Show that $ \int_{0}^{\infty} \lambda^{a} x^{a - 1} e^{-\lambda x} dx = \Gamma(a) $

Question

Show that $ \int_{0}^{\infty} \lambda^{a} x^{a - 1} e^{-\lambda x} dx = \Gamma(a) $, where $\lambda > 0$ and $a > 0$.

Reminder: $\Gamma(a) = \int_{0}^{\infty} x^{a -1}e^{-x} dx $.

Attempt:

I tried to do integration by parts, using $f = x^{a -1}; ~~ g'=e^{-\lambda x} \Rightarrow f' = (a-1)x^{a-2};~~ g = \frac{e^{-\lambda x}}{-\lambda}$.

Proceeding with the calculations, I got

$$ \int_{0}^{\infty} \lambda^{a} x^{a - 1} e^{-\lambda x} dx = \lambda^{a -1}(a-1) \int_{0}^{\infty} x^{a - 2} e^{-\lambda x} dx $$

And that's about it. I don't know how to continue or if I'm on the right path.

Answer 1

$$ \int_0^\infty (\lambda x)^{a-1} e^{-\lambda x} (\lambda\, dx) = \int_0^\infty u^{a-1} e^{-u} \, du \quad \text{where } u = \lambda x \text{ so } du= \lambda\,dx. $$

Answer 2

If one knows that $$ \Gamma(a) = \int_{0}^{\infty} x^{a -1}e^{-x} \, dx, $$ what about the following change of variable? $$ x=\lambda u, \quad dx=\lambda \, du. $$