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

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?