How to evaluate this notoriously hard exponential integral I am trying to evaluate the following integral
$$I = \int_{T}^{\infty} \exp\left[\beta t^{1-2H} - \gamma t^{2 - 2H} \right]t^{H-2} \log ^{\alpha}(t) \mbox{d}t$$
or alternatively if it is simpler in anyway
$$J = \int_{T}^{\infty} \exp\left[\beta t^{1-2H} - \gamma t^{2 - 2H} \right]t^{H-2} \left[(1-H) \log ^{\alpha}(t) - \alpha \log ^{\alpha - 1}(t)\right] \mbox{d}t$$
where $\alpha, \beta, \gamma$ are all non-zero constants and $H \in \left(0, 1\right)$
I have tried integrating by parts, differentiating under the integral sign (Feynman's technique), tried to substitute in $y = t^{2H}$ but have not had much luck.
Another thought i had was to try complete the square which gave an additional exp term.
Is there a way to compute this tricky integral. 
Your help is greatly appreciated.
edit
Upon Substituting $\log (t) = y$, I get the following representation
$$\int_{e^T}^{\infty} y^{\alpha } \exp\left[\beta e^{(1-2 H) y}-\gamma  e^{(2-2 H) y}+(H-1) y\right] \mbox{d}y$$
not sure if it is much easier to work with.
edit2
How about an asymptotic form for the integral, will that be easier to derive ?
 A: $\textbf{Ansatz}$ I was able to make some progress in calculating the antiderivative
$$I=\int \exp \left( -\beta \exp \left( t~\delta \right) \right) t^{\alpha }dt$$ 
where $\delta $ may set to 
$$\delta =1-2H\text{ or }\delta =2-2H$$
The integral may be expressed by the infinity sum
$$I\left( \alpha ,\beta ,\delta ,t\right) =\frac{1}{\delta ^{1+\alpha }}%
\sum_{k=0}^{\infty }\frac{\beta ^{k}\left( -1\right) ^{k-\alpha }}{%
k!k^{1+\alpha }}\Gamma \left( 1+\alpha ,-k~\delta ~t\right) +C$$
This can be simply proved by the differentiation. For some special values e.g. $\alpha=n>0$ explicit expressions can be obtained stocha. A general expression is in work. For that an intensity study of similar methods, published in the standard publication of [MITTAG-LEFFLER FUNCTIONS] (https://www.researchgate.net/publication/45870179_Mittag-Leffler_Functions_and_Their_Applications) has to be done.  Note that $k=0$ is a
problem, which has to be taken care of. 
In case of the integration limits $%
[0~~1]$ the sum can be split. If e.g. $\beta =\delta =1$    
$$I\left( \alpha ,t\right) =\sum_{k=1}^{\infty }\frac{\left( -1\right)
^{k-\alpha }}{k!k^{1+\alpha }}\Gamma \left( 1+\alpha ,-k~t\right) $$
$$\int_{0}^{1}\exp \left( -\exp \left( t\right) \right) t^{\alpha }dt=\frac{1}{%
\alpha +1}+\left( I\left( 1\right) -I\left( 0\right) \right) $$
For other integration limits, one has to expand the integral e.g. 
$$\lim_{s\rightarrow 0}\int \exp \left( t~s-\exp \left( t\right) \right)
t^{\alpha }dt=\lim_{s\rightarrow 0}\sum_{k=0}^{\infty }\frac{\left(
-1\right) ^{1+k}}{k!\left( s-k\right) ^{1+\alpha }}\Gamma \left( 1+\alpha
,\left( s-k\right) ~t\right) $$
Note: This is the Laplacetransform of:
$$L\left[ \exp \left( -\beta \exp \left( t~\delta \right) \right) t^{\alpha }%
\right] $$
which may be calculated by the convolution theorem´tattvamasi.
$\textbf{Addendum}$ I finally managed to identify the infinity sum for $T=0$. In this case the solution may be expressed in terms of a $\lambda$-Generalized Hurwitz-Lerch- Zeta Function, which is defined e.g. in Srivastava.
