Question

Consider the famous Gamma function, defined by Γ(x) = R [infinity] 0 t x−1 e −t dt. Using mathematical induction, prove that Γ(n + 1) = n! for any n ∈ N ∪ {0}, where n! = n(n − 1)· · · 3 · 2 · 1 is the usual factorial function

Answer 1

Answer:

See proof below

Step-by-step explanation:

The inductive proof consists on the following steps:

1) Base case: for n=0, we will prove that Γ(1)=0!=1. We have that

$\gamma(1)=\int_{0}^{\infty}t^{1-1}e^{-t}dt=\int_{0}^{\infty}t^{1-1}e^{-t}dt$

$=-e^{-t}|^{\infty}_{0}=0+e^{0}=1$

Hence the base case holds.

2) Inductive step: suppose that Γ(n + 1) = n! for some natural number n. We will prove that Γ((n + 1)+1) = (n+1)!

$\gamma(n+2)=\int_{0}^{\infty}t^{n+2-1}e^{-t}dt=\int_{0}^{\infty}t^{n+1}e^{-t}dt$

Use integration by parts, with the following parts:

u=t^{n+1}, du=(n+1)t^n

dv=e^{-t}, v=-e^{-t}

$\gamma(n+2)=-e^{-t}t^{n+1}|^{\infty}_{0}+(n+1)\int_{0}^{\infty}t^{n}e^{-t}dt$

$=(n+1)\gamma(n+1)=(n+1)n!=(n+1)!$

and we used the induction hypotheses on this last line. Also, -t^n e^-t tends to zero as n tends to infiity (the exponential decays faster than any polynomial).

We have proved the statement for n+1, and by mathematical induction, the statement holds for all n.