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

https://arbital.com/p/5c9

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?