Confusion over Gamma distribution CDF

Question

Here's a probability question (probably really simple) I'm not sure how to solve:

Gamma distribution $X\sim \mathcal{G}(\alpha,\beta)$ with $\mu = 20$ and $\sigma^2 = 80$
$P(X \le 24)$ = ?

The previous question was finding the values of $\alpha$ and $\beta$, which I did using $\mu$ = $\alpha$$\beta$ and $\sigma^2$ = $\alpha$$\beta^2$.

For the gamma distribution cdf, my textbook says $P(X \le x) = F(x; \alpha, \beta) = F(x / \beta; \alpha,1)$ where $F(x / \beta; \alpha,1)$ is the standard gamma distribution cdf $$ F(x;\alpha,1) = \frac{1}{\Gamma(\alpha)}\int_0^x {y^{\alpha-1}e^{-y}} \text{d}y $$

To integrate that, it looks like I need to use the chain rule, but our professor never did an example. Is there a shortcut method? We've never used integration in a real example, only to define the pdf and get the cdf for different distributions.

Edit

The examples in my textbook involving standard gamma distribution problems say to look up the values for $F(x;\alpha)$ in Table A.4 of the appendix. When I looked, Table A.4 was missing, which really disappoints me. Are there any standard gamma distribution tables online that I can print out and hand in with the assignment? I checked Wolfram Alpha but they didn't have one. Casio has something, but I'm not sure what the shape and scale parameters are.

Edit 2

Found that table. In the front of the book, it Table A.5 came right after A.3, which is why I thought A.4 was missing. I went to the library to see if they had the same textbook; they did, and someone had the common sense (which I didn't have) to look in the back of the book, and there it was. No more help is needed.

Answer 1

As suggested by probabilityislogic, my comment is converted to an answer.

You need to integrate by parts repeatedly beginning with $u=y^{\alpha-1}$, $v=−e^{−y}$, $\mathrm dv=e^{−y}\mathrm dy$, and using $$\int_0^x u\ \mathrm dv= uv\biggr|_0^x − \int_0^x v\ \mathrm du.$$ Since $\mathrm du = (\alpha-1)y^{\alpha-2}\mathrm dy$, each time you do an integration by parts, you will get an integral with a smaller exponent for $y$ on the right hand side. If $\alpha$ is an integer (as it is in this particular case), you will be able to finish up the process with a $\int_0^x e^{-y}\mathrm dy$. If $\alpha$ is not an integer, things are more complicated because there is no general closed-form expression for $\int_0^x y^{\gamma}e^{-y}\mathrm dy$ where $0 < \gamma < 1$. As noted by Xi'an, the cdf is the incomplete gamma function, and its numerical values have been tabulated.

If integration by parts is not the point of this exercise as suggested in Elvis's comment, you may want to check if your professor wants you to think of the value of a gamma random variable as an arrival time in a Poisson random process and solve the problem from that viewpoint.