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  title: '"Consider using [3jp] for the proof?"',
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  anchorContext: '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:\n$$x! = \\Gamma (x+1),$$\nwhere $\\Gamma $ is the gamma function:\n$$\\Gamma(x)=\\int_{0}^{\\infty}t^{x-1}e^{-t}\\mathrm{d} t$$\nWhy does this correspond to the factorial function as defined previously? We can prove by induction that for all positive integers $x$:\n$$\\prod_{i=1}^{x}i = \\int_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$$\nFirst, we verify for the case where $x=1$\\. Indeed:\n$$\\prod_{i=1}^{1}i = \\int_{0}^{\\infty}t^{1}e^{-t}\\mathrm{d} t$$\n$$1=1$$\nNow we suppose that the equality holds for a given $x$:\n$$\\prod_{i=1}^{x}i = \\int_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$$\nand try to prove that it holds for $x + 1$:\n$$\\prod_{i=1}^{x+1}i = \\int_{0}^{\\infty}t^{x+1}e^{-t}\\mathrm{d} t$$\nWe'll start with the induction hypothesis, and manipulate until we get the equality for $x+1$\\.\n$$\\prod_{i=1}^{x}i = \\int_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$$\n$$(x+1)\\prod_{i=1}^{x}i = (x+1)\\int_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$$\n$$\\prod_{i=1}^{x+1}i = (x+1)\\int_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$$\n$$= 0+\\int_{0}^{\\infty}(x+1)t^{x}e^{-t}\\mathrm{d} t$$\n$$= \\left (-t^{x+1}e^{-t})  \\right]_{0}^{\\infty}+\\int_{0}^{\\infty}(x+1)t^{x}e^{-t}\\mathrm{d} t$$\n$$= \\left (-t^{x+1}e^{-t})  \\right]_{0}^{\\infty}-\\int_{0}^{\\infty}(x+1)t^{x}(-e^{-t})\\mathrm{d} t$$\nBy the product rule of integration:\n$$=\\int_{0}^{\\infty}t^{x+1}e^{-t}\\mathrm{d} t$$\nThis completes the proof by induction, and that's why we can define factorials in terms of the gamma function\\.',
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