Help with proof of expected value of gamma distribution I am struggling with this proof of the expected value for the gamma distribution.
I need help with the step indicated by the red arrow. Could someone please break it down for me.

Thanks.

The argument is direct if one knows that every gamma function is a PDF. Namely, call $$g_a(x)=\frac{\lambda^a}{\Gamma(a)}x^{a-1}\mathrm e^{-\lambda x},$$ then, for every positive $a$, $g_a$ is a PDF hence $$\int_0^\infty g_a(x)\mathrm dx=1.$$ In your case, $f_X=g_\alpha$ and you are after $$E(X)=\int_0^\infty xf_X(x)\mathrm dx=\int_0^\infty xg_\alpha(x)\mathrm dx.$$ The key observation is that, for every positive $a$, $$xg_a(x)=\frac{\Gamma(a+1)}{\Gamma(a)\lambda}g_{a+1}(x)=\frac{a}{\lambda}g_{a+1}(x),$$ hence $$\int_0^\infty xg_a(x)\mathrm dx=\frac{a}{\lambda}\int_0^\infty g_{a+1}(x)\mathrm dx=\frac{a}{\lambda},$$ in particular, $$E(X)=\frac{\alpha}{\lambda}.$$ I believe this is the argument the red arrow is referring to.
