Question: "The factorial n! of a non-negative integer n is the product of all positive integers less than or equal to n. The gamma function is an extension of the factorial function to real numbers. It is defined by:

$\displaystyle \Gamma(x)=\int_0^\infty{e^{-t}}{t^{x-1}}{dt}, {x>0}$

(a) Evaluate $\displaystyle \Gamma(1)$

Assuming the limits represent t, if I evaluate $\displaystyle \Gamma(1)$, does the function become $\displaystyle \int_0^\infty{e^{-t}{dt}$? In which case when I integrate do I get: $\displaystyle {-e^{-t}}$ with limits between 0 and infinity, which equals 1?

Part (b):

"Show that for x>0:

$\displaystyle \Gamma(x+1)={x\Gamma(x)}$

So I thought that $\displaystyle \Gamma(x+1)=\int_0^\infty{e^{-t}}{t^x}{dt}$. If I let $\displaystyle u=t^x$, then $\displaystyle u'={xt^{x-1}}$, and if I let $\displaystyle {v'=e^{-t}}$, then $\displaystyle {v=-e^{-t}}$. Then I can use 'by parts' to show this. Do I treat x as some constant, like n?

Finally:

"(c) Show that for n=1,2,3...

$\displaystyle \Gamma(n)=(n-1)!$

That is, the gamma function generalizes the factorial, with its argument shifted down by 1"

I am entirely unsure about how to work this one out. Any tips/hints? Many thanks!

CP