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This is styled as an economics problem, but it's very mathy so I put it here.
Wages are higher in cities. The reason is that people are more productive when they are in close proximity to other productive people. So I modeled this as $$P = P_0 + \int_A P f(r) dA $$ where $r$ is the distance to other people and $f(r)$ is a decreasing function. To make it simple I assume everyone is the same, i.e., the same $P_0$ and $f(\cdot)$, and the city is infinite so we can ignore boundary conditions.

I tried to solve this (in polar coordinates), but the problem is $f(r)$. I tried $\frac{1}{r}$ and $\frac{1}{r^2}$, but they blow up when $r\to0$. So any help would be appreciated.

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  • $\begingroup$ If $P$ is a constant across all people (and therefore with respect to $r$), then you really just have $P = P_0 + P \int_{r=0}^\infty 2 \pi r f(r) \, dr = P_0 + 2 \pi P \int_{r=0}^\infty r f(r) \, dr$. You just need that definite integral to be bounded, so you can use $1/(r^k+1)$ for some $k$. But it might be useful to prove something for any valid $f(r)$. $\endgroup$ – Brian Tung Oct 23 '20 at 16:44
  • $\begingroup$ Sorry about edit noise, a bit of brain fade, I guess. :-) $\endgroup$ – Brian Tung Oct 23 '20 at 16:46
  • $\begingroup$ Thanks. I got antiderivatives for k=1 and k=2 but they blew up at infinity. Couldn't find anything for k=3 (not sure I wanted to). So I just used a linear function that goes to zero at some radius $r_L$ and the result was an uninspiring $P = P_0/(1-bunchastuff)$. $\endgroup$ – Daniel Oct 23 '20 at 22:05
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You could move this to the economics stack exchange, but a traditional functional form would be something like $Ae^{-pr}$ to get around that problem, where $A$ and $p$ are some weights that could be 0. Also makes the integration particularly easy.

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    $\begingroup$ Thanks (can't believe I didn't think of that). I ran it through, it yielded a more 'pleasing' solution. $\endgroup$ – Daniel Oct 26 '20 at 13:33

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