# 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.

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Forget the $\inf$ for a moment. Lets compute it for any $(a,b)$.$$I(a,b) = \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\exp\left(-((x-a)^2+‌​(y-b)^2)^{1/2}\right)dxdy$$Let $x = a + r\cos(\theta)$ and $y = b + r \sin(\theta)$. This gives us$$I(a,b) = \int_{0}^{2 \pi} \int_{0}^{\infty} \exp(-r)rdr d \theta = 2 \pi \int_0^{\infty} r \exp(-r) dr =2 \pi$$My hunch is that the final answer should also be $2 \pi$ as well. The integrand far away from the polygon doesn't get affected much and nearby there should be some nice cancellations occurring to give a value independent of polygon. – user17762 Jun 27 '12 at 1:28
Presumably in the region where $(x, y)$ is inside the polygon the infimum would be zero and the exponential thus one? If so this means that Marvis's hunch can't be right, since just taking any polygon with an area greater than $2\pi$ will give a value larger than that... – Steven Stadnicki Jun 27 '12 at 1:45
@StevenStadnicki By $P$, I thought $P$ is the set of vertices of the polygon in which case my hunch seems to be reasonable. – user17762 Jun 27 '12 at 1:52
@Marvis I think if P is just the set of vertices then the integral has to be larger than $2\pi$, because then the external 'sectors' (see my answer for what I mean by this) would contribute a $2\pi$ factor by themselves, with the internal regions contributing even more. I think the problem may not have an explicit solution if you use just the vertices, but it has a clean answer if we take the distance-to-polygon as a whole. – Steven Stadnicki Jun 27 '12 at 2:14

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.)

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+1. Beautiful! Wish I could up-vote it multiple times! Wow! this is indeed very very nice! – user17762 Jun 27 '12 at 2:41
@Marvis If I could give you half the points, I would - I knew how to break the integral down into regions, but without your comment I wouldn't have remembered the explicit evaluation on the angular arcs (and piggybacked that onto the straight segments). Many thanks! – Steven Stadnicki Jun 27 '12 at 3:29
A poor man's generalization of this integral to $n$-dimensional case where $P$ is a convex polyhedron would be "Volume of the $n$ dimensional convex polyhedron + Surface area of the $n$ dimensional convex polyhedron + $\Gamma(n) \times$Surface area of the $n$-dimensional unit ball". – user17762 Jun 27 '12 at 3:55
@Marvis I don't know if that actually works - not only do you have the $n$-volume and the $(n-1)$-faces to go with the $0$-vertices (the latter contributing your last term), you also have the various intermediate-dimensional facets of the skeleton, and those seem to contribute in non-obvious ways in higher dimensions... – Steven Stadnicki Jun 27 '12 at 4:07
This very original. Thanks for your solution! – user34574 Jun 27 '12 at 17:10