In my professor's notes I read the following.

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


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

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.

  • 1
    $\begingroup$ Change of variables. $\endgroup$ – user9464 Oct 23 '19 at 21:00

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


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