evaluate the integral \[\int_{a>0}^{\infty}\frac{1}{1+x^{\alpha /2}}dx\]

Question

Could you please help to evaluate the above integral A form based on hypergeometric functions will be considerered closed form

Answer 1

Using Mathematica you can get the result for special values of $\alpha$ or $a$:
Assuming: $$ \left(1\leq a ~~ \text{Or} ~~ \alpha < \frac{2\pi}{\arccos(a)} ~~ \text{Or} ~~ \exp\left(\frac{2i\pi}{\alpha}\right) \notin \text{I}\!\text{R}\right) ~~ \text{And} ~~ \alpha>2 $$ you get $$ \int_{a>0}^{\infty}\frac{1}{1+x^{\alpha /2}}\,\text{d}x= \frac{2a^{1-\alpha/2}}{\alpha-2}\, _0F_2\left(1;\frac{b-2}{\alpha};2-\frac{2}{\alpha};-a^{-b/2}\right) $$