Evaluating $\int_{T}^{\infty} \exp\left[\beta - \phi t \right]t^{-\frac{3}{2}} \log ^{\kappa}(t) \mbox{d}t$

Question

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

Answer 1

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.