Evaluating $\int_{T}^{\infty} \exp\left[\beta - \phi t \right]t^{-\frac{3}{2}} \log ^{\kappa}(t) \mbox{d}t$ 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
 A: Your integral $I\left( \phi ,\beta ,\kappa \right) $ can be evaluated for $\kappa =n\in N>0$:
$$I\left( \phi ,\beta ,n\right) =\exp \left( \beta \right) \int_{T}^{\infty
}\exp \left( -\phi \,t\right) t^{-\frac{3}{2}}\log \left( t\right) ^{n}dt$$
with the trick of [Jack D'Aurizio] (Integrating $\int_0^{\frac{\pi}{2}} x (\log\tan x)^{2n+1}\;dx$)
$$I\left( \phi ,\beta ,n\right) =\exp \left( \beta \right) \underset{\alpha
\rightarrow 0}{\lim }\frac{d^{n}}{d\alpha ^{n}}\int_{0}^{\infty }u^{\alpha -%
\frac{3}{2}}\exp \left( -\phi \,t\right) dt$$
Performing the integration:
$$I\left( \phi ,\beta ,n\right) =\exp \left( \beta \right) \sqrt{\phi }%
\underset{\alpha \rightarrow 0}{\lim }\frac{d^{2n+1}}{d\alpha ^{2n+1}}\phi
^{-\alpha }\Gamma \left( \alpha -\frac{1}{2},T~\phi \right)$$
the differentiation and take the limit as $\alpha \rightarrow 0$ leads to:
$$I\left( \phi ,\beta ,n\right) =\exp \left( \beta \right) \sqrt{\phi }%
\sum_{k=0}^{n}\binom{n}{k}\left( -1\right) ^{n-k}\left( \log [\phi ]\right)
^{n-k}\times $$
$$\times \Gamma \left( -\frac{1}{2},T~\phi \right) \left( \log [T~\phi
]\right) ^{k}+k!\sum_{m=1}^{k}\frac{\left( \log \left( T~\phi \right)
\right) ^{k-m}}{\left( k-m\right) !}G_{m+1,m+2}^{m+2,0}\left( T~\phi
\left\vert 
\begin{array}{c}
1,1,...,1 \\ 
0,0,...,0,-\frac{1}{2}%
\end{array}%
\right. \right)$$
where the general Leibniz rule (https://en.wikipedia.org/wiki/Product_rule) and the Wolfram Function side is used.
