Original proof:

Let's consider the second integral with $\alpha = 0$. Noting, on the one hand, that the integrand is not real for $t \gt 1$ if, for example, $\gamma=2.5$, and, on the other hand, observing its relation to the density function of the ${\rm Beta}(\beta,\gamma-1)$ distribution, it seems likely that the integral (with $\alpha = 0$) is actually
$$
I(s;\beta,\gamma) = \int_0^1 {e^{ - st} t^{\beta - 1} (1 - t)^{\gamma - 2} \,{\rm d}t},
$$
where $\beta \gt 0$ and $\gamma \gt 1$.
Now, $\frac{{\Gamma (\beta + \gamma - 1)}}{{\Gamma (\beta )\Gamma (\gamma - 1)}} I(s;\beta,\gamma)$, for $s \geq 0$, is the Laplace transform of the
${\rm Beta}(\beta,\gamma-1)$ distribution. The moment-generating function is given by
$$
m(s) = 1 + \sum\limits_{k = 1}^\infty {\bigg(\prod\limits_{r = 0}^{k - 1} {\frac{{\beta + r}}{{\beta + \gamma - 1 + r}}} \bigg)\frac{{s^k }}{{k!}}}.
$$
Writing
$$
\frac{{\Gamma (\beta + \gamma - 1)}}{{\Gamma (\beta )\Gamma (\gamma - 1)}} I(s;\beta,\gamma) = m(-s)
$$
gives
$$
I(s;\beta,\gamma) = \frac{{\Gamma (\beta )\Gamma (\gamma - 1)}}{{\Gamma (\beta + \gamma - 1)}}\bigg[1 + \sum\limits_{k = 1}^\infty {\bigg(\prod\limits_{r = 0}^{k - 1} {\frac{{\beta + r}}{{\beta + \gamma - 1 + r}}} \bigg)\frac{{(-s)^k }}{{k!}}}\bigg].
$$

## It turns out that this result is well known (cf. Eq. (3) here):

Using the standard notation in the theory of special functions (in particular the hypergeometric function),
$I(s;\beta,\gamma)$ can be written as
$$
I(s;\beta,\gamma) = \frac{{\Gamma (\beta )\Gamma (\gamma - 1)}}{{\Gamma (\beta + \gamma - 1)}} \bigg[ 1 + \sum\limits_{k = 1}^\infty {\frac{{(\beta )_k }}{{(\beta + \gamma - 1)_k }}\frac{{( - s)^k }}{{k!}}} \bigg]
$$
($(x)_n$ represents the rising factorial). Finally, in terms of the confluent hypergeometric function of the first kind,
$$
I(s;\beta,\gamma) = \frac{{\Gamma (\beta )\Gamma (\gamma - 1)}}{{\Gamma (\beta + \gamma - 1)}} {}_1F_1 (\beta ;\beta + \gamma - 1; - s).
$$