 Solution

At $x=0$, both denominators are nonzero because $a\notin 2\pi \mathbb{Z}$; they remain nonzero for every real $x\ge 0$. The integrand is therefore bounded on each finite interval. At infinity its terms decay respectively as $O(e^{-(1+b)x})$ and $O(e^{-(1-b)x})$, so $0<b<1$ gives absolute convergence.

Set $t=e^{-x}$. The two integrals becomeBy the Euler integral for the hypergeometric function and the Gamma function recurrence, ${\int }_0^1{t}^{s-1}/(1-zt)dt={}_2{F}_1(1,s;s+1;z)/s$ for $s>0$ and the indicated unit-circle values $z\ne 1$. HenceThe Gauss hypergeometric functions here use their continuation from the Euler integral, so no pole crosses the integration interval.