Integral $\int_{0}^{\infty}e^{-u}e^{-u^{\alpha}x}\mathrm{d}u$

Question

I want to know the value of this integral:

$$\int_{0}^{\infty}e^{-u}e^{-u^{\alpha}x}\mathrm{d}u$$

where $\alpha>0$, $x>0$.

Thank you.

Answer 1

There doesn't seem to be any general form. However, for $\alpha=0$ we have $I(0)=e^{-x}$, for

$\alpha=1$ we have $I(1)=\dfrac1{1+x}$ , for $\alpha=2$ we have $I(2)=\sqrt\pi\cdot e^{^\tfrac1{4x}}\cdot\dfrac1{2\sqrt x}\cdot\text{erfc}\bigg(\dfrac1{2\sqrt x}\bigg)$,

and for $\alpha\!=\!\dfrac12$ we have $I\bigg(\dfrac12\bigg)=1-\sqrt\pi\cdot~e^{^\tfrac{x^2}4}\cdot\dfrac x2\cdot\text{erfc}\dfrac x2\quad$ Deriving the first two results

should be trivial. For the latter two, see error function and Gaussian integral. The cases

$\alpha=3$ and $\alpha=\dfrac13$ involve Airy functions, Anger functions, Bessel functions, and Weber

functions. The rest, hypergeometric series. Negative values of $\alpha$ involve Meijer G-functions.

As I said, I see no discernible pattern.

Answer 2

Case $1$: $\alpha\leq1$

Then $\int_0^\infty e^{-u}e^{-u^\alpha x}~du$

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^nu^{\alpha n}e^{-u}}{n!}du$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^n\Gamma(\alpha n+1)}{n!}$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#definite_integrals)

Case $2$: $\alpha\geq1$

Then $\int_0^\infty e^{-u}e^{-u^\alpha x}~du$

$=\int_0^\infty e^{-u^\frac{1}{\alpha}}e^{-xu}~d\left(u^\frac{1}{\alpha}\right)$

$=\dfrac{1}{\alpha}\int_0^\infty u^{\frac{1}{\alpha}-1}e^{-u^\frac{1}{\alpha}}e^{-xu}~du$

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nu^{\frac{n+1}{\alpha}-1}e^{-xu}}{\alpha n!}du$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\Gamma\left(\dfrac{n+1}{\alpha}\right)}{\alpha x^\frac{n+1}{\alpha}n!}$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#definite_integrals)