If you don't mind writing out steps I'm lost :) 5. The gamma function, which plays an important role in advanced applications, is definedfor n ≥ 1 byT(n) =(a) Show that the integral defining I(n) converges for n ≥ 1.(Hint: Show that t-le-t

Given, The gamma function for n≥1 by Γ(n) = ∫0∞ tn-1 e-t dt.(a) Γ(n) = ∫0∞ tn-1 e-t dt Let, consider f(t) = tn+1 and g(t) = et Taking logaritham,logf(t) = (n+1) logt , logg(t) = t·logeWe already know that,nlogt < tSo, (n+1) logt < t·logelogf(t) < logg(t)Taking antilog both the side,f(t) < g(t) (∴ t>0)tn+1 < etMultiply by t-2 both the side,tn+1·t-2< et·t-2tn-1 < et·t-2tn-1 < t-2e-ttn-1·e-t < t-2So, if t→∞ then, t-2 becomes 0.Hence, It is proved that the integral defining Γ(n) converges for n≥1.(b) Given, Γ(n) = ∫0∞ tn-1 e-t dtLet, n = n+1Γ(n+1) = ∫0∞ tn·e-t dtTaking integration by parts,Γ(n+1) = tn∫0∞e-t dt - ∫0∞ ddt( tn) ∫e-t dt dt = -tne-t0∞ - ∫0∞ ntn-1×(-e-t) dt = -tn0 +n∫0∞ tn-1e-t dt = 0 +n∫0∞ tn-1e-t dt = n·Γ(n) (From given, ∫0∞ tn-1e-t dt = Γ(n) )Hence, It is proved that Γ(n+1) = n·Γ(n)(c) Γ(n+1) = n·Γ(n)n·Γ(n) = nΓ(n-1)*Γ(n-2)*Γ(n-2).........Γ(1) = n!Now,Γ(0) = ∫0∞e-t dtΓ(0) = -e-t0∞Γ(0) = -0+1Γ(0) = 1Hence, It is proved that Γ(n+1) = n! ; if n≥1 is an integer.