Show $\int_0^1 (-\ln t)^{x-1} \, dt = \int_0^\infty t^{x-1}e^{-t}\,dt$

Question

$$\Gamma(x)=\int_0^1 (-\ln(t))^{x-1} \, dt = \int_0^\infty t^{x-1}(e^{-t})\,dt$$

How do I show that these two integrals are equal to one another? Is there a way to use one of the integrals to derive the other?

Answer 1

Hint:

In the l.h.s. integral, use substitution: set $$u=-\ln t\iff t=\mathrm e^{-u}, \quad \mathrm d t= -\mathrm e^{-u}\,\mathrm d u,$$ and calculate the bounds for $u$ corresponding to the bounds $0$ and $1$ for $t$.