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 ?

  • 3
    $\begingroup$ could you please redeem us from some of the notational clutter (= unnecessary many constants). thanks! $\endgroup$ – tired Jan 1 '18 at 14:00
  • $\begingroup$ 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. $\endgroup$ – Comic Book Guy Jan 1 '18 at 14:09
  • $\begingroup$ @tired I have removed the constants, I hope it helps $\endgroup$ – Comic Book Guy Jan 1 '18 at 14:13
  • $\begingroup$ Where did this thing appear? @ComicBookGuy $\endgroup$ – rae306 Jan 1 '18 at 14:18
  • 5
    $\begingroup$ What makes this "notoriously" hard? Is this integral know/used/referenced a lot? $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.