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Let's introduce a general integral with 5 parameters:

$$I(p,q,r,s;z)=\int_0^1(1-t^p)^q(1+z~t^r)^sdt$$

In case $p=r$ it's easy to see that $I$ can be expressed as hypergeometric function. However, if $p \neq r$, it becomes more complicated.

I have tried to use the series approach for the case $|z|<1$:

$$I(p,q,r,s;z)=\sum_{j=0}^\infty \frac{(s)_j}{j!} z^j \int_0^1(1-t^p)^qt^{rj}dt$$

The integral inside the sum is Beta function:

$$\int_0^1(1-t^p)^qt^{rj}dt=\frac{1}{p} B \left( \frac{rj+1}{p},q+1 \right)=$$

Using the relation to binomial coefficients:

$$=\frac{1}{rj+pq+1} \left( \begin{array}( \frac{rj+1}{p}+q-1 \\ ~~~~\frac{rj+1}{p}-1 \end{array} \right)^{-1}$$

Or using relation to gamma function:

$$=\frac{q}{rj+pq+1} \frac{\Gamma \left(\frac{rj+1}{p} \right)\Gamma \left(q \right)}{\Gamma \left(\frac{rj+1}{p}+q \right)}$$

Unless $p=r$, I don't see how to express the series in the form suitable for converting to hypergeometric function. Most likely ${_2F_1}$ doesn't cover this integral, however some ${_nF_m}$ might?


Can $I(p,q,r,s;z)$ be expressed in terms of generalized hypergeometric functions?


Maybe a more simple case would be $r=l p$ where $l \in \mathbb{N}$?

Update: Mathematica solves cases $r=2p$, $r=3p$ and even $r=p/2$ in terms of generalized hypergeometric functions of different orders, for example $r=2p$ gives ${_3F_2}$ while $r=3p$ gives ${_4F_3}$, so I don't have much hope for the general case.

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    $\begingroup$ Perhaps it should be $r=lp\implies~_{l+1}F_l$? $\endgroup$ Commented Oct 15, 2017 at 21:28
  • $\begingroup$ @SimplyBeautifulArt, it looks like it, yes, that's why I haven't checked further, I was interested in the general case, but if only integer is possible, I'll try to get the general formula for this case $\endgroup$
    – Yuriy S
    Commented Oct 15, 2017 at 21:43
  • $\begingroup$ Perhaps $l$ needn't be an integer and that this holds for any $l$! Though ofc, I doubt it :-/ $\endgroup$ Commented Oct 15, 2017 at 21:44
  • $\begingroup$ @SimplyBeautifulArt, I tried and failed to imagine what a hypergeometric function of fractional indices looks like $\endgroup$
    – Yuriy S
    Commented Oct 15, 2017 at 21:48

1 Answer 1

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I think I figured out the anwer to this question. At least for $|z|<1$ we have a concrete definition of a generalized hypergeometric function: the ratio of consecutive coefficients should be a rational function of $j$. Let's rewrite the integral as a series with Gamma functions:

$$I(p,q,r,s;z)= \frac{1}{p}\sum_{j=0}^\infty \frac{\Gamma(s+1)}{j!~\Gamma(s-j+1)} \frac{\Gamma(q+1)\Gamma \left(\frac{rj+1}{p} \right)}{\Gamma \left(\frac{rj+1}{p}+q+1 \right)} z^j$$

Now it's a simple matter to write the ratio:

$$\frac{C_{j+1}}{C_j}=-\frac{(-s+j)\Gamma \left(\frac{rj+1}{p}+\color{blue}{\frac{r}{p}} \right)\Gamma \left(\frac{rj+1}{p}+q+1 \right)}{(1+j)\Gamma \left(\frac{rj+1}{p} \right)\Gamma \left(\frac{rj+1}{p}+q+1+\color{blue}{\frac{r}{p}} \right)}$$

Since we have $y \Gamma(y)=\Gamma(y+1)$, for the ratio to be a rational function we really need $\frac{r}{p}$ to be an integer. However, it might be possible to do something with rational values, since Mathematica solves the case of $\frac{r}{p}=\frac{1}{2}$ in terms of 2 hypergeometric functions.


Let's pick a simple case: $\frac{r}{p}=2$. We have $\Gamma(x+2)=x(x+1)\Gamma(x)$.

$$\frac{C_{n+1}}{C_n}=-\frac{(-s+j)\left(\frac{1}{2p}+j \right) \left(\frac{p+1}{2p}+j \right)}{(1+j) \left(\frac{pq+p+1}{2p}+j \right)\left(\frac{pq+2p+1}{2p}+j \right)}$$

By definition our series represents:

$${_3F_2} \left(-s,\frac{1}{2p},\frac{p+1}{2p};\frac{pq+p+1}{2p},\frac{pq+2p+1}{2p};-z \right)$$

Or finally (unless I made some algebraic mistake, I'll check later):

$$I(p,q,2p,s;z)=\frac{\Gamma(s+1)\Gamma(q+1)}{p} {_3F_2} \left(-s,\frac{1}{2p},\frac{p+1}{2p};\frac{pq+p+1}{2p},\frac{pq+2p+1}{2p};-z \right)$$


It's easy to see why the order of the hypergeometric function increases for $\frac{r}{p}=3,4,5,\dots$, since we get more and more factors in the gamma function.


Update: I don't know what kind of mistake I made, but let's correct this, obviously:

$$C_0=\left( \begin{array}( s \\ 0 \end{array} \right) \frac{1}{p} B \left( \frac{1}{p},q+1 \right)=\frac{1}{p} B \left( \frac{1}{p},q+1 \right)$$

Which means the correct answer is:

$$I(p,q,2p,s;z)=\frac{B \left( \frac{1}{p},q+1 \right)}{p} {_3F_2} \left(-s,\frac{1}{2p},\frac{p+1}{2p};\frac{pq+p+1}{2p},\frac{pq+2p+1}{2p};-z \right)$$


In the same way for $r=3p$:

$$I(p,q,3p,s;z)=\frac{B \left( \frac{1}{p},q+1 \right)}{p} {_4F_3} \left(-s,a,a+\frac{1}{3},a+\frac{2}{3};b,b+\frac{1}{3},b+\frac{2}{3};-z \right)$$

$$a=\frac{1}{3p}, \quad b=\frac{pq+p+1}{3p}$$

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  • $\begingroup$ Accepted for the lack of a better answer $\endgroup$
    – Yuriy S
    Commented Oct 19, 2017 at 15:37
  • $\begingroup$ hi,@Yuriy, could you help to take a look at this integral: bit.ly/2DuMZKx and see how to lead to the final solution if you are available. thanks a lot. $\endgroup$
    – Vic
    Commented Jan 20, 2018 at 19:26

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