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The characteristic function of the Fisher$(1,\alpha)$ distribution is: $$C(t)=\frac{\Gamma \left(\frac{\alpha +1}{2}\right) U\left(\frac{1}{2},1-\frac{\alpha }{2},-i t \alpha \right)}{\Gamma \left(\frac{\alpha }{2}\right)}$$

What is the Fourier transform, $\mathcal {F} _ {t,x}^{-1}$ , of $\lim_{n \to \infty} (C(t))^n$ Where $U$ is the confluent hypergeometric function. I did some simplification in the hope of making things simpler:

$$C(t) = \frac{\Gamma \left(\frac{\alpha}{2}+ \frac{1}{2}\right) U\left(\frac{1}{2},1-\frac{\alpha }{2},-i t \alpha \right)}{\Gamma \left(\frac{\alpha }{2}\right)}$$ $$= \frac{\Gamma \left(\frac{\alpha}{2}+ \frac{1}{2} \right) \Gamma \left( 1-\frac{\alpha}{2} \right) }{\Gamma \left(\frac{\alpha }{2}\right) \Gamma \left( 1 -\frac{\alpha}{2} - \frac{1}{2} \right) \Gamma \left( \frac{1}{2} \right)}.\int^1_0 e^{-i t^2} t^{-\frac{1}{2}} (1-t)^{-\frac{\alpha }{2}-\frac{1}{2}}dt \;\;(1)$$ Using the properties of the gamma function and this post here `

Rewrite $\Gamma(-z)$ in terms of $\Gamma(z)$

I was able to simplify (If my math is correct) , $$\frac{\Gamma \left(\frac{\alpha}{2}+ \frac{1}{2} \right) \Gamma \left( 1-\frac{\alpha}{2} \right) }{\Gamma \left(\frac{\alpha }{2}\right) \Gamma \left( 1 -\frac{\alpha}{2} - \frac{1}{2} \right) \Gamma \left( \frac{1}{2} \right)} = \frac{-2 *\pi * \sin(\pi(1+\alpha)) (\Gamma \left( \alpha \right) )^2}{\sin(\pi(1+\frac{\alpha}{2}))}$$ Combining this with (1), We get a product of the gamma function with the integral in (1). Any ideas?


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