How to evaluate $\int_0^\infty e^{-x^a}dx$

Question

please help me how to do it.I have no clue about how to begin.I attempted to replicate how the Gaussian integral is solved but that would require 'a' number of equations relating spherical and Cartesian co ordinates which I'm not even sure exist.Then I attempted the substitution $x^a=t$ which yielded $\Gamma(2-1/a)=(1-1/a)!$ a result from the gamma function. Is this correct? How do we find the value of the gamma function for fractional values?

Answer 1

The Gamma function is defined by $$\Gamma(x) = \int_0^\infty t^{x-1}e^{-t}dt$$ and we have $\Gamma(x) = (x-1)!$. This can be written a bit simpler as $$ x! = \int_0^\infty t^{x}e^{-t}dt$$ The approach you tried works. Substitute $t = x^a$, $dt = ax^{a-1}$ to get $$ \begin{aligned} \int_0^\infty e^{-x^a} dx &= \int_0^\infty \frac{e^{-t}}{ax^{a-1}}dt\\ &=\frac{1}{a}\int_0^\infty x^{1-a}e^{-t}dt\\ &=\frac{1}{a}\int_0^\infty t^{\frac{1}{a}-1}e^{-t}dt\\ &= \frac{1}{a}(\frac{1}{a}-1)!\\ &= \frac{1}{a}! \end{aligned} $$

In the last line I used the defining property of the factorial function $x! = x(x-1)!$.

For many values this has a closed form, but for most values it does not, take a look at https://en.wikipedia.org/wiki/Gamma_function#Particular_values