"Consider using [3jp] for the proof?"


by Eric Bruylant Jul 13 2016

The factorial function can be defined in a different way so that it is defined for all real numbers \(and in fact for complex numbers too\)\. We define $~$x!$~$ as follows: $$~$x! \= \\Gamma (x+1),$~$$ where $~$\\Gamma $~$ is the gamma function: $$~$\\Gamma(x)\=\\int\_{0}^{\\infty}t^{x-1}e^{-t}\\mathrm{d} t$~$$ Why does this correspond to the factorial function as defined previously? We can prove by induction that for all positive integers $~$x$~$: $$~$\\prod\_{i\=1}^{x}i \= \\int\_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$~$$ First, we verify for the case where $~$x\=1$~$\. Indeed: $$~$\\prod\_{i\=1}^{1}i \= \\int\_{0}^{\\infty}t^{1}e^{-t}\\mathrm{d} t$~$$ $$~$1\=1$~$$ Now we suppose that the equality holds for a given $~$x$~$: $$~$\\prod\_{i\=1}^{x}i \= \\int\_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$~$$ and try to prove that it holds for $~$x + 1$~$: $$~$\\prod\_{i\=1}^{x+1}i \= \\int\_{0}^{\\infty}t^{x+1}e^{-t}\\mathrm{d} t$~$$ We'll start with the induction hypothesis, and manipulate until we get the equality for $~$x+1$~$\. $$~$\\prod\_{i\=1}^{x}i \= \\int\_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$~$$ $$~$(x+1)\\prod\_{i\=1}^{x}i \= (x+1)\\int\_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$~$$ $$~$\\prod\_{i\=1}^{x+1}i \= (x+1)\\int\_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$~$$ $$~$\= 0+\\int\_{0}^{\\infty}(x+1)t^{x}e^{-t}\\mathrm{d} t$~$$ $$~$\= \\left (-t^{x+1}e^{-t}) \\right\]\_{0}^{\\infty}+\\int\_{0}^{\\infty}(x+1)t^{x}e^{-t}\\mathrm{d} t$~$$ $$~$\= \\left (-t^{x+1}e^{-t}) \\right\]\_{0}^{\\infty}-\\int\_{0}^{\\infty}(x+1)t^{x}(-e^{-t})\\mathrm{d} t$~$$ By the product rule of integration: $$~$\=\\int\_{0}^{\\infty}t^{x+1}e^{-t}\\mathrm{d} t$~$$ This completes the proof by induction, and that's why we can define factorials in terms of the gamma function\.

Consider using Arbital hidden text for the proof?


Michael Cohen

I'm having difficulty figuring out how to do that.

Eric Bruylant

Added, feel free to alter.