How to do the change of variables in this integral

Question

Using the change of variables $t=sx$ for the single definite integral:

$$\Gamma \left( x\right) =\int _{0}^{\infty }e^{-t}t^{x-1}dt$$

what magical process, in full, do you use to get to:

$$\Gamma \left( x\right) \sim x^{x}\int _{0}^{\infty }e^{-x\left( s-\log s\right) }\dfrac {ds} {s}$$

Answer 1

We consider the integration variables $s$ and $t$. Substitution yields \begin{align*} t&=sx\\ dt&=x ds \end{align*}

We obtain \begin{align*} \Gamma(x)&=\int_{0}^{\infty }e^{-t}t^{x-1}dt\\ &=\int_{0}^{\infty }e^{-sx}(sx)^{x-1}xds\\ &=x^x\int_{0}^{\infty }e^{-sx}s^{x}\frac{ds}{s}\\ &=x^x\int_{0}^{\infty }e^{-sx}e^{x\log s}\frac{ds}{s}\\ &=x^x\int_{0}^{\infty }e^{-x(s-\log s)}\frac{ds}{s}\\ \end{align*} and the claim follows.