Answer Happy

prove equation 4 ,. it is already proved. but you have to solve it in more steps
APP-4 But and so Hence, The Gamma Function Euler's integral definition of the gamma function is To 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, 4 and thus (2) gives [T [₁²0 = [orde du e-²2² dv T'(x) = Γ(1) = By letting t = u², we can write (3) as e-(²+²) du dv = 4 T(4) = 3r(3) = 3.2.1, and so on. In this manner it is seen that when. s 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: [TF = πT [r]² ² - ( ² [*e-² du ) ( ² [~ ₁² dv) - 4 [" ["ded. du or T(2) = 1T(1) = 1 tedt. T(3) = 21(2) = 2.1 T()=√T. e'dt = 1, Switching to polar coordinates u = r cos 0, v = r sin 0 enables us to evaluate the double integral: π/2 +00 T(4) = [*1-¹²e-²¹ dt. *This function was first defined by Leonhard Euler in his text Institutiones Calculi Integralis published in 1768. T(63) = 2 * er dr d0 = π. e-² du. (1) (4) T(x) (3)