Does the gamma function depend on lambda?

Question

In my professor's notes I read the following.

The gamma-distribution is given by: $$g_a(x)=\frac{\lambda^a}{\Gamma(a)}x^{a-1}\mathrm e^{-\lambda x},$$

Where

$$ \Gamma(a) = \int_0^{\infty}\,e^{-x\lambda}\lambda^ax^{a-1}dx \,. $$

Quiz: does the above constant depends on $\lambda$?

How can this integral not depend on $\lambda$? However I read elsewhere that $\int_0^{\infty}\,e^{-x}x^{a-1}dx \,. $ So I am puzzled.

Answer 1

With the substitution $u = \lambda x$ we obtain $$\int_0^\infty e^{-x\lambda} \lambda^a x^{a-1} \, dx = \int_0^\infty e^{-u} u^{a-1} \, du.$$