## Need some help integrating gamma distribution

Hello,

I am trying to prove that the gamma function is a proper probability distribution (integrates to one).

The gamma function is given as gamma(x) = integral{(a^(x-1))*(e^(-a))da} where the lower bound on the integral is 0 and the upper bound is positive infinity.

I start the problem using integration by parts. u = (a^(x-1)), du = (x-1)*a^(x-2) dv = e^(-a), v = -e^(-a)

I obtain: gamma(x) = (-e^(-a))*a^(x-1) - integral{(-e^(-a))*(x-1)*a^(x-2)da}

I use integration by parts again on the right hand integral. u = ((x-1)*a^(x-2)), du = (x-2)*(x-1)*a^(x-3) dv = e^(-a), v = -e^(-a)

I obtain:

gamma(x) = (-e^(-a))*a^(x-1) + (-e^(-a))*((x-1)*a^(x-2)) - integral{(-e^(-a))*(x-2)*(x-1)*a^(x-3)da}

I see this integral as a series but am having trouble writing out the series form and going about evaluating the series between 0 and positive infinity.