# Does the gamma function depend on lambda?

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.

• Change of variables. – user9464 Oct 23 '19 at 21:00

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