Expected lifespan implied by the Lindy Effect Sorry in advance, the maths in this one is pretty basic but I'm fairly rusty at this point so I just wanted to check my reasoning.
I came to the idea of the Lindy effect via Nassim Nicholas Taleb and I found a good write up of how J. Richard Gott arrived at a technique for estimating the future lifespan of pretty much anything here (https://fs.blog/2012/06/copernican-principle/).
Suppose you observe something that has been around N years and you assume that you are observing that thing at a random point in its' life then there is a 50% chance that you are observing it somewhere between 75% and 25% of its' lifespan. Therefore there is a 50% chance that it will continue to exist between N/3 and 3N years.
Using a similar argument I can say there is a

*

*25% chance it will exist shorter than another $N/3$ years

*50% chance it will exist shorter than another $N$ years

*66.66..% chance it will exist shorter than another $2N$ years

*75% chance it will exist shorter than another $3N$ years

..etc.
Now, supposing I want to find the expected future lifespan given this approach. Would I be correct in saying that it would be
$$\int_0^1\frac{Nx}{1 - x} \, dx = N\int_0^1\left(\frac{1}{1-x} - 1\right) \, dx$$
$$=N\Big[-\ln(1-x)-x\Big]_0^1$$
$$=-N(\ln(0) -1 + \ln(1) + 0)$$
$$=\infty$$
If so, I think I'd find it a little disappointing. Gotts' argument is so elegant I feel like it should produce something better than everything has the same expected lifespan no matter how long it has existed.
EDIT:
I've had a go with doing this with a proper probability density function
$$p(x) = \frac{N}{(N+x)^2}$$
Now we have
$$\int_0^\infty p(x)\,dx = \int_0^\infty \frac{N}{(N+x)^2}\,dx$$
$$ = \left[ \frac{-N}{N+x} \right]_0^\infty$$
$$ = 1$$
and furthermore
$$\int_\frac{N}{3}^{3N} p(x)\,dx = \left[ \frac{-N}{N+x} \right]_\frac{N}{3}^{3N}$$
$$=\frac{-N}{N+3N}-\frac{-N}{N+\frac{N}{3}}$$
$$=\frac{1}{2}$$
so it fits with Gott's observation
This gives us
$$\int_0^\infty xp(x)\,dx = \int_0^\infty \frac{Nx}{(N+x)^2}\,dx$$
$$=\int_0^\infty \frac{Nx+N^2}{(N+x)^2}-\frac{N^2}{(N+x)^2}\,dx$$
$$=\int_0^\infty \frac{N}{N+x}-\frac{N^2}{(N+x)^2}\,dx$$
$$ = \left[ N\ln(N+x)+\frac{N^2}{N+x} \right]_0^\infty$$
$$=\infty$$
 A: As you’ve noticed, Gott’s model with the assumption that you are equally likely to observe an event over any point in it’s lifetime and that you have an uninformed uniform prior (whatever that means...) has a fat tail, so it doesn’t ever give a finite expected value.
It’s more useful and typical to think of the median distribution which Gott clearly predicts is $2N$  You could also try transforming it by taking the $\log$ and taking the mean of that.
A: Let $T$ be the unknown lifespan and $X$ the present age of the phenomenon being observed. Gott's distribution$^\dagger$ for the future duration ($T-X$), given $X=x$, is then
$$\begin{align}P(T-X \le rX\mid X=x)={r\over r+1}\,[r>0]\tag{1}\\
\end{align}$$
which can be rewritten as
$$\begin{align}P(T \le t\mid X=x)=(1-{x\over t})\,[t>x].\tag{2}
\end{align}$$
(Note that (2) shows $(x/T\mid X=x)\sim\text{Uniform}(0,1)$.) From (2) we have immediately
$$E(T\mid X=x) = \int_0^\infty P(T>t\mid X=x)\,dt = \infty
$$
(or we could obtain the same result by differentiating (2) to obtain the pdf and using that to find the expectation).  Consequently, we have also $E(T-X\mid X=x)=E(T\mid X=x) - x=\infty$.
A third alternative is the OP's recent edit, which is verified by noting that from (1) we have $$P(T-X\le y\mid X=x)={y\over y+x}[y>0]$$
so by differentiating we obtain the pdf used by the OP:
$$p_{T-X\mid X}(y\mid x)={x\over (x+y)^2}(y>0)$$
which yields $$E(T-X\mid X=x)=\int_0^\infty{x\over (x+y)^2}dy=\infty. $$

Although these distributions have no finite expected value, they are easily- and well-described using quantiles: For any r.v. $Z$, let $Z_p$ denote the value such that $P(Z\le Z_p)=p$. Then from (1) and (2) we have
$$(T\mid X=x)_p = {1\over 1-p}\,x$$
$$(T-X\mid X=x)_p ={p\over 1-p}\,x$$
E.g., the medians are $$(T\mid X=x)_{1/2} =2\,x$$ $$(T-X\mid X=x)_{1/2} =x$$ and $95$% posterior probability intervals are
$$\big((T\mid X=x)_{.025},\ \ \ (T\mid X=x)_{.975}\big) = \left({40\over 39}\,x,\ {40}\,x\right)$$
$$\big((T-X\mid X=x)_{.025},\ \ \ (T-X\mid X=x)_{.975}\big)=\left({1\over 39}\,x,\ {39}\,x\right).\tag{3}$$

$^\dagger $ This describes Gott's distributions in Bayesian terms, but he doesn't use those terms himself. In his model, $X$ is the only "random" quantity, $T$ being an unknown constant, and he does not explicitly use conditional probabilities. His version of (1) is, before observing $X=x$:
$$P(T-X\le rX) = {r\over r+1}\,[r>0]$$
which is a correct equation for the unconditional distribution of $X\sim\text{Uniform}(0,T)$). However, he asserts that after observing $X=x$ the same equation continues to hold, with $x$ replacing $X$ (which effectively becomes a posterior distribution for a certain prior on $T$, i.e. my equation (1)):
$$P(T-x\le rx) = {r\over r+1}\,[r>0]$$
He proceeds to use this latter equation to make inferences about the unknown constant $T$ -- precisely the inferences (e.g. (3) above) that one would make if the latter equation were regarded as a posterior distribution for $T$ conditional on the observation $X=x$.
