I've come across this integral:
\begin{equation} I(x)=\frac{1}{\Gamma(a)}\int_x^{\infty}t^{a-1}(1+t)^{b-a-1}e^{-zt}dt \end{equation}
a few times now, but I cant seem to get anywhere with it. I've called it the "Incomplete Tricomi Hypergeometric Function," since $I(0)=U(a,b;z)$, in analogy with the Incomplete Gamma Function.
Specifically, I'm looking at the case $I(1)$ near $z\rightarrow0^+$. Does anyone have any ideas? I've been stumped all day.
Let $J(z) = \Gamma(a) I(1)$. If $b < 1$, the integral $J(0)$ converges and $$\lim_{z \to 0^+} J(z) = J(0) = \frac {{_2 F_1}(1 - b, 1 + a - b; 2 - b; -1)} {1 - b}.$$ If $b \geq 1$, subtract $t^{b - 2} e^{-z t}$ from the integrand and prove that the resulting integral is asymptotically smaller than $J(z)$. Then $$J(z) \sim \int_1^\infty t^{b - 2} e^{-z t} dt = z^{1 - b} \Gamma(b - 1, z) \sim \cases {\Gamma(b - 1) z^{1 - b} & $b > 1$ \\ -\ln z & $b = 1$}, \quad z \to 0^+.$$ If the question is about getting the expansion for the case $b = n \in \mathbb N$ down to the $o(1)$ term, that can be done by taking the binomial expansion of $t^{n - 2} (1 + 1/t)^{n - a - 1}$ and repeating the process. We obtain $$J(z) = \sum_{k = 0}^{n - 2} c_k \Gamma(n - k - 1) z^{-n + k + 1} - c_{n - 1} (\gamma + \ln z) + \\ {_3 F_2}(1, 1, 1 + a; 2, n + 1; -1) c_n - {_3 F_2}(1, 1, 2 - n; 2, 2 - a; -1) c_{n - 2} + O(z), \\ c_k = \binom {n - a - 1} k, \quad z \to 0^+.$$