1

I am trying to fined a closed form for this integration

$$\int_{0}^{\infty}x^{a}e^{x^{b}}e^{-(e^{x^{b}}-1)^c}dx,$$

where $a,b,c>0$

I think the generalized integro-exponential ($E_{s}^{r}(z)=\frac{1}{r!} \int_{1}^{\infty}(\log t)^{r}t^{-s}e^{-zt}dt$) function can be used here.

ELAWADY
  • 101
  • I post a solution of a similar integral on https://math.stackexchange.com/questions/2596374/evaluating-int-t-infty-exp-left-beta-phi-t-rightt-frac32?rq=1 and https://math.stackexchange.com/questions/2586898/how-to-evaluate-this-notoriously-hard-exponential-integral – stocha Aug 17 '19 at 10:03
  • Sorry, i could not relate this integral with the answer you provided. I think, a solution to this integral can be obtained when $c=1$. – ELAWADY Aug 17 '19 at 10:24
  • Yes, it looks like this. – stocha Aug 17 '19 at 16:06
  • 1
    For c=1, the integral has closed form in terms of generalized integro-exponential function. The posts you suggested did not provide a soulution to this integral. – ELAWADY Aug 18 '19 at 08:44
  • @Elawady. It would be nice for the community, if you could post your solution for the case c=1. – stocha Aug 18 '19 at 09:29

1 Answers1

3

For c=1, If we consider THE following integral

$I=\int_{0}^{\infty}x^{a}e^{x^{b}}e^{-\lambda(e^{x^{b}}-1)}dy$

then by letting $y=x^{b}$ we get

I=$\frac{e^{\lambda}}{b^{a+1}}\int_{1}^{\infty} (log~y)^{\frac{a}{b}}e^{-\lambda y}dy$

if we consider $\frac{a}{b}$ is an integer then

$I=\frac{e^{\lambda}}{b^{a+1}} (\frac{a}{b})! E_{0}^{\frac{a}{b}}(\lambda)$

where $E_{s}^{r}(z)=\frac{1}{r!} \int_{1}^{\infty}(log t)^{r}t^{-s}e^{-zt}dt$ is the generalized integro-exponential function, (Milgram 1985).

Note that, $E_{0}^{r}(z)=\frac{E_{1}^{r-1}(z)}{z}$ and $E_{1}^{0}(z)=E_{1}(z)=\Gamma(0,z)$, where $\Gamma(s,z)=\int_{z}^{\infty}t^{s-1} e^{-t}dt$ is the complementary incomplete gamma function.

Also, it can be written as a Meijer G-function as

$E_{s}^{r}(z)=G_{r+1,r+2}^{r+2,0}\left(z \mid \begin{array}{c} ;s,\dots,s \\ 0,s-1,\dots,s-1; \end{array} \right)$

For real values of $\frac{a}{b}>-1$ we may consider an extension of generalized integro-exponential function (Pogany et al. 2017)

ELAWADY
  • 101