6
$\begingroup$

Euler gave the following well-known integral representations for the Gauss hypergeometric function ${_2F_1}$ and the generalized hypergeometric function ${_3F_2}$: for $0<\Re{\left(\beta\right)}<\Re{\left(\gamma\right)}$,

$$\small{{_2F_1}{\left(\alpha,\beta;\gamma;z\right)}=\frac{1}{\operatorname{B}{\left(\beta,\gamma-\beta\right)}}\int_{0}^{1}\frac{t^{\beta-1}\left(1-t\right)^{\gamma-\beta-1}}{\left(1-zt\right)^{\alpha}}\,\mathrm{d}t};\tag{1}$$

and for $0<\Re{\left(\mu\right)}<\Re{\left(\nu\right)}$,

$$\small{{_3F_2}{\left(\alpha,\beta,\mu;\gamma,\nu;z\right)}=\frac{1}{\operatorname{B}{\left(\mu,\nu-\mu\right)}}\int_{0}^{1}t^{\mu-1}\left(1-t\right)^{\nu-\mu-1}{_2F_1}{\left(\alpha,\beta;\gamma;zt\right)}\,\mathrm{d}t}.\tag{2}$$

I'm curious to learn if there is a way evaluate the following integral (possibly in terms of higher order generalized hypergeometric functions or the two-variable Appell functions?):

$$\small{\mathcal{I}{\left(\alpha,\beta,\gamma,z;\mu,\nu,\rho,w\right)}=\int_{0}^{1}\frac{t^{\mu-1}\left(1-t\right)^{\nu-\mu-1}}{\left(1-wt\right)^{\rho}}{_2F_1}{\left(\alpha,\beta;\gamma;zt\right)}\,\mathrm{d}t}.\tag{3}$$

Now, this integral $\mathcal{I}$ is a straightforward generalization of $(3)$, and it seems only natural to me that there is a paper on this integral out there somewhere. But if it exists it has eluded me, despite most furious Googling on my part.

If any of our resident master integrators have any insight to offer, I'd be very grateful. I'd also welcome any niche references that might be relevant here if someone happens to have any.

Cheers!

$\endgroup$
4
$\begingroup$

In the special case $\mu=\gamma$ Prudnikov-Brychkov-Marychev (Vol. III, formula 2.21.1.20) gives an evaluation in terms of Appell's $F_3$ (of four Appel's functions, this is the one with the maximal number of parameters): $$\mathcal I\left(\alpha,\beta,\gamma,z;\gamma,\nu,\rho,w\right)= \frac{B\left(\gamma,\nu-\gamma\right)}{(1-w)^{\rho}}{}F_3\left(\rho,\alpha,\nu-\gamma,\beta,\nu;\frac{w}{w-1};z\right)$$ Since all arguments are ''free'' (there is no relation between them), your generalization is yet one more step beyond Appell. A relation to generalized hypergeometric functions for generic $w,z$ would be extremely surprising.

$\endgroup$
  • $\begingroup$ Thank you very much for your response. The only reason I hesitate to accept it is that you leave open the question about whether the general case is unsolvable or just merely incredibly difficult. That said, the special case you give does in fact suffice for my underlying interest in this question. If no other responses appear in the next couple of days, then I'll accept this answer. $\endgroup$ – David H Dec 27 '15 at 6:11
  • $\begingroup$ Specifically, I'm very interested in the special case $F_3{\left(\frac12,\frac12,\frac12,\frac12,2;x;y\right)}$ and whether it can be reduced to elliptic integrals or the Appell $F_1$ function. If you can help me with this special case, I'll not only accept your response, I'd be inclined to award it the maximum bounty. $\endgroup$ – David H Dec 27 '15 at 6:18
  • $\begingroup$ Dear @DavidH sorry that the answer sounds negative. It might happen that there exists some further generalization of Appell's functions which incorporates your integral (Kampé de Fériet or maybe some multivariate hypergeometrics?) but I am afraid that in any case it will be not more than a notation. As for the specific $F_3$ from your comment, does it come from something like $\int_0^1 \frac{ \arcsin \sqrt{a t} \,dt}{ \sqrt{ (1-t) (1-bt) }}$ ? $\endgroup$ – Start wearing purple Dec 27 '15 at 10:31
  • 2
    $\begingroup$ Indeed, it comes from an integral I asked about here many months ago. I've found that this integral is the lynchpin in obtaining a closed form expression for a wide class of integrals I've encountered in certain mathematical physics problems, hence my interest. $\endgroup$ – David H Dec 27 '15 at 11:00
  • $\begingroup$ How $B(\gamma,\nu-\gamma)\mathrm{F}^{1:2;1}_{1:1;0}\Bigg(\begin{matrix}\gamma&:&\alpha,\beta&;&\rho&\\\nu&:&\gamma&;&-&\end{matrix}\Bigg|z,w\Bigg)$ can reduces to $\dfrac{B\left(\gamma,\nu-\gamma\right)}{(1-w)^{\rho}}F_3\left(\rho,\alpha,\nu-\gamma,\beta,\nu;\dfrac{w}{w-1};z\right)$ ? $\endgroup$ – Harry Peter Aug 30 '16 at 12:05
2
$\begingroup$

$\int_0^1\dfrac{t^{\mu-1}(1-t)^{\nu-\mu-1}}{(1-wt)^\rho}{_2F_1}(\alpha,\beta;\gamma;zt)~dt$

$=\int_0^1\sum\limits_{n=0}^\infty\dfrac{(\alpha)_n(\beta)_nz^nt^{n+\mu-1}(1-t)^{\nu-\mu-1}}{(\gamma)_nn!(1-wt)^\rho}dt$

$=\sum\limits_{n=0}^\infty\dfrac{(\alpha)_n(\beta)_nz^nB(n+\mu,\nu-\mu){_2F_1}(\rho,n+\mu;n+\nu;w)}{(\gamma)_nn!}$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\Gamma(n+\mu)\Gamma(\nu-\mu)(\alpha)_n(\beta)_n(\rho)_k(n+\mu)_kz^nw^k}{\Gamma(n+\nu)(\gamma)_n(n+\nu)_kn!k!}$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\Gamma(n+k+\mu)\Gamma(\nu-\mu)(\alpha)_n(\beta)_n(\rho)_kz^nw^k}{\Gamma(n+k+\nu)(\gamma)_nn!k!}$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{\Gamma(\mu)\Gamma(\nu-\mu)(\mu)_{n+k}(\alpha)_n(\beta)_n(\rho)_kz^nw^k}{\Gamma(\nu)(\nu)_{n+k}(\gamma)_nn!k!}$

$=B(\mu,\nu-\mu)\mathrm{F}^{1:2;1}_{1:1;0}\Bigg(\begin{matrix}\mu&:&\alpha,\beta&;&\rho&\\\nu&:&\gamma&;&-&\end{matrix}\Bigg|z,w\Bigg)$

$\endgroup$
  • 1
    $\begingroup$ Ok, I've seen hypergeometric functions using notation similar to the one you give in the last line above, but I'm having trouble finding a definition of what the indices represent. Can you explain, or redirect me to a complete definition? $\endgroup$ – David H Aug 29 '16 at 4:50
2
$\begingroup$

This answer is meant to connect the ones given by @Harry Peter and @Start wearing purple, clarifying a few questions emerged in the comments.

The integral of interest can be evaluated in the way pointed out by @Harry Peter, without forgetting to set some conditions on the parameters. First of all, for $\left|z\right|<1$ we can use the power series representation of the Gauss hypergeometric fucntion ${}_2F_1$ $$\begin{align*}\mathcal{I}(\alpha,\beta,\gamma,z;\mu,\nu,\rho,w)&=\int_0^1\frac{t^{\mu-1}(1-t)^{\nu-\mu-1}}{(1-wt)^{\rho}}{}_2F_1(\alpha,\beta,\gamma,zt)\,\mathrm{d}t\\[6pt]&=\int_0^1\sum_{n=0}^{\infty}\frac{t^{\mu+n-1}(1-t)^{\nu-\mu-1}}{(1-wt)^{\rho}}\frac{(\alpha)_n(\beta)_n}{(\gamma)_n}\frac{z^n}{n!}\,\mathrm{d}t, \end{align*}$$ where $(d)_n$ is the (rising) Pochhammer symbol, defined by $$(d)_n=\begin{cases} 1 &\;n=0\\ d(d+1)\cdots(d+n-1) &\;n>0. \end{cases}$$ The integral can now be performed using Euler representation, which in our case holds for $\Re(\nu+n)>\Re(\mu+n)>0$ and $\left|\mathrm{arg}(1-w)\right|<\pi$, $$\begin{align*}\mathcal{I}(\alpha,\beta,\gamma,z;\mu,\nu,\rho,w)&=\sum_{n=0}^{\infty}\frac{(\alpha)_n(\beta)_n}{(\gamma)_n}\frac{z^n}{n!}B(n+\mu,\nu-\mu){}_2F_1(\rho,n+\mu;n+\nu;w)\\[6pt]&=\sum_{n,m=0}^{\infty}\frac{\Gamma(n+\mu)\Gamma(\nu-\mu)}{\Gamma(n+\nu)}\frac{(\alpha)_n(\beta)_n}{(\gamma)_n}\frac{(\rho)_m(n+\mu)_m}{(n+\nu)_m}\frac{z^n}{n!}\frac{w^m}{m!}, \end{align*}$$ valid for $\left|w\right|<1$. Considering that $$(d)_n=\frac{\Gamma(d+n)}{\Gamma(d)}\quad\text{for}\;\;d\neq 0,-1,-2,\dots$$ when $\mu,\nu\neq 0,-1,-2,\dots$ we can write $$\begin{align*}\mathcal{I}(\alpha,\beta,\gamma,z;\mu,\nu,\rho,w)&=\sum_{n,m=0}^{\infty}\frac{\Gamma(n+\mu+m)\Gamma(\nu-\mu)}{\Gamma(n+\nu+m)}\frac{(\alpha)_n(\beta)_n}{(\gamma)_n}(\rho)_m\frac{z^n}{n!}\frac{w^m}{m!}\\[6pt]&=\frac{\Gamma(\mu)\Gamma(\nu-\mu)}{\Gamma(\nu)}\sum_{n,m=0}^{\infty}\frac{(\mu)_{n+m}(\alpha)_n(\beta)_n(\rho)_m}{(\nu)_{n+m}(\gamma)_n}\frac{z^n}{n!}\frac{w^m}{m!}\\[6pt]&=B(\mu,\nu-\mu)\,\mathrm{F}^{1:2;1}_{1:1;0}\left(\left.\begin{matrix}\mu&:&\alpha,\beta&;&\rho&\\\nu&:&\gamma&;&-&\end{matrix}\right|z,w\right). \end{align*}$$ $\mathrm{F}^{p:q;k}_{l:m;n}$ denotes Kampé de Fériet's double hypergeometric function in the (modified) notation of Burchnall and Chaundy [see Srivastava and Panda - "An integral representation for the product of two Jacobi polynomials", Eq. (26)] $$\begin{align*}&\mathrm{F}^{p:q;k}_{l:m;n}\left(\left.\begin{matrix}(a_p)&:&(b_q)&;&(c_k)&\\(\alpha_l)&:&(\beta_m)&;&(\gamma_n)&\end{matrix}\right|x,y\right)\\[6pt]&\quad=\sum_{r,s=0}^{\infty}\frac{\prod_{j=1}^p(a_j)_{r+s}\prod_{j=1}^q(b_j)_r\prod_{j=1}^k(c_j)_s}{\prod_{j=1}^l(\alpha_j)_{r+s}\prod_{j=1}^m(\beta_j)_r\prod_{j=1}^n(\gamma_j)_s}\frac{x^r}{r!}\frac{y^s}{s!}, \end{align*}$$ where $(d_h)$ denotes the sequence of $h$ parameters $d_1,\dots,d_h$. In general, convergence of this double series is assured if one of the following conditions holds

i) $p+q<l+m+1$, $p+k<l+n+1$ and $\left|x\right|<\infty$, $\left|y\right|<\infty$

ii) $p+q=l+m+1$, $p+k=l+n+1$ and

$ \begin{align*}\;\;\begin{cases}\left|x\right|^{\frac{1}{p-l}}+\left|y\right|^{\frac{1}{p-l}}<1 &\text{if}\;\;p>l\\[5pt]\max\left\{\left|x\right|,\left|y\right|\right\}<1 &\text{if}\;\;p\le l.\end{cases}\end{align*} $

In our case we have $p=1$, $q=2$, $k=1$, $l=1$, $m=1$ and $n=0$, so we are in (ii) and the double series converges only if $\max\left\{\left|z\right|,\left|w\right|\right\}<1$. Collecting all the constraints introduced we finally have $$\mathcal{I}(\alpha,\beta,\gamma,z;\mu,\nu,\rho,w)=B(\mu,\nu-\mu)\,\mathrm{F}^{1:2;1}_{1:1;0}\left(\left.\begin{matrix}\mu&:&\alpha,\beta&;&\rho&\\\nu&:&\gamma&;&-&\end{matrix}\right|z,w\right),$$ if $\max\left\{\left|z\right|,\left|w\right|\right\}<1$, $\left|\text{arg}(1-w)\right|<\pi$ and $\Re(\nu)>\Re(\mu)>0$.



In the case $\mu=\gamma$ the general result reduces to the one given by @Start wearing purple $$\begin{align*}\mathcal{I}(\alpha,\beta,\gamma,z;\gamma,\nu,\rho,&w)=B(\gamma,\nu-\gamma)\,\mathrm{F}^{1:2;1}_{1:1;0}\left(\left.\begin{matrix}\gamma&:&\alpha,\beta&;&\rho&\\\nu&:&\gamma&;&-&\end{matrix}\right|z,w\right)\\[6pt]&=\int_0^1\frac{t^{\gamma-1}(1-t)^{\nu-\gamma-1}}{(1-wt)^{\rho}}{}_2F_1(\alpha,\beta,\gamma;zt)\,\mathrm{d}t\\[6pt]&=\frac{1}{(1-w)^{\rho}}\int_0^1t^{\gamma-1}(1-t)^{\nu-\gamma-1}\left(1-\frac{w}{w-1}+\frac{wt}{w-1}\right)^{-\rho}{}_2F_1(\alpha,\beta,\gamma;zt)\,\mathrm{d}t, \end{align*}$$ where we have multiplied and divided by $(1-w)^{\rho}$. According to the single integral representation of Appell series $F_3$, the last expression is $$\mathcal{I}(\alpha,\beta,\gamma,z;\gamma,\nu,\rho,w)=\frac{B(\gamma,\nu-\gamma)}{(1-w)^{\rho}}F_3\left(\rho,\alpha,\nu-\gamma,\beta;\nu;\frac{w}{w-1};z\right).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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