I need help with this integration: $\int_{0}^{\infty}x^{a}e^{x^{b}}e^{-(e^{x^{b}}-1)^c}dx$

Question

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.

Answer 1

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)