Show that for every \(\alpha>0\), \[ \int_0^\infty e^{-x^\alpha}\,dx =\frac1\alpha\,\Gamma\!\left(\frac1\alpha\right). \]
Solution
1. Make the substitution\[ t=x^\alpha. \]
Since \(\alpha>0\), the map \(x\mapsto x^\alpha\) is an increasing homeomorphism of \((0,\infty)\) onto itself. Its inverse is \[ x=t^{1/\alpha}, \] and therefore \[ dx=\frac1\alpha t^{1/\alpha-1}\,dt. \]
2. The integral becomes\[ \int_0^\infty e^{-x^\alpha}\,dx =\frac1\alpha \int_0^\infty e^{-t}t^{1/\alpha-1}\,dt. \]
1. The endpoints remain \(0\) and\(\infty\) because \(x\mapsto x^\alpha\) maps those endpoints to themselves.
3. The remaining integral is the Gamma function at \(1/\alpha\).
Euler’s integral gives, for \(s>0\), \[ \Gamma(s)=\int_0^\infty e^{-t}t^{s-1}\,dt. \] Taking \(s=1/\alpha>0\) in 2 gives \[ \int_0^\infty e^{-x^\alpha}\,dx =\frac1\alpha\Gamma\!\left(\frac1\alpha\right). \]