Integral with polygon I've got another one weird integral
$$
\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\exp\left(-\inf\limits_{(a,b)\in P}((x-a)^2+(y-b)^2)^{1/2}\right)dxdy
$$
where $P$ is a convex polygon. Help me please, I don't know how to compute infimum.
 A: Expanding into an answer since it's a bit too long for a comment: you should be able to break the integral into several pieces; the chunk inside the polygon (which will be just the area of the polygon) a sector for each vertex of the polygon (representing the areas of the plane that are closer to that vertex than to any other or to any edge), and a semi-infinite rectangle for each edge of the polygon (representing the areas of the plane that are closer to points on that edge than to any other edge or to any vertex).  For instance, consider the square:

Here the black region is the (square) polygon itself; each of the peach regions is a sector closer to its nearest vertex than to any of the edges; and each of the blue regions is part of the rectangle closer to the nearest edge than to any other part of the polygon.
The key is that the integrals over each separate region can be calculated explicitly by something close to Marvis's method from the comments, since they can be converted into double-integrals where the 'inner' line integral is a constant independent of the outer parameter.  For instance, consider the top right peach area here (assuming that the square spans $[-1..1]^2$); we can break up the integral into a set of line integrals over the rays at angle $\theta$ from the (nearest) vertex $(1, 1)$, getting the value $\int_0^{\pi/2}\int_0^\infty r e^{-r} dr d\theta$ (the factor of $r$ here coming from the Jacobian, as he mentions).  In fact, since the total angular span of the peach areas will be $2\pi$, we can see that the full contribution from those areas will be $2\pi$ as well (just as in his comments).
The blue regions can be handled similarly; for instance, the one on the right is $\int_{-1}^1\int_1^\infty e^{-(x-1)}\ dx\ dy = \int_{-1}^1\int_0^\infty e^{-x}\ dx\ dy =\int_{-1}^1 1\ dy = 2$.  Generically, the total contribution from these regions will be equal to the perimeter of the polygon (since each one will be the integral of 1 over a length that corresponds to one side of the polygon); and assuming then that my presumption about the value of the integrand inside the polygon being equal to $1$ is correct, the contribution from that region is then just the area of the polygon, so the overall value should be $A+P+2\pi$.
(The dimensionality on this is a little bit wonky, but so is the dimensionality in the integral itself - we're exponentiating a distance, so it's inherently unit-dependent and thus it shouldn't be surprising that the answer mixes dimensional values.)
