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I have a task - to plot graphics of the function:

$$ I(E) = \frac{16i \pi k \mu}{(\beta - ik)^{4}} \frac{1}{1 + i\eta} {}_2 F_1(1+i\eta, 2; 2 + i \eta; x) $$

where

$$ x = \left( \frac{\beta + ik}{\beta - ik} \right)^2 $$ $$ k = \sqrt{\frac{\mu E}{20.9006}} $$ $$ \eta = 0.15748 \sqrt{\frac{\mu}{E}} $$ $$ \mu = \frac{m_d m_t}{m_d + m_t} \approx 1.2 $$

For task simplicity, $\beta = 1$. But in future I have to find it from the next equation:

$$ \left( \frac{\beta - ik}{\beta + ik} \right)^{2i\eta} = e^{4 \eta arctg{\frac{k}{\beta}}}, \beta \in \mathbb{R} $$

Energy parameter $E \in (0, 1]$ MeV, $i$ - is the imaginary unit.

Now my problem is the numerical calculation of the hypergeometric function ${}_2 F_1 (a, b; c; x)$. It is known that the hypergeometric function has a lot of representations. For example, there is a formula of Euler. But it requires that $Re (c) > Re (b) > 0$ and $|x| < 1$.

Question: What method or representation should I use to compute such type of hypergeometric function?

Service wolfram-alpha computes this function with no error...

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    Numerical recipes advice to solve the hypergeometric differential equation along a path (which does not intersect the branch cut at $x\in[1,\infty)$) starting from the circle $|x|=1/2$ where the series converges.2012-05-14
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    Your numerator and denominator parameters are complex? Interesting. Where did this come from?2012-05-14
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    @Fabian: to quote their words, "it's like swatting a fly with a golden brick". It works, but I'm sure there are more efficient ways of going about this. I'll have to do some digging...2012-05-14
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    @J.M. for sure. But then one has to use different methods depending on $x$ (which I guess the OP wanted to avoid).2012-05-14
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    @J.M. no, they are real(if you speak about $\eta, \beta, k, mu$). this is from physics - it is some kind of differential cross section for Helium. or similar.2012-05-14
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    @J.M. You say that direct numerical solving of hypergeometric equation hard task, isn't it?2012-05-14
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    No, I meant the arguments of your hypergeometric function. $\eta$ is real, so indeed $1+i\eta$ is complex. And yes, the numerics will be tough. As I said, I need to think about this...2012-05-14
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    In the meantime, you might appreciate the use of [Pfaff's transformation](http://dlmf.nist.gov/15.8.E1) as a starting point: $${}_2 F_1\left({{1+i\eta,2}\atop{2+i\eta}}\mid x\right)=\frac1{(x-1)^2}{}_2 F_1\left({{1,2}\atop{2+i\eta}}\mid \frac{x}{x-1}\right)$$2012-05-14

1 Answers 1

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Since $E > 0$, $x \not= 1$. Thus, the hypergeometric function ${}_2 F_1(1+i \eta, 2 ; 2+ i \eta ; x)$ is defined. Using Euler's integral representation with $a=2$, $b=1+ i \eta$ and $c = 2 + i \eta$, we get: $$ {}_2 F_1(1+i \eta, 2 ; 2+ i \eta ; x) = (1+ i \eta) \int_0^1 t^{i \eta} (1- t x)^{-2} \mathrm{d} t $$ The integral is convergent for real $\eta$ and complex $x$ not on $[1, \infty)$.

Apart from the integral representation, you may find the work by Joshua Willis on "Acceleration of generalized hypergeometric functions through precise remainder asymptotics" right on for your problem.

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    Thanks a lot. I use exactly this integral to compute hypergeometric function. I compute it using Simpson method, and all works for me. My problem was in computation of Gamma functions which have to stand before the integral - but you have reduce the fraction and simplified my task!2012-05-15