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The Gamma Function Γ(x)   The gamma function is used extensively in analysis. What appears below is an answer to a question posed on Yahoo Answers (although before I had time to write it up, the problem had already been solved by another visitor). As there is much that can be proven or shown with regard to the gamma function, more will be added as additional questions appear on Yahoo.   Proof of the convergence of the integral form of the gamma function Γ(x) for x > 0: (∀x  ∈ (0,∞))(∫[0,∞]tx-1e-tdt converges.) We have: ∫[0,∞]tx-1e-tdt. First, we divide the integral into two parts: ∫[0,1]tx-1e-tdt + ∫[1,∞]tx-1e-tdt. Part 1. As (1) tx-1e-t > 0 for t ∈ (0,1), (2) e-t < 1 for t ∈ (0,1) and consequently tx-1e-t < tx-1 for t ∈ (0,1), and (3) ∫[0,1]tx-1dt = tx /x |[0,1] = 1x /x ∈ ℝ for all x > 0, we have that ∫[0,1]tx-1e-tdt converges for all x > 0. Part 2. To show that ∫[1,∞]tx-1e-tdt converges for x > 0, we use the limit test and the convergence of the integral ∫[1,∞]t -2dt. We have: lim [t→∞] tx-1e-t / t -2 = lim [t→∞] tx+1/et . We apply L'Hospital's Rule n times until x+1-n ≤ 0. Then lim [t→∞] ((x+1)(x)(x-1)∙∙∙(x+1-(n-1))tx+1-n)/et  = 0; hence, for x > 0, lim [t→∞] tx-1e-t / t -2 = 0; hence, as ∫[1,∞]t -2dt converges, we have that, for x > 0, ∫[1,∞]tx-1e-tdt converges as well. Combining parts (1) and (2), we have for x > 0, ∫[0,∞]tx-1e-tdt converges.

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