Integration of the $\int_{0}^{1}e^{\frac{\kappa}{\alpha} y^{\alpha}}y^{\beta}(1-y^2)^{\frac{d-3}{2}}\text{d}y$

Question

Does one calculate the following integral: $$B_{d}(\kappa, \alpha, \beta) :=\int_{0}^{1}e^{\frac{\kappa}{\alpha} y^{\alpha}}y^{\beta}(1-y^2)^{\frac{d-3}{2}}\text{d}y$$

where $d,\kappa, \beta >0$ and $\alpha \in (0,2]$

Thanks in advance:

Answer 1

It can be evaluated in terms of the Wright function ${}_p \Psi_q$ as follows: \begin{align*} \int_0^1 {\mathrm{e}^{(\kappa /\alpha )y^\alpha } y^\beta (1 - y^2 )^{(d - 3)/2} \mathrm{d}y} & = \frac{1}{2}\int_0^1 {\mathrm{e}^{(\kappa /\alpha )t^{\alpha /2} } t^{(\beta - 1)/2} (1 - t)^{(d - 3)/2} \mathrm{d}t} \\ &= \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{(\kappa /\alpha )^n }}{{n!}}\int_0^1 {t^{(\alpha n + \beta - 1)/2} (1 - t)^{(d - 3)/2} \mathrm{d}t} } \\ &= \frac{1}{2}\Gamma\! \left( {\frac{{d - 1}}{2}} \right)\sum\limits_{n = 0}^\infty {\frac{{(\kappa /\alpha )^n }}{{n!}}\frac{{\Gamma\! \left( {\frac{{\alpha n + \beta + 1}}{2}} \right)}}{{\Gamma\! \left( {\frac{{\alpha n + \beta + d}}{2}} \right)}}} \\ &= \frac{1}{2}\Gamma\! \left( {\frac{{d - 1}}{2}} \right){}_1\Psi _1 \!\left( {\frac{\alpha }{2},\frac{{\beta + 1}}{2};\frac{\alpha }{2},\frac{{\beta + d + 1}}{2};\frac{\kappa }{\alpha }} \right), \end{align*} provided that $\operatorname{Re}(d)>1$, $\operatorname{Re}(\alpha)>0$ and $\operatorname{Re}(\beta)>-1$.