Closed form for the hypergeometic function $\,_{4}F_{3}\left(1,-k,k+\frac{3}{2},\frac{1}{2};\frac{1}{2}-k,k+2,\frac{3}{2};1\right)$

Question

Let $K(x)$ be the complete elliptic integral of the first kind with the following convention $$K(x):=\int_{0}^{\pi/2}\frac{\mathrm{d}\theta}{\sqrt{1-x\sin^{2}(\theta)}}.$$ For a research work, I would like to compute the Fourier-Legendre coefficients of $K(x)^{2}$, that is, integrals of the type $$\int_{0}^{1}K(x)^{2}P_{n}\left(2x-1\right)\mathrm{d}x$$ where $P_{n}(x)$ are the Legendre polynomials. For now, I'm only able to prove the following representation for odd $n$ $$\int_{0}^{1}K(x)^{2}P_{2k+1}(2x-1)\mathrm{d}x=\frac{1}{(2k+1)(k+1)}\,_{4}F_{3}\left(\left.{1,-k,k+\frac{3}{2},\frac{1}{2}\atop\frac{1}{2}-k,k+2,\frac{3}{2}}\right|1\right).$$ I tried to search some identities and to apply some classical results to this $_{4}F_{3}$ but I'm not able to find a closed form of such function (for closed form I intend some representation in terms of ratio of Gamma functions or in terms to functions strictly related to Gamma function).

So my question is:

does anyone know if this hypergeometric function admits a closed form?

Bonus question: for even $n$, for now, I have no idea to attack this problem, so any suggestions are welcome.

Answer 1

Let $f(n) = \int_0^1 K(x)^2 P_n(2x-1) dx$. It's well-known that $\int_0^1 x^n K(x)^2 dx \in \mathbb{Q} + \mathbb{Q}\zeta(3)$, so $f(n) \in \mathbb{Q} + \mathbb{Q}\zeta(3)$ as well, in particular $f(0) = \frac{7}{2}\zeta(3), f(1)=1$.

We prove below that $$\tag{$*$}-(n+1)^4 f(n)+(n+2)^4 f(n+2)-2 n-3=0$$ So this implies $f(2n+1)\in \mathbb{Q}$, already observed by OP.

Let $g(n)=f(2n+1), h(n) = f(2n)$, then \begin{aligned}-5-4n-16(1+n)^4 g(n) + (3+2n)^4 g(n+1) &= 0 \\ -3-4n-(2n+1)^4h(n)+(2+2n)^4h(n+1)&=0 \end{aligned}

For both difference equations, it can be proved they don't have hypergeometric solution, so in this sense, $f(n)$ and OP's $_4F_3$ $\color{red}{\text{have no closed-form}}$.

Since they're only of 1st order, we can still down $g(n), h(n)$ explicitly using a single summation:

$$f(2n) = \frac{\Gamma \left(n+\frac{1}{2}\right)^4}{16 \Gamma (n+1)^4}\sum _{m=0}^{n-1} \frac{(3+4 m) \Gamma (m+1)^4}{\Gamma \left(\frac{3}{2}+m\right)^4}+\frac{7 \zeta (3) \Gamma \left(n+\frac{1}{2}\right)^4}{2 \pi ^2 \Gamma (n+1)^4}$$

for $f(2n+1)$, it's essentially the terminating $_4F_3$ given by OP, so omitted here.

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Proof of $(*)$: this is entirely algorithmic. Let $S_n$ be shift in $n$, i.e. $S_n f(n) = f(n+1)$ and $D_x$ be derivative.

Let $F(n,x) = K(x)^2 P_n(2x-1)$ and creative telescoping algorithm says for $G(n,x)$ equals $$G(n,x) = \left[-(2 (2 n+3) (x-1)^2 x^2) D_x^2 S_n + (2 n+3) (3 n+1) (x-1) x (2 x-1) D_x S_n - 3 (n+1) (2 n+3) (x-1) x D_x - 2 (2 n+3) \left(3 n^2+6 n+4\right) (x-1) x S_n \right] F(n,x)$$

we have $[(n+2)^4 S_n^2-(n+1)^4]F + D_x G = 0$ (which can be checked), so $$\begin{aligned} \left[(n+2)^4 S_n^2-(n+1)^4 \right] f(n) &= \int_0^1 [(n+2)^4 S_n^2-(n+1)^4] F(n,x) dx \\ &= -\int_0^1 D_x G(n,x) dx = -G(n,1)+G(n,0)\end{aligned}$$ which then easily computes to be $3+2n$, this proves $(*)$.

Answer 2

$K(x)$ = Complete Elliptic Integral of the First Kind. $$K(x)=\int_{0}^{\pi/2}\frac{1}{\sqrt{1-x\sin^2\theta}}d\theta$$

$f(n)$ = Fourier-Legendre Coefficients of $K(x)^2$ $$f(n)=\int_{0}^{1}K(x)^2P_n(2x-1)dx$$

Series Representation of $P_n(x)$ : $$P_n(x)=\frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}^2(x-1)^{n-k}(x+1)^k$$

Replacing $x\to2x-1$: $$P_n(2x-1)=(x-1)^n\sum_{k=0}^{n}\binom{n}{k}^2(\frac{x}{x-1})^k$$ Now, $$f(n)=\int_{0}^{1}K(x)^2(x-1)^n\sum_{k=0}^{n}\binom{n}{k}^2(\frac{x}{x-1})^kdx$$ Replace $\frac{x}{x-1}\to u$, Limits change from $x:0 \to u:0$, $x:1 \to u:-\infty$ $$f(n)=\int_{-\infty}^{0}K(\frac{x}{x-1})^2\frac{1}{(x-1)^{n+2}}\sum_{k=0}^{n}\binom{n}{k}^2x^kdx$$ Also, $$K(\frac{x}{x-1})=\sqrt{1-x}K(x)$$ Therefore, $$f(n)=-\int_{-\infty}^{0}K(x)^2\frac{1}{(x-1)^{n+1}}\sum_{k=0}^{n}\binom{n}{k}^2x^kdx$$

$\frac{1}{(x-1)^{n+1}}\sum_{k=0}^{n}\binom{n}{k}^2x^k$ can be decomposed into partial fractions as follows: $$\frac{1}{(x-1)^{n+1}}\sum_{k=0}^{n}\binom{n}{k}^2x^k=\sum_{k=0}^{n}\frac{(n+k)!}{(k!)^2(n-k)!}\frac{1}{(x-1)^{k+1}}$$ I am not sure how to prove the above equation, as it is purely experimental work (I was able to find out the coefficients using Sequence A063007)

Now, $$f(n)=-\int_{-\infty}^{0}K(x)^2\sum_{k=0}^{n}\frac{(n+k)!}{(k!)^2(n-k)!}\frac{1}{(x-1)^{k+1}}dx$$ $$f(n)=-\sum_{k=0}^{n}\frac{(n+k)!}{(k!)^2(n-k)!}\int_{-\infty}^{0}\frac{K(x)^2}{(x-1)^{k+1}}dx$$

We define a new function as follows: $$Q(n)=\int_{-\infty}^{0}\frac{K(x)^2}{(x-1)^{n}}dx$$ Upon observation: $$Q(n)=(-1)^{n}(\alpha_n\zeta(3)-\beta_n)$$ The coefficients $\alpha_n$ & $\beta_n$ follow the following recurrence relation: (Found out using this post Conjectured closed form for $\int_0^1x^{2q-1}K(x)^2dx$ where $K(x)$ is the complete elliptical integral of the 1ˢᵗ kind) $$r_n=\frac{(2n-3)(2n^2-6n+5)}{2(n-1)^3}r_{n-1}-(\frac{n-2}{n-1})^3r_{n-2}$$ Where the initial conditions are as follows: $$\alpha_1=7/2, \alpha_2=7/4$$ $$\beta_1=0, \beta_2=1/2$$ $\alpha,\beta$ Graphed as $x, y$

Now, $$f(n)=-\sum_{k=0}^{n}\frac{(n+k)!}{(k!)^2(n-k)!}Q(k+1)$$ $$f(n)=\sum_{k=0}^{n}\frac{(n+k)!}{(k!)^2(n-k)!}[(-1)^{k}(\alpha_{k+1}\zeta(3)-\beta_{k+1})]$$ $$\color{red} {f(n)=\left(\sum_{k=0}^{n}\frac{(n+k)!}{(k!)^2(n-k)!}(-1)^k\alpha_{k+1}\right)\zeta(3)-\left(\sum_{k=0}^{n}\frac{(n+k)!}{(k!)^2(n-k)!}(-1)^k\beta_{k+1}\right)}$$ Let: $$\mathrm{A_n}=\left(\sum_{k=0}^{n}\frac{(n+k)!}{(k!)^2(n-k)!}(-1)^k\alpha_{k+1}\right)$$ $$\mathrm{B_n}=\left(\sum_{k=0}^{n}\frac{(n+k)!}{(k!)^2(n-k)!}(-1)^k\beta_{k+1}\right)$$

Conjecture through Experimental Observation: $$\mathrm{A_{2n-1}}=0$$ $$f(2n-1)=-\mathrm{B_{2n-1}}$$ Values of $f(2n-1)$: $$f(1)=1$$ $$f(3)=\frac{7}{27}$$ $$f(5)=\frac{407}{3375}$$ $$f(7)=\frac{81621}{1157625}$$ $$f(9)=\frac{39333249}{843908625}$$

By Observation: $$f(2n-1)=\frac{C}{(\prod_{r=1}^{n}(2r-1))^3}$$ $$f(2n-1)=(\frac{\sqrt{\pi}}{2^n\Gamma(n+\frac{1}{2})})^3C$$ Still searching about C.

Answer 3

Consider

$${}_4F_3\left(\begin{matrix}a_1,a_2,a_3,a_4 \\ b_1,b_2,b_3\end{matrix}~\bigg |~1\right)$$ When $\sum_i a_i-\sum_ib_i=-1$, Wolfram Research gives the formula

$${}_4F_3\left(\begin{matrix}a_1,a_2,a_3,a_4 \\ b_1,b_2,b_3\end{matrix}~\bigg |~1\right) \\ =\frac{\mathrm e^{-\mathrm i\pi(b_1+1)} \Gamma(b_2)\Gamma(b_3)\prod_{i=1}^4 \sqrt{\Gamma(1-a_i)~\Gamma(1-b_1+a_i)}}{\Gamma(1-b_1) \prod_{i,j~;~ j\neq 1}\sqrt{\Gamma(b_j-a_i)}}~\begin{Bmatrix}\frac{b_3-a_1-a_4-1}{2} & \frac{b_2-a_1-a_3-1}{2} & \frac{a_1+a_2-b_1-1}{2} \\ \frac{b_3-a_2-a_3-1}{2} & \frac{b_2-a_2-a_4-1}{2} & \frac{a_3+a_4-b_1-1}{2}\end{Bmatrix}$$

Where $$\begin{Bmatrix} \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot\end{Bmatrix}$$ Is a Wigner 6j.

In your case you want to calculate $${}_4F_3\left(\begin{matrix}1,-k,k+\frac{3}{2},\frac{1}{2} \\ \frac{1}{2}-k,k+2,\frac{3}{2}\end{matrix}~\bigg|~1\right)$$ You can check that $$1-k+k+3/2+1/2-(1/2-k)-(k+2)-3/2=-1$$ And so the linked formula is valid. Of course you will need to plug all the values into the formula and simplify.

Answer 4

Motivated by @Black Emperor's answer $$Q(5)=\frac{12509 }{16384}\zeta (3)-\frac{82085}{221184}$$ $$Q(6)=\frac{18509269}{55296000}-\frac{21329 }{32768}\zeta (3)$$ $$Q(7)=\frac{298417 }{524288}\zeta (3)-\frac{89992679}{294912000}$$ $$Q(8)=\frac{56796554519}{202309632000}-\frac{531559}{1048576} \zeta (3)$$ $$Q(9)=\frac{491564381 }{1073741824}\zeta (3)-\frac{53919965143021}{207165063168000}$$ $$Q(10)=\frac{8150801125136869}{33560740233216000}-\frac{897566789}{2147483648} \zeta(3)$$ $$Q(11)=\frac{13226643619}{34359738368} \zeta (3)-\frac{24471879971879527}{107394368746291200}$$ $$Q(12)=\frac{61408867748616923207}{285883809602627174400}-\frac{24539981809 }{68719476736}\zeta(3)$$

Answer 5

For more, define $$ K(k)=\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-\color{red}{k^2}t^2} } \text{d}t. $$ Then $$ \int_{0}^{1}K(k)^2\text{d}k =\frac{\pi^4}{32} {}_7F_6\left ( \frac12,\frac12,\frac12,\frac12,\frac12,\frac12,\frac54; \frac14 ,1,1,1,1,1;1\right ), $$ $$ \int_{0}^{1}k^2K(k)^2\text{d}k =1+\frac{3\pi^4}{512} {}_7F_6\left ( \frac12,\frac12,\frac12,\frac12,\frac32,\frac32,\frac74; \frac34 ,1,2,2,2,2;1\right ). $$ Consequently, all $\int_{0}^{1}k^{2n}K(k)^2\text{d}k$ have similar expressions due to the recurrence relation.