Prove that $\Gamma(1+1/x)=\int_{0}^{\infty} e^{{-t}^{x}} dt$

Question

Please help me how to do it. I have no clue about how to begin.

Answer 1

Let $ x\in\left(0,+\infty\right) $, we have :

\begin{aligned}\Gamma\left(1+\frac{1}{x}\right)&=\frac{1}{x}\Gamma\left(\frac{1}{x}\right)\\ &=\int_{0}^{+\infty}{\frac{1}{x}y^{\frac{1}{x}-1}\,\mathrm{e}^{-y}\,\mathrm{d}y}\end{aligned}

Then substituting $ \small\left\lbrace\begin{aligned}t&=y^{\frac{1}{x}}\\ \mathrm{d}t&=\frac{1}{x}y^{\frac{1}{x}-1}\,\mathrm{d}y\end{aligned}\right. $, we get : \begin{aligned}\Gamma\left(1+\frac{1}{x}\right)=\int_{0}^{+\infty}{\mathrm{e}^{-t^{x}}\,\mathrm{d}t}\end{aligned}