Image integration along a pathline

There are some ways, outlined in posts that I linked to in a comment, to get pixels along a line in a given image, notably Brenseham's algorithm.

Since you are looking to analyze a rectangular segment around a line I think, however, that rotating the image may be the way to go.

Consider two points that form a line as follows:

img = Binarize@Import["https://i.sstatic.net/K8u7A.png"]; {p0, p1} = {{403, 528}, {613, 248}}; HighlightImage[ img, Graphics[{ Yellow, Thickness[0.002], Dynamic@Line[{p0, p1}] }] ] 

Now, take the angle between the line and the horizontal axis and rotate the image by that:

alpha = VectorAngle[{1, 0}, p1 - p0]; center = ImageDimensions[img]/2.; rotatedP0 = center + RotationMatrix[alpha].(p0 - center); HighlightImage[ ImageRotate[img, alpha, Full], Graphics[{ Yellow, Thickness[0.002], Line[{ rotatedP0, rotatedP0 + RotationMatrix[alpha].(p1 - p0) }] }] ] 

It is possible to extract the desired rectangle from the new image by simply using ImageTake.

The complicated looking arithmetic in the answer deals with the fact that ImageRotate rotates the image about the center of the image. We need to account for that to properly rotate the chosen start and end points in the corresponding way.

You could use a Karhunen Loeve Decomposition to get the rotation matrix like in C.E's answer:

img = Import["https://i.sstatic.net/K8u7A.png"]; pts = ImageValuePositions[img, 1, 0.99]; kld = KarhunenLoeveDecomposition[Transpose[pts], Standardized -> True]; (* kld[[2]] contains the orthonormal rotation matrix *) With[{vecs = kld[[2]]}, With[{m = Mean[pts], v1 = vecs[[1]], v2 = vecs[[2]]}, ListPlot[pts, PlotRange -> ({0, #} & /@ ImageDimensions[img]), Epilog -> {Red, Arrow[{m, m + 200*v1}], Arrow[{m, m + 100*v2}]}, PlotStyle -> {White, PointSize[Tiny]}, Background -> Black ]]] proj = Transpose@First@kld; ListPlot[proj] resampled = TimeSeriesResample[SortBy[proj, First]][[All, 2]]; ListLinePlot[MovingMap[MeanDeviation, resampled, 30], PlotRange -> All] 

We first get the picture and change it into an array of 0 and 1's.:

pic = Binarize@Import["https://i.sstatic.net/K8u7A.png"]; dat = ImageData[pic]; 

We then approximately determine by inspection point 1 and 2 by:

{p1, p2} = {{120, 403}, {400, 625}}; 

Of course you may play with this and determine more accurate points.

To make the arithmetic easier, we define new coordinates where p1 is at {0,0} and p2-p1 along the x-axis. The transformation from old to new and back is done by:

newcoord[p_] := RotationMatrix[{p2 - p1, {Max[p2 - p1], 0}}].(p - p1); oldcoord[p_] := p1 + RotationMatrix[{{Max[p2 - p1], 0}, p2 - p1}].(p); 

And the new coordinates of p2:

newp2x = newcoord[p2][[1]]; 

Now we can get the intensities along the line from p1 to p2 by the points along the new x axis:

intensities = dat[[##]] & @@@ Round[oldcoord /@ Table[{i, 0}, {i, 0, newp2x + 0.5}]]; ListLinePlot[intensities] 

To get the sum of pixels in an interval perpendicular to p2-p1 of length sigma we may add up pixels along the new y axis:

sigma = 5; bins = Total /@ Table[dat[[##]] & @@@ (Round[ oldcoord /@ Table[{i, j}, {j, -sigma, sigma}]]), {i, 0, newp2x + 0.5}]; ListLinePlot[bins]