What is this Function/Distribution? I'm trying to read this paper and they mention this "distribution". 
The paper is very, very vague. I can't
make heads or tails out of it. I think it might be a "deterministic"
distribution, but I am not certain. I also believe it's a Laplace transform 
of sorts.
The questions are:
1) Is this a correct distribution?
2) How do you integrate this?
"For the distribution of task times, with $X$ as 
the random variable representing a successful task
and $Y$ as the random variable representing the processor
failure, the probability that the job finishes before the
failure is..."
$$q = \int_{0}^{\infty} e^{-\alpha x} f(x) dx $$
where
$$f(x) = \begin{cases}
      0 & \text{everywhere}\  \\
      \infty & \text{at $x = \tau$}
    \end{cases}. $$
 A: 
1) Is this a correct distribution? 

Yes. An expression for a Dirac delta "function", such as $$f(x) = \delta(x-\tau)=\begin{cases}\infty, & x=\tau\\ 0, & \text{elsewhere.}\end{cases}$$ does not define a probability density function, but is a common abuse of notation serving to specify that the random variable $X$ has a probability distribution defined by the Dirac probability measure $\delta_\tau,$ which puts all the probability mass on the constant $\tau$; i.e., for any Borel subset $A\subseteq\mathbb{R},$ $$\delta_\tau(A)=\mathbb{1}_A(\tau)$$ and in particular $\delta_\tau(\{\tau\})=1$, so $X$ is said to be almost surely equal to the constant $\tau$ (i.e. $X\overset{\text{a.s.}}{=}\tau$). This is certainly a legitimate probability distribution, but it does not have a probability density function because the cumulative distribution function is not continous:
$$F(x)=P(X\le x)=\delta_\tau((-\infty,x]) = \begin{cases}0, &x<\tau\\1, &x\ge \tau\end{cases}$$
so the derivative of $F$ vanishes everywhere except at $x=\tau$, where it is undefined.

2) How do you integrate this?

Again, the expression $$\int_{0}^{\infty} e^{-\alpha x} f(x)\, \mathrm dx$$ is not to be taken literally, but rather as the Lebesgue integral suggested by $f(x)\,\mathrm dx\leftrightarrow \mathrm d\delta_\tau(x):$
$$\int_{0}^{\infty} e^{-\alpha x}\,\mathrm d\delta_\tau(x)=e^{-\alpha \tau},$$
where the RHS follows from the fact (proved, e.g. here) that for any $g:\mathcal{X}\rightarrow\mathbb{R}$ and $\tau\in\mathcal{X},$
$$\int_{\mathcal{X}} g(x)\,\mathrm d\delta_\tau(x)=g(\tau).$$ 
NB: A more elementary approach sometimes used (e.g. here) is to define, for any $g:\mathcal{X}\to\mathbb{R}$ with $\mathcal{X}$ a real interval, and $\tau\in\mathcal{X},$
$$\int_{\mathcal{X}} g(x)\,\delta(x-\tau)\,\mathrm dx\ ::=\ \lim_{\epsilon\to 0}\int_{\mathcal{X}} g(x)\, \delta_\epsilon(x-\tau) dx\quad\quad\text{}$$ 
where $\delta_\epsilon(x-\tau)$ is a rectangular density function of width $\epsilon$ centered on $x=\tau.$ (Note that as $\epsilon\to 0,$ the probability mass becomes more and more concentrated nearer and nearer to $x=\tau$, but $\delta_\epsilon(x-\tau)$ does not converge to a valid probability density function.) Now it can be shown (as at the above link) that if $g$ is continous, then 
$$\int_{\mathcal{X}} g(x)\,\delta(x-\tau)\,\mathrm dx\ \ =\ g(\tau),$$
giving the same result as before.

Aside: It might be worth noting that the quantity $q$ is defined as the probability that the processor-lifetime $Y$ exceeds the task-lifetime $X$,  $$q::=P(Y>X),$$
and it is an easily proved and highly intuitive theorem that for jointly distributed random variables $X$ and $Y$,
$$X\overset{\text{a.s.}}{=}\tau\ \implies\ P(Y>X)=P(Y>\tau).$$ 
Hence, we have immediately the result
$$q=P(Y>\tau)=e^{-\alpha\tau}$$
because the authors have evidently assumed that $Y$ has an exponential distribution with parameter $\alpha.$ (This assumption is the reason for the exponential term in the given integral.)
