# Equality involving Appell hypergeometric function

After some algebra, Wolfram online integrator gave me the following: $$\tag{1} \int (1-a-t)^{N-2}\ \sqrt{2t-t^2}\ \text{d} t = c\ \cdot t^{3/2}\ \operatorname{F}_1 \left( \frac{3}{2}; -\frac{1}{2}, 2-N; \frac{5}{2}; \frac{t}{2}, \frac{t}{1-a}\right)+C$$ where:

• $\operatorname{F}_1$ is the Appell hypergeometric function of two variables defined by the expansion: $$\operatorname{F}_1(\alpha; \beta, \beta^\prime ; \gamma; x,y):= \sum_{m,n=0}^\infty \frac{(\alpha)_{m+n}\ (\beta)_m\ (\beta^\prime)_n}{m!\ n!\ (\gamma)_{m+n}}\ x^m\ y^n$$ ($(a_k):=\frac{\Gamma (a+k)}{\Gamma(a)}$ is the Pochhammer symbol), which converges in the region $|x|,|y|<1$;
• $N\geq 3$ an integer, $a\in ]0,1[$ and $t\in [0,1-a]$ (so the RHside makes sense);
• $c=c(a,N)$ is a known constant and $C$ is an arbitrary constant (coming from indefinite integration).

My question is:

How can I get (1) without using any software?

I suppose a series expansion of the integrand and term by term integration should be used, but I could not figure out how to do explicit computations. Neither I succeeded in simplifying the derivative of the LHside to get the integrand...

Any hint will be appreciated.

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It is an application of following integral formula for $\operatorname{F}_1$ : $$\operatorname{F}_1(a;b_1,b_2;c;z_1,z_2)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1 x^{a-1}(1-x)^{c-a-1}(1-x z_1)^{-b_1}(1-x z_2)^{-b_2}\,dx$$

giving : $$\operatorname{F}_1 \left( \frac 32; -\frac 12, 2-N; \frac 52; \frac t2, \frac t{1-a}\right)=\frac{\Gamma(\frac 52)}{\Gamma(\frac 32)}\int_0^1 x^{\frac 12}\left(1-x \frac t2\right)^{\frac 12}\left(1-x \frac t{1-a}\right)^{N-2}\, dx$$

$$=\frac 32 t^{-\frac 12-1}\int_0^t y^{\frac 12}\left(1-\frac y2\right)^{\frac 12}\left(1-\frac y{1-a}\right)^{N-2}\, dy$$

$$=t^{-\frac 32} C_1\int_0^t \sqrt{2y-y^2}\left(1-a-y\right)^{N-2}\, dy$$

i.e. your result (with $\frac 1c=C_1$ 'absorbing' the constants involved).

Hoping this helped,

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Thanks a lot Raymond... As you can see below I found also a different solution. –  Pacciu May 31 '12 at 17:59
@Pacciu: Fine! So that you'll have a proof from integrals and sums point of view! –  Raymond Manzoni May 31 '12 at 19:51

I found also the following term-by-term integration solution.

One has: $$(1-a-t)^{N-2}\ \sqrt{2t-t^2} = \sqrt{2}\ (1-a)^{N-2}\ \sqrt{t}\ \left( 1-\frac{t}{1-a}\right)^{N-2}\ \sqrt{1-\frac{t}{2}}$$ and by the binomial theorem: $$\begin{split} \left( 1-\frac{t}{1-a}\right)^{N-2} &= \sum_{k=0}^{N-2} (-1)^k\ \binom{N-2}{k}\ \left(\frac{t}{1-a}\right)^k\\ &= \sum_{k=0}^{N-2} \frac{(2-N)_k}{k!}\ \left(\frac{t}{1-a}\right)^k\\ &= \sum_{k=0}^\infty \frac{(2-N)_k}{k!}\ \left(\frac{t}{1-a}\right)^k\\ \sqrt{1-\frac{t}{2}} &= \sum_{k=0}^\infty (-1)^k\ \binom{1/2}{k}\ \left( \frac{t}{2}\right)^k\\ &= \sum_{k=0}^\infty \frac{(-1/2)_k}{k!}\ \left( \frac{t}{2}\right)^k\; ; \end{split}$$ now, since both series in the rightmost sides converge absolutely, Mertens theorem applies and one can take the Cauchy product: $$\begin{split} \left( 1-\frac{t}{1-a}\right)^{N-2}\ \sqrt{1-\frac{t}{2}} &= \left( \sum_{k=0}^\infty \frac{(2-N)_k}{k!}\ \left(\frac{t}{1-a}\right)^k\right)*\left( \sum_{k=0}^\infty \frac{(-1/2)_k}{k!}\ \left( \frac{t}{2}\right)^k\right)\\ &= \sum_{k=0}^\infty \sum_{h=0}^k \frac{(2-N)_h}{h!}\ \left(\frac{t}{1-a}\right)^h\ \frac{(-1/2)_{k-h}}{(k-h)!}\ \left( \frac{t}{2}\right)^{k-h}\\ &= \sum_{k=0}^\infty \sum_{h=0}^k \frac{(2-N)_h\ (-1/2)_{k-h}}{h!\ (k-h)!}\ \frac{1}{2^h\ (1-a)^{k-h}}\ t^k \end{split}$$ with the rigthmost side converging absolutely and uniformly on compact subsets. Therefore one gets: $$(1-a-t)^{N-2}\ \sqrt{2t-t^2} = \sqrt{2}\ (1-a)^{N-2}\ \sum_{k=0}^\infty \sum_{h=0}^k \frac{(2-N)_h\ (-1/2)_{k-h}}{h!\ (k-h)!}\ \frac{1}{2^h\ (1-a)^{k-h}}\ t^{k+1/2}$$ and a rearrangement of the RHside yields: $$(1-a-t)^{N-2}\ \sqrt{2t-t^2} = \sqrt{2}\ (1-a)^{N-2}\ \sum_{m,n=0}^\infty \frac{(2-N)_m\ (-1/2)_n}{m!\ n!}\ \frac{1}{2^m\ (1-a)^n}\ t^{m+n+1/2}\; ;$$ thus term-by-term integration shows that: $$\begin{split} \int (1-a-t)^{N-2}\ \sqrt{2t-t^2}\ \text{d} t &= \sqrt{2}\ (1-a)^{N-2}\ \sum_{m,n=0}^\infty \frac{(2-N)_m\ (-1/2)_n}{m!\ n!}\ \frac{1}{2^m\ (1-a)^n}\ \int t^{m+n+1/2}\ \text{d} t\\ &= \sqrt{2}\ (1-a)^{N-2}\ \sum_{m,n=0}^\infty \frac{(2-N)_m\ (-1/2)_n}{m!\ n!\ (3/2 +m+n)}\ \frac{1}{2^m\ (1-a)^n}\ t^{m+n+3/2}\ + C\\ &= \frac{2\sqrt{2}}{3}\ (1-a)^{N-2}\ \sum_{m,n=0}^\infty \frac{(3/2)_{m+n}\ (2-N)_m\ (-1/2)_n}{m!\ n!\ (5/2)_{m+n}}\ \frac{1}{2^m\ (1-a)^n}\ t^{m+n+3/2}\\ &\phantom{=} +C\\ &= \frac{2\sqrt{2}}{3}\ (1-a)^{N-2}\ t^{3/2}\ F_1\left( \frac{3}{2}; -\frac{1}{2}, 2-N; \frac{5}{2}; \frac{t}{2}, \frac{t}{1-a}\right) +C \end{split}$$ because: $$\frac{3}{2}\ \frac{1}{m+n+3/2} = \frac{(3/2)_{m+n}}{(5/2)_{m+n}}\; .$$

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