# Exponential integral with power in square root

I have been trying to solve an integral. I know that the solution exists for the form

$$1- \dfrac{2}{\mathcal{R}^2_{\mathcal{G}}}\int_{0}^{\mathcal{R}_{\mathcal{G}}} \exp (-\Phi r^{\alpha}) r\, {\rm d}r,$$

where $$\alpha>0, \mathcal{R}_\mathcal{G}>0$$. However, I want to solve

$$1- \dfrac{2}{\mathcal{R}^2_{\mathcal{G}}}\int_{0}^{\mathcal{R}_{\mathcal{G}}} \exp \left(-\Phi \left(\sqrt{r^2+h^2}\right)^{\alpha}\right) r\, {\rm d}r,$$

where $$\Phi>0$$, $$\alpha>0, \mathcal{R}_\mathcal{G}>0$$and $$h>0$$. Any clue please?

• Please tell us your effort. Have you tried $t=\sqrt{r^2-h^2}$? – Dinesh Shankar Sep 18 '18 at 1:30
• yes I have tried, but on substituting this, the lower limit of integral becomes -h and I checked Table of Integrals (3.351) for an equivalent solution but no hope. Infact the equations in 3.351 are for integer values of n but in my case the substitution may result in non negative values. – hakkunamattata Sep 18 '18 at 2:32