I was looking to show the result for the MGF of the Erlang Distribution, and got stuck whilst computing the integral:
$$\\ \int e^{-st} \frac{\lambda^k t^{k-1}e^{-\lambda t}}{(k-1)!} dt$$
I know the result should evaluate to
$$\\ \left( \frac{\lambda}{\lambda+s}\right)^k $$ but I am unsure as to how to go about tackling the initial integral. I initially started by integrating by parts, but whilst the (k-1)! term appeared nicely I didn't see how the other terms would cancel/rearrange as required. Any help would be appreciated. Thanks.
Best Answer
Let $I_k$ be the integral in question, that is, $$ I_k = \int_0^\infty e^{-st} \frac{\lambda^k t^{k-1} e^{-\lambda t}}{(k-1)!} \, dt $$ and use induction on $k$: For $k=1$, we have $$ I_1 = \lambda \int_0^\infty e^{-(s+\lambda)t} \, dt = \lambda \left[-\frac{e^{-(s+\lambda) t}}{s+\lambda}\right]_0^\infty = \frac{\lambda}{s+\lambda} $$ assuming that $s+\lambda>0$. Then, using integration by parts with $u = t^k$ and $dv = e^{-(s+\lambda) t} dt$, we obtain \begin{align} I_{k+1} &= \frac{\lambda^{k+1}}{k!} \int_0^\infty t^k e^{-(s+\lambda)t} \, dt \\ &= \frac{\lambda^{k+1}}{k!} \left[ \left. t^{k-1} e^{-(s+\lambda) t} \right|_0^\infty + \frac{k}{s+\lambda} \int_0^\infty t^{k-1} e^{-(s+\lambda)t} \, dt \right] \\ &= \frac{\lambda}{s+\lambda} \int_0^\infty e^{-st} \frac{\lambda^k t^{k-1} e^{-\lambda t}}{(k-1)!} \, dt \\ &= \frac{\lambda}{s+\lambda} I_k \end{align} which completes the inductive step.