I am trying to evaluate the following integral \begin{eqnarray} I(\phi, \beta, \kappa) = \int_{T}^{\infty} \exp\left[\beta - \phi t \right]t^{-\frac{3}{2}} \log ^{\kappa}(t) \mbox{d}t \end{eqnarray} where $\phi$ and $\beta$ are positive constants and $\kappa > -4$.
Integration by parts for $I$
Let $u = t^{-\frac{3}{2}} \log ^{\kappa}(t)$ and $\frac{dv}{dt} = \exp\left[\beta - \phi t \right]$ then using integration by parts formula I got \begin{eqnarray} - t^{-\frac{3}{2}} \log ^{\kappa}(t) \phi^{-1} \exp\left[\beta - \phi t \right] \mid_T^{\infty} + \phi^{-1} \int_{T}^{\infty} \exp\left[\beta - \phi t \right]\frac{(2 \kappa -3 \log (t)) \log ^{\kappa -1}(t)}{2 t^{5/2}} \mbox{d}t \end{eqnarray} I do n't think this new integral is simpler.
Taylor Series expansion attempt for $I$
Using Taylor Series expansion for the exponential term I get \begin{eqnarray}\nonumber &=& \int_{T}^{\infty} \sum_{n=0}^{\infty} \frac{\left[\beta - \phi t \right]^n}{n!} t^{-\frac{3}{2}} \log ^{\kappa}(t) \mbox{d}t\\\nonumber &=& \int_{T}^{\infty} \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{1}{n!} {{n}\choose{k}} \beta^{n - k} (- \phi t)^k t^{-\frac{3}{2}} \log^{\kappa}(t) \mbox{d}t\\ &=& \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{1}{n!} {{n}\choose{k}} \beta^{n - k} (- \phi)^k \int_{T}^{\infty} t^{-\frac{3}{2} + k} \log^{\kappa}(t) \mbox{d}t \end{eqnarray} For $k \neq 0$ the last integral does not converge, but i think my original integral should converge, even if conditionally.
Any suggestions how to evaluate this integral? Thank you