Expected value of $e^{vS}$, where $S$ is an exponential I am studying queueing theory and in particular I am dealing with priority queues with preemption.
I found this very interesting paper that treats various topics of interest.
The system is composed of two queues and only one service. One of the two queues has higher priority and preempt the other. So the competition time of a job on the lower queue is:
$$C = S + \sum_{i=0}^{N}S'(i) + \sum_{i=0}^{N}D(i)$$
where $S$ is the service time without interruption from the higher priority jobs, $D$ is the duration of the interruption and $S'$ is the portion of a potential service time duration (since after the interruption, the job begins from the begin and a new independent service time is added).
The server has a service time $S$.
The jobs with higher priority appear with a Poisson process with rate $\upsilon$
In the paper the expected value of the competition time is: 
$$
E(C) = (E(e^{\upsilon S})-1 )(E(D) + \frac{1}{\upsilon})$$
where $E(D)$ is known.
My problem is with $E(e^{\upsilon S})-1$ . In my calculations, $S$, the service time, is a random variable like $\lambda e^{-\lambda x} \frac{1}{C}$, since the jobs length is exponential and where $C$ is the server speed.
To compute the expected value I used this formula (present in my notes): given X a RV and Y = g(X), then $E[Y] = E[g(X)] = \int_{-\infty}^{\infty}g(x)f(x) dx$, so $$ \int_{0}^{\infty}e^{\upsilon x}\lambda e^{-\lambda x}\frac{1}{C} dx = \frac{1}{C}\frac{\lambda}{\lambda -\upsilon}$$since $\upsilon - \lambda < 0$. But this quantity, given fixed values for $\upsilon, \lambda$ and $C$ is less than $1$ so the final result would be negative!
Where am I wrong?
 A: I am not going to wade through the initial queueing model, so I'll just take your word for it that you want to find the expected value of $e^{vS}$ where $S$ is an exponential random variable.  This being the case, what you are looking for here is the moment generating function of an exponential random variable.  From what you say in your comments, it appears to me that you have a random variable $S \sim \text{Exp}(\lambda/C)$ with rate parameter $\lambda/C$.  (You have stated the PDF for this random variable incorrectly.)
Using the correct density function for this random variable, and applying the law of the unconscious statistician gives:
$$\begin{equation} \begin{aligned}
M_S(v) \equiv \mathbb{E}(\exp(vS))
&= \int \limits_0^\infty \exp(vs) \cdot \frac{\lambda}{C} \exp \Big( - \frac{\lambda}{C} \cdot s \Big) \ ds \\[6pt]
&= \frac{\lambda}{C} \int \limits_0^\infty \exp \Big( - \frac{\lambda-vC}{C} \cdot s \Big)  \ ds \\[6pt]
&= \begin{cases}
\frac{\lambda}{\lambda-vC} & & & \text{for } v < \lambda / C, \\
\infty & & & \text{otherwise}. \\ 
\end{cases} \\[6pt]
\end{aligned} \end{equation}$$
In your working you have made a few errors.  Firstly, you have written the density function incorrectly.  More importantly, the integral you are calculating diverges when $v \geqslant \lambda /C$, so it is not correct to assert the same formula for the MGF in this case.
