I really need some help integrating a gamma function

[jrkapoor]

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. Help on this would be great as I am also trying to prove that the poisson, and weibull distributions are proper. I see the use of L'hopital's rule with my result from the gamma distribution to prove the Poisson is proper and see the Weibull distribution is a variant case of the gamma distribution. From my perspective, proof of Poisson and Weibull as proper hinge directly on understanding the proof of gamma as proper.

 

[galactus]

If we have the probability distribution:

\(\displaystyle \int_{0}^{\infty}kx^{a-1}e^{\frac{-x}{b}}dx\)

Make the sub\(\displaystyle y=\frac{x}{b}\)

\(\displaystyle =kb^{a}\int_{0}^{\infty}y^{a-1}e^{-y}dy\)

This defines Gamma: \(\displaystyle \int_{0}^{\infty}y^{a-1}e^{-y}dy\)

\(\displaystyle \int_{0}^{\infty}kx^{a-1}e^{\frac{-x}{b}}dx=kb^{a}{\Gamma}(a)=1\)

Therefore, \(\displaystyle k=\frac{1}{b^{a}{\Gamma}(a)}\)

This leads us to the Gamma distribution:

\(\displaystyle g(x;a,b)=\frac{1}{b^{a}{\Gamma(a)}}x^{a-1}e^{\frac{-x}{b}}dx, \;\ x>0\)

\(\displaystyle 0, \;\ elsewhere\)

where a>0, b>0