Integral
I=\int t^{2/3} \exp^{a t} dt, where a < 0
If the integral is from $\displaystyle 0$ to $\displaystyle + \infty$, then setting $\displaystyle s=-a$ You have...
$\displaystyle \displaystyle \int_{0}^{+\infty} t^{\frac{2}{3}}\ e^{-s t}\ dt = \mathcal{L} \{t^\frac{2}{3}\}= \frac{\Gamma(\frac{5}{3})}{s^{\frac{5}{3}}}$ (1)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
In that case the integral is not an elementary function but it can be represented as Taylor series...
$\displaystyle \displaystyle I(z)= \int_{0}^{z} t^{\frac{2}{3}} e^{a t} dt = \sum_{n=0}^{\infty} \frac{a^{n}}{n!}\ \int_{0}^{z} t^{n+\frac{2}{3}}\ dt= z^{\frac{5}{3}}\ \sum_{n=0}^{\infty} \frac{a^{n}}{(n+\frac{5}{3})\ n!}\ z^{n} $
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$