Why doesn't integrating infinitesimally small likelihoods work in this sense? We know that something is going to happen after $x$ amount of time, but the exact time at which the event occurs is random within $x$ time. (Like, say we did it a bunch of times where it happened in $x-y$ time, and all the $y$s were uniformly distributed reals $0$ to $x$.)  
It is easy to see that you can compute the likelihood that the event has occurred by using $t/x$ with $t$ being the time that has elapsed and $x$ being the total amount of time.
It's also easy to see the likelihood of it occurring in the next $n$ amount of time by computing $n/(x-t)$ (With $t$ again being the amount of time elapsed)
As $n$ gets very small, the likelihood of it happening in the next $n$ amount of time also clearly also becomes very small. This can be expressed by realizing that:
$$\lim_{n \to 0}\frac{n}{x-t}=0$$
But now, if you take that limit and integrate it over some period of $t$ s.t. $0<t<x$, you should get $t/x$ again, since we're essentially adding up all the likelihoods of it happening at each instant in the interval:
$$\int_{0}^{x-a} \lim_{n \to 0} \frac{n}{x-t} dt$$ 
Resolving the limit first clearly doesn't yield. Does that mean that the integral I have constructed does not represent what I think it does?
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I considered using arclength in the integral but stopped myself in realizing that we are concerned with the values the function outputs, not the length of the curve.
 A: There are two problems:

*

*First, you didn't try to integrate the likelihood over an interval of infinitesimally small length— you tried to integrate the likelihood over an interval of zero length


*Second, the probability you tried to compute was "The odds of it happening in $[t, t+n)$ conditioned on the assumption it did not happen in $[0, t)$", but what the integral actually needs is "The odds of it happening in $[t, t+n)$".
Setting it up via non-standard analysis, let $H$ be transfinite and $\epsilon = (x-a)/H$. The odds of the event happening in $[t, t + \epsilon)$ is $\epsilon / x$. The integral you were trying to set up is then approximated by the sum
$$ \sum_{i=0}^{H-1} \frac{\epsilon}{x} \approx \int_{0}^{x-a} \frac{dt}{x} = \frac{x-a}{x}$$
In some cases, it's easier to just use standard means, though: if the probability of the event happening in $[0, t)$ is $t/x$, then the differential form is just
$$ d \left(\frac{t}{x} \right) = \frac{dt}{x} $$
(no $dx$ term because $x$ is a constant) and so the corresponding probability density is $\frac{1}{x}$.
A: The integral you have:
$$
\int_{0}^{x-a}\lim_{n \to 0}\frac{n}{x-t} dt \ \ \ \ \ \ \ \  \ \ \ \  (1)
$$
is very similar to doing the following.
Let $f \text{ and } g$ be two integrable functions $\ni f,\ g: \mathbb{R} \rightarrow \mathbb{R},\ x \mapsto y$. Then:
$$
\int_a^b f(x) + g(x)\ dx = \int_a^b f(x)\ dx + \int_a^b g(x)\ dx \ \ \ \ \ \  \ \ \ \ (2)
$$
So, you're simply figuring out what to integrate before doing it.
A: First notice that 
$$\int_{0}^{x-a} \lim_{n \to 0} \frac{n}{(x-t)} dt$$
is not dimensionally correct.
It has the dimensions of time. 
There is also a confusion about what is $t$. 
You are confounding probability and the probability density function (pdf).
If the event didn't happen up to time $t_0$, the pdf is $1/(x-t_0)$, not $n/(x-t_0)$. 
If we want to use this pdf to find the probability of the event occurring between time $0$ and some other time we must use the pdf $1/(x-t_0)$ with $t_0=0$, since $t_0$ here is the time before which the event had not occurred.
Thus, the pdf is $1/x$ as expected and the probability is $t/x$. 
