Answer Happy

prove equation 4 ,. it is already proved. but you have to solve it in more steps
APP-4 But and so The Gamma Function Euler's integral definition of the gamma function* is T(x) = tedt. Hence, Convergence of the integral requires that x-1 > -1, or x > 0. The recurrence relation T(x + 1) = x(x) (2) that we saw in Section 5.3 can be obtained from (1) by employing integration by parts. Now when x = 1, Lear Γ(2) = 1T(1) = 1 T(3) 2T (2) = 2.1 T(4) 31(3) = 3.2.1, and so on. In this manner it is seen that when n is a positive integer, T(n + 1) = n!. For this reason the gamma function is often called the generalized factorial function. Although the integral form (1) does not converge for x < 0, it can be shown by means of alternative definitions that the gamma function is defined for all real and complex numbers except x = -n, n = 0, 1, 2, .... As a consequence, (2) is actually valid for x # -n. Considered as a function of a real variable x, the graph of I'(x) is as given in FIGURE A.1. Observe that the nonpositive integers correspond to the vertical asymptotes of the graph. In Problems 31 and 32 in Exercises 5.3, we utilized the fact that I() = V. This result can be derived from (1) by setting x = 1: ro) = [ and thus (2) gives By letting t = u², we can write (3) as T(1) = Le du - Ledv = JO [T₁+² = (2[*e* d) (2 [*e* dx) = 4 ["[" = du dv e 'dt = 1, tedt. rd)-2[e-du. = e-(²+²) du dv. *This function was first defined by Leonhard Euler in his text Institutiones Calculi Integralis published in 1768. Switching to polar coordinates u = r cos 0, v=r sin 0 enables us to evaluate the double integral: + [" ["e-²-³ du dv = 4 ["²" ["e²³rdr do = m. 4 (1) [r]= π or T()=√T. (4) In view of (2) and (4) we can find additional values of the gamma function. For example, when x = -1, it follows from (2) that I() = -(-). Therefore, I(-) = -21() = -2√T. [(x) (3) ▬▬▬▬▬▬▬▬▬▬x FIGURE A.1 Graph of gamma function