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
[tex]\gamma(1)=\int_{0}^{\infty}t^{1-1}e^{-t}dt=\int_{0}^{\infty}t^{1-1}e^{-t}dt[/tex]
[tex]=-e^{-t}|^{\infty}_{0}=0+e^{0}=1[/tex]
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)!
[tex]\gamma(n+2)=\int_{0}^{\infty}t^{n+2-1}e^{-t}dt=\int_{0}^{\infty}t^{n+1}e^{-t}dt[/tex]
Use integration by parts, with the following parts:
u=t^{n+1}, du=(n+1)t^n
dv=e^{-t}, v=-e^{-t}
[tex]\gamma(n+2)=-e^{-t}t^{n+1}|^{\infty}_{0}+(n+1)\int_{0}^{\infty}t^{n}e^{-t}dt[/tex]
[tex]=(n+1)\gamma(n+1)=(n+1)n!=(n+1)![/tex]
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.
