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I need to compute integrals involving spherical Bessel functions, which led me to compute the following hypergeometric function :

$_1F_2((\alpha+1)/2;\beta+3/2,(\alpha+3)/2;z)$

Also, in my particular problem :

$\bullet$ $\alpha$ and $\beta$ are always integers

$\bullet$ $\alpha \geq \beta \geq 2$

$\bullet$ $z$ is always real

$\bullet$ $z \leq 0$ and can be anywhere between 0 and $\sim -10^{13}$

I have tried to use a couple of math libraries (Arb in C, mpmath in python) that implement the $_1F_2$ function. Unfortunately, I always end up with the same problem : for very large values of $|z|$, the computation fails. I assume this is due to convergence issues, as the code is trying to brute force the computation (I think) instead of using asymptotic approximations. Arb has those approximations, but only for other hypergeometric functions, such as $_0F_1$. Also, for mpmath the computation time also becomes an issue for large $|z|$.

Since my mathematical skills are quite limited, my question is to know whether it is possible (given the specific parameters I'm using) to simplify the $_1F_2$ function or maybe express it with other functions (hypergeometric or not) whose computation is more robust in the libraries I mention ?

Thanks !

Edit :

It might be important to mention that the function I'm trying to evaluate is actually $_1F_2((\alpha+1)/2;\beta+3/2,(\alpha+3)/2;-X^2)$, with $X$ being a real between 0 and $\sim 10^6 - 10^7$

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  • $\begingroup$ For integer values of $\alpha$ and $\beta$ the given hypergeometric function can we written as $$\frac{S(z) \sinh(2\sqrt{z})+ C(z)\cosh(2\sqrt{z})}{z^m}$$ with $C,S$ being polynomials and $m\in\mathbb{Z}+\frac{1}{2}$. $\endgroup$ – Jack D'Aurizio Jun 30 '17 at 16:16
  • $\begingroup$ @JackD'Aurizio Sounds good, thanks ! Do you per chance have a reference where I could find an analytic expression for $S(z)$ and $C(z)$ ? $\endgroup$ – S.I. Jun 30 '17 at 16:26
  • $\begingroup$ Such polynomials can be found by exploiting the contiguity relations for $\phantom{}_1 F_2$. A starting point is given by $$\phantom{}_2 F_1\left(\frac{1}{2};\frac{3}{2},\frac{3}{2},z\right) = \frac{1}{2\sqrt{z}}\int_{0}^{2\sqrt{z}}\frac{\sinh(t)}{t}\,dt $$ and you just have to apply differentiation or integration by parts the correct number of times. (+1) by the way, interesting question. $\endgroup$ – Jack D'Aurizio Jun 30 '17 at 16:49
  • $\begingroup$ Thanks again @JackD'Aurizio (also for the +1 !) I'm slightly confused though : after playing around with Mathematica for a bit, I obtained the following result when fixing $\alpha=3$ and $\beta=1$ : $$ \frac{-3(-1+\cos{(2 X)}+X \sin{(2 X)})}{2 X^4} $$ Is this coherent with what you said earlier ? Another thing : when I fix $\alpha=2$, Mathematica doesn't give me any analytical form... any idea why ? $\endgroup$ – S.I. Jun 30 '17 at 17:33
  • $\begingroup$ For $\alpha=3$ and $\beta=1$ I got $\frac{3 \left(1-\text{Cosh}\left[2 \sqrt{z}\right]+\sqrt{z} \text{Sinh}\left[2 \sqrt{z}\right]\right)}{2 z^2}$. Maybe I am dealing with $\beta+\frac{3}{2}$ and you with $\frac{\beta+3}{2}$? $\endgroup$ – Jack D'Aurizio Jun 30 '17 at 17:36
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We have $$\begin{eqnarray*}\phantom{}_1 F_2\left(\frac{a+1}{2};b+\frac{3}{2},\frac{a+3}{2}; z\right)=\sum_{n\geq 0}\frac{(a+1)}{(a+1+2n)}\cdot\frac{\left(b+\frac{1}{2}\right)!}{\left(b+n+\frac{1}{2}\right)!}\cdot \frac{z^n}{n!}\end{eqnarray*}$$

$$\begin{eqnarray*}\phantom{}_1 F_2\left(\frac{a+1}{2};b+\frac{3}{2},\frac{a+3}{2}; -z^2\right)&=&\sum_{n\geq 0}\frac{(a+1)}{(a+1+2n)}\cdot\frac{\left(b+\frac{1}{2}\right)!}{\left(b+n+\frac{1}{2}\right)!}\cdot \frac{(-1)^n z^{2n}}{n!}\\[0.2cm]&=&\frac{a+1}{z^{a+1}}\int_{0}^{z}x^a\,\phantom{}_0 F_1\left(;\frac{3}{2}+b;-x^2\right)\,dx\end{eqnarray*}$$ where $g_b(x)=\phantom{}_0 F_1\left(;\frac{3}{2}+b;-x^2\right)$, for small values of $b\in\mathbb{N}$, is given by $$ g_0(x)=\frac{\sin(2x)}{2x},\quad g_1(x)=\frac{2\sin(2x)-6x\cos(2x)}{8x^3},\\ g_2(x)=\frac{15(3-4x^2)\sin(2x)-90x\cos(2x)}{32x^5} $$ For a specific value of $b$ it is not difficult (by induction) to compute the closed form of $g_b(x)$, then apply integration by parts to compute the closed form of the wanted hypergeometric function.

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  • $\begingroup$ Hello again @JackD'Aurizio ! Correct me if I'm wrong, but shouldn't the $g_1(x)$ function read : $$ g_1(x)=\frac{3\sin(2x)-6x\cos(2x)}{8x^3} $$ instead of : $$ g_1(x)=\frac{2\sin(2x)-6x\cos(2x)}{8x^3} $$ A minor typo, I admit :-) $\endgroup$ – S.I. Jul 17 '17 at 14:10

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