I don't know how to arrive to (4) in which $0<s<1$ and (5) identities [edit: 5th identity here ][2] in https://arxiv.org/abs/1808.05159v1. My attempt to (4) was using $\frac{1}{\Gamma(-s)}=\frac{\Gamma(1-s)}{-s}$ and rewriting the right hand side as a double integral $$ \lambda^s = \int_{0}^\infty \int_{0}^\infty( x^{-s}e^{-x} + \frac{e^{-\lambda y}-1}{y^{1+s}}) dx dy $$ but I don't think that gets me anywhere. Thank you for the help
Assume that $\lambda > 0$ and $0 < s <1$.
Integrating by parts, we have $$I = \int_{0}^{\infty} \left(e^{- \lambda t}-1 \right) \frac{\mathrm dt}{t^{1+s}} = \frac{\left(1-e^{ -\lambda t} \right)}{st^{s}} \Bigg|_{0}^{\infty} - \frac{\lambda}{s} \int_{0}^{\infty} \frac{e^{- \lambda t}}{t^{s}} \, \mathrm dt. $$
Since $\lambda > 0$ and $s>0$, the boundary term vanishes at infinity.
And since $0 < s <1$, the boundary terms also vanishes as $t \to 0^{+}$.
So we have $$I = - \frac{\lambda}{s} \int_{0}^{\infty} \frac{e^{- \lambda t}}{t^{s}} \, \mathrm dt.$$
Now make the substitution $u = \lambda t$.
$$I = - \frac{\lambda^{s}}{s} \int_{0}^{\infty} \frac{e^{-u}}{u^{s}} \, \mathrm du = - \frac{\lambda^{s}}{s} \, \Gamma(1-s) = \lambda^{s} \, \Gamma(-s)$$