# 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.

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 ?

• could you please redeem us from some of the notational clutter (= unnecessary many constants). thanks! – tired Jan 1 '18 at 14:00
• i only kept the constants so i could easily present the square completed integral. If you think the constants are too much nuisance, i'll get rid of them but it will only reduce the integral by1 constant. – Comic Book Guy Jan 1 '18 at 14:09
• @tired I have removed the constants, I hope it helps – Comic Book Guy Jan 1 '18 at 14:13
• Where did this thing appear? @ComicBookGuy – rae306 Jan 1 '18 at 14:18
• What makes this "notoriously" hard? Is this integral know/used/referenced a lot? – alex.jordan Jan 8 '18 at 4:53

$\textbf{Ansatz}$ 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.