I have been trying to solve an integral. I know that the solution exists for the form
$$1- \dfrac{2}{\mathcal{R}^2}\int_{0}^{\mathcal{R}_{\mathcal{G}}} \exp (-\Phi r^{\alpha}) r\, {\rm d}r,$$
where $\alpha>0, \mathcal{R}>0$. However, I want to solve
$$1- \dfrac{2}{\mathcal{R}^2}\int_{0}^{\mathcal{R}} \exp \left(-\Phi \left(\sqrt{r^2+h^2}\right)^{\alpha}\right) r\, {\rm d}r,$$
where $\Phi>0$, $\alpha>0, \mathcal{R}>0$. I have tried, but on substituting $t=r^2+h^2$, the lower limit of integral becomes h and I checked Table of Integrals (3.351) for an equivalent solution but no hope. Any help on this please?
Doing the same as Dinesh Shankar in his/her answer. $$f(R)=1- \dfrac{2}{{R}^2}\int_{0}^R \exp \left(-\Phi \left(\sqrt{r^2+h^2}\right)^{\alpha}\right)\, r\, dr$$ $$f(R)=1-\frac{2 }{\alpha\, R^2\,\Phi ^{2/\alpha }}\left( \Gamma \left(\frac{2}{\alpha },\left(h^2\right)^{\alpha /2} \Phi \right)-\Gamma \left(\frac{2}{\alpha },\left(h^2+R^2\right)^{\alpha /2} \Phi \right)\right)$$