the gamma function r(x) is defined by r(x)= integral from 0 to infinity of x^n-1(e^-1)dx where n>0. Use integration by parts to show that r(n+1)=nr(n).
To demonstrate that r(n+1) = n*r(n) using integration by parts, we need to evaluate the integral of the gamma function, r(x), for x = n+1.
The formula for integration by parts is as follows:
∫(u dv) = uv - ∫(v du)
Let's choose u = x^(n), and dv = e^(-x) dx. By differentiating u and integrating dv, we can determine du and v, respectively:
Differentiating u:
du = d/dx(x^(n)) = nx^(n-1)
Integrating dv:
v = ∫(e^(-x) dx) = -e^(-x)
Now, we can apply the formula for integration by parts:
∫(u dv) = uv - ∫(v du)
Using our chosen values for u, du, v, and dv, we get:
∫(x^n * e^(-x) dx) = -x^n * e^(-x) - ∫(-e^(-x) * nx^(n-1) dx)
Simplifying, we have:
∫(x^n * e^(-x) dx) = -x^n * e^(-x) + n * ∫(x^(n-1) * e^(-x) dx)
Next, we evaluate the integral on the right side of the equation. This is r(n):
∫(x^(n-1) * e^(-x) dx) = r(n)
Substituting this result back into our equation, we have:
∫(x^n * e^(-x) dx) = -x^n * e^(-x) + n * r(n)
Now, let's evaluate the integral for x = n+1:
r(n+1) = ∫((n+1)^n * e^(-(n+1)) dx)
Using the equation we derived earlier, we substitute x = n+1:
r(n+1) = -[(n+1)^n * e^(-(n+1))] + n * r(n)
Simplifying further, we have:
r(n+1) = -[(n+1)^n * e^(-n-1)] + n * r(n)
The term e^(-n-1) can be factored out:
r(n+1) = (n * r(n)) - (n+1)^n * e^(-n-1)
Since r(n+1) = (n * r(n)) - (n+1)^n * e^(-n-1), we have successfully demonstrated that r(n+1) = n*r(n) using integration by parts.