Re: Differentiation of Line Integral

"Kaba" <none@xxxxxxxx> wrote in message
news:MPG.1f45d840e6b5203a989781@xxxxxxxxxxxxxxxxx

That is the same formula as given by Hans Lamecker. It is missing the
absolute value over the Jacobian 1/(a-b).

No, it is not missing the Jacobian. See below.

Huh? What is a determinant doing in this problem?

Well, that is how the change of variables generalizes to higher
dimensions, giving the volume of the volume-element (not sure of the
right translation) in a given point. Included out of a habit, no effect
in dimension 1.

You are confusing change of variables in an iterated
integral with change of variable in a line integral.

If you have the iterated integral I = int_{R} f(x,y) dx dy,
for some region R in the xy-plane, then you can make a
change of variables: x = g(u,v), y = h(u,v). The area
elements are related by
dx dy = |J(u,v)| du dv
where J(u,v) is the determinant of the matrix of
first-order partial derivatives of g and h. Then
I = int_{S} f(g(u,v),h(u,v)) |J(u,v)| du dv
where S is the set of (u,v) corresponding to the
points (x,y) in R.

The OP has a *line* integral:
F(a,b) = int_{0}^{1} f(b+k*(a-b)) k dk
where dk is an infinitesimal in *one* dimension. I have
stressed here that a and b are parameters, but k is the
variable of integration. You can make a change of
variables in the plane (i.e., map a and b to two other
points), but this has nothing to do with changing the
variable of integration.

For example, if you have an affine transformation T,
let a = T(a') and b = T(b'). Then
G(a',b') = F(T(a'),T(b'))
= int_{0}^{1} f(T(b')+k*(T(a')-T(b'))) k dk
= int_{0}^{1} f(T(b'+k*(a'-b'))) k dk
= int_{0}^{1} g(b'+k*(a'-b')) k dk
where g is the composition of f and T. Now you can
attempt to measure how G varies with a' or b', but
this is the same problem as before. You have to
compute the gradient of G, which requires you to
compute the gradient of g.

--
Dave Eberly
http://www.geometrictools.com


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