}_4F_3\left(\frac{1}{2},\frac{3}{2},\frac{5}{2},n+1;2,3,n+2;x\right)\right|_{x=1-u^2}^{x=1}, $$ where $(\alpha)_n=\alpha(\alpha+1)\ldots(\alpha+n-1)$ is the Pochhammer symbol. Unfortunately, I do not know how to sum this.

Another ansatz is to use the identity $$ {}_3F_2\left(\frac{1}{2},\frac{3}{2},\frac{5}{2};2,3;1-v^2\right)=\frac{64 \left((1+v) K\left(\frac{1-v}{2}\right)-2vE\left(\frac{1-v}{2}\right)\right)^2}{9 \pi^2 (1-v^2)^2}, $$ where $K$ and $E$ are the complete elliptic integrals of the first and second kind. However, I also do not know how to integrate these terms with squared complete elliptic integrals.

$\endgroup$
2
  • $\begingroup$ The squared hypergeometric function under the integral is in fact the Legendre function, may be it will help? $\endgroup$
    – Sergei
    Aug 23, 2016 at 20:03
  • $\begingroup$ @Sergei Can you say how to write it as a Legendre function? Thank you $\endgroup$
    – physicus
    Aug 25, 2016 at 22:38

1 Answer 1

6
$\begingroup$

Using Maple to play with the equation (details) suggests that $$ \int_0^u {}_2F_1\left(\frac{1}{4},\frac{5}{4},2,1-v^2\right)^2 dv = -\frac{32}{\pi} + \frac83 u \cdot {}_3F_2\!\left(\frac12, \frac12, \frac52, 1, 2, 1-u^2\right). $$ You can almost prove this equality by observing that both sides satisfy $$ \left( u-1 \right) ^{2} \left( u+1 \right) ^{2}{\frac {{\rm d}^{4}}{ {\rm d}{u}^{4}}}y \left( u \right) +12\,u \left( u-1 \right) \left( u +1 \right) {\frac {{\rm d}^{3}}{{\rm d}{u}^{3}}}y \left( u \right) + \left( 33\,{u}^{2}-9 \right) {\frac {{\rm d}^{2}}{{\rm d}{u}^{2}}}y \left( u \right) +15 u \, {\frac {\rm d}{{\rm d}u}}y \left( u \right)= 0 $$ and their series expansions at $u=1$ agree. More precisely, $u=1$ is a regular singular point of the differential equation, with indicial polynomial $n(n+1)(n-1)$, so that solutions of this equation are characterized by the coefficients of $(u-1)^{-1}$, $1$, $\log(u-1)$ and $(u-1)$ in their generalized series expansions at $1$. The coefficients of $(u-1)^{-1}$, $\log(u-1)$, and $(u-1)$ are (respectively) $0$, $0$, and $1$ for both sides. Therefore, proving the above equality for all $u$ reduces to proving that it holds for $u=1$, i.e. that $$\int_0^1 {}_2F_1\left(\frac{1}{4},\frac{5}{4},2,1-v^2\right)^2 dv = \frac83 - \frac{32}{9 \pi}.$$

$\endgroup$

Your Answer

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

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