Finding a moment-generating function of a continuous r.v. given its pdf

Hi,

I'm trying to make sense of a problem we saw in my probability class. mr. fantastic helped me dust off my calculus skills for the integration part but now I'm having trouble understanding the very last part. Here's the problem and its solution:

The probability density of a continuous random variable Y is given by

$\displaystyle f(y) = \begin{cases}\alpha{e^{-\alpha y}}&\text{ for 0} < y < \infty \text{ for some integer k},\\0&\text{otherwise},\end{cases}$

where $\displaystyle \alpha >0$

Prove that the moment generating function, m(t), of Y is given by

$\displaystyle

m(t) = \frac{\alpha}{\alpha - t}$ for $\displaystyle t<\alpha$

Solution:

$\displaystyle m(t) = E[e^{ty}] = \int^\infty_0 e^{ty}\alpha{e^{-\alpha y}} dy$

$\displaystyle = \lim_{k\to\infty}\int^k_0 {\alpha}e^{\left({t-\alpha}\right)y}dy

$

$\displaystyle

= \lim_{k\to\infty}[\frac{\alpha}{t-\alpha}e^{\left({t-\alpha}\right)y}]|(y=k,y=0)$

$\displaystyle

= \frac{\alpha}{t-\alpha}\lim_{k\to\infty}[e^{\left({t-\alpha}\right)k}-1] = \begin{cases}\frac{\alpha}{\alpha -t}&\text{ if t} < \infty ,\\\infty &\text{if t} >= \infty,\end{cases}

$

Can someone please explain

$\displaystyle

= \frac{\alpha}{t-\alpha}\lim_{k\to\infty}[e^{\left({t-\alpha}\right)k}-1] = \begin{cases}\frac{\alpha}{\alpha -t}&\text{ if t} < \infty ,\\\infty &\text{if t} >= \infty,\end{cases}

$ to me? I'm confused by the very end. If k is the one being taken to $\displaystyle \infty$ then why should the outcome depend on t? Please explain how we get $\displaystyle \frac{\alpha}{\alpha -t}$ and $\displaystyle \infty$.

Thanks