Help Euler defined the function and proved G(n) = n! = 1 middot 2 middot 3 middo

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Euler defined the function and proved G(n) = n! = 1 middot 2 middot 3 middot middot middot n for all positive integers n. Redo Euler's work by first calculating G(0) = 1 and then use integration by parts to show G(n) = n G(n - 1). That says G(n) = n! by recursion. Using standard methods to reduce to single integrals, evaluate the following double integral in terms of G(m) and G(n): Calculate the Jacobian of the two-variable substitution What is the region S in the ut-plane that maps onto [0, infinity) times [0, infinity) in the xy-plane under the map (u, v) (x, y) = (tu, t(1 - u))? Hint: simplify x + y. Using the substitution in parts (c) and (d) and the change-of-variables theorem, show that Euler used the G(n) integral to invent the definition of n! for non-integers n. Use the result of parts (b) and (e) to find the value of (1/2)!. You may need to evaluate the integral The easiest way to do that is to see that is a very simple curve, whose area is known.

Explanation / Answer

(a) All integral bounds here are between 0 and +inf and i'll ignore them in notation

G(0) = int(e^-x) = [-e^(-x)] = -lim(x->+inf)(e^-x) + e^0 = 0+1=1
G(n) = int(x^ne^-x) = [-e^(-x)x^n] + n int(x^(n-1)e^(-x))
Since lim(x->+inf) e^(-x)x^n = 0, then G(n) = n G(n-1)

So G(n)=n!

(b) f(m,n) = int(x^m e^(-x) dx) * int(y^n e^(-y) dy) = G(n)G(m)

(c)
Let (x,y)=f(u,t) = (tu,t(1-u))

dx/du = t
dx/dt = u
dy/du = -t
dy/dt = (1-u)

So the jacobian is t*(1-u)-u*(-t)=t

(d)
By pluggin x+y=tu+t(1-u)=t in y=t(1-u), then y=(x+y)(1-u) => 1-u=y/(x+y) => u=x/(x+y)
So we have the inverse transform g(x,y)=(u,t)=(x/(x+y),x+y)
So for 0<x<+inf and 0<y<+inf we have :

0 <= u = x/(x+y) <= 1 (since x+y >= x)
0 <= v = x+y < +inf

So the domain will be [0,1] x [0,+inf]


(e)
By using the change of variable theorem we find :
f(m,n) = int(0<u<1) int(0<t<+inf) t * (tu)^m (t(1-u))^n e^(-t) du dt
f(m,n) = int(0<u<1) int(0<t<+inf) t^(m+n+1)e^(-t) u^m (1-u)^m du dt
f(m,n) = int(0<t<+inf) t^(m+n+1)e^(-t) dt * int(0<u<1) u^m (1-u)^m du

So f(m,n) = G(m+n+1)int(0<u<1) u^m (1-u)^m du


(f)

(1/2)!^2 = (1/2)!*(1/2)! = f(1/2,1/2)=G(2)*int(0<u<1) sqrt(u(1-u)) du

So (1/2)! = sqrt(2I) where I = int(0<u<1) sqrt(u(1-u))du

Notice that y = sqrt(u(1-u))
=>y^2 = u-u^2
=> (u-1/2)^2 + y^2 = (1/2)^2
So we are integrating the half circle of center (1/2,0) and radius 1/2 , so I=1/2(Pi/4)=Pi/8

Finally (1/2)! = sqrt(Pi/4)=sqrt(Pi)/2

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