# Proof verification: $\displaystyle\int_0^\infty e^{-x^\alpha/\alpha}\,\mathrm dx =\alpha^{(1-\alpha)/\alpha}\Gamma (1/\alpha)$

Proposition 0.1 Let $\alpha\in\mathbb R_{\neq 0}$ $$\int_0^\infty e^{-x^\alpha/\alpha}\,\mathrm dx =\alpha^{(1-\alpha)/\alpha}\Gamma (1/\alpha)$$

When $\alpha=2$, the integral is half of the Gaussian integral, which the equality is verified to hold. It turns out that it is possible to find out the properties of the generalized Gaussian integral with the help of what we have studied on Gamma function. Or, to study Gamma function with the generalized Gaussian integral.

Proof $$\int_0^\infty e^{-x^\alpha/\alpha}\,\mathrm dx$$ Do the change of variables $y=x^\alpha/\alpha\implies x=\alpha^{1/\alpha}y^{1/\alpha}$ as we are just considering $\mathbb R_+$. $$=\int_0^\infty e^{-y}\frac{\mathrm dx}{\mathrm dy}\,\mathrm dy$$ $$=\int_0^\infty e^{-y}\alpha^{1/\alpha}\frac1\alpha y^{(1/\alpha)-1}\,\mathrm dy$$

$$=\alpha^{(1-\alpha)/\alpha}\int_0^\infty y^{(1/\alpha)-1}e^{-y}\,\mathrm dy$$ $$=\alpha^{(1-\alpha)/\alpha}\Gamma (1/\alpha)$$ And we are done. $\square$

The result could help us set up a lot of interesting result. For example, by let $\alpha\to 0$ at both sides, we could establish a nice little-$\mathcal o$ approximation of $\Gamma$ (however, it is not very attractive given the Stirling's formula), and the property $\Gamma(x+1)=x\Gamma(x)$ could also help us evaluate integrals in the form $$\int_0^\infty e^{-x^{\alpha/(1-\alpha)}}\,\mathrm dx$$ Given that we know $$\int_0^\infty e^{-x^\alpha}\,\mathrm dx$$ But before all these, I'd like to further verify that this proof is valid and the result is true.