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By inspecting more carefully nicoguaro's solution, I realize that in general: $$v_n(b) = (-1)^{n+1} \frac{\Gamma(n+1,b) - \Gamma(n+1,-b)}{2b^{n+1}}$$ It's straightforward to confirm that $v_0 = \frac{\sinh(b)}b$ and that $\frac{\partial}{\partial b} v_n = v_{n+1}$ for all $n \in \mathbb{N}$, so this is the same function! So we can use the Taylor expansion (...
I think that a numerical integration might work in your case, I don't see why it should not. Besides numerically integrating your function, you could try an asymptotic expansion for small $a$. Since you are using Mathematica, I tried to compute the integral using Wolfram Language with the function AsymptoticIntegrate, specifically I used AsymptoticIntegrate[...