## Multiple logarithmic integral

Prove that

Solution

First of all we observe that the integral as well as the integral diverge whereas the proposed integral converges which is an interesting fact. Now,

since is determined by the inequalities

## Conservative field

(i) Let such that . Let be a normal area. Prove that

(ii) Examine if is a conservative field. If so, fiend a force of it.

(i) Using Green’s theorem we have that

(ii) Yes, it is. We note that

and of course a force is .

## Multiple integral on a zero measured set

Let be a Jordan measurable set of zero measure. Prove that .

Solution

Since there exists a sequence of closed rectangles of such that and it is where is the volume of the rectangle . Then foreach

## Green’s Theorem and Area of Polygons

In this post we are presenting a simple formula for the area of any simple polygon that only requires knowledge of the coordinates of each vertex. It is as follows:

(1)

where is the number of vertices, is the -th point when labelled in a counter-clockwise manner and the starting vertex is found both at the start and end of the list of vertices meaning  .

Solution

The derivation is quite simple and it makes use of Green’s theorem. Green’s Theorem states that, for a “well-behaved” curve forming the boundary of a region

(2)

Since the area of is equal to we can use Green’s Theorem to calculate area  by choosing and that satisfy

Thus letting be the area of region we have that

(3)

Now, consider the polygon below, bordered by the piecewise-smooth curve where starts at the point and ends at the “next” point along the polygon’s edge when proceeding counter-clockwise .

Since line integrals over piecewise-smooth curves are additive over length, we have that:

(4)

To compute the -th line line integral above, parametrise the segment from to . Hence

(5)

Substituting this parametrisation into the integral, we find:

Summing all of the ‘s we then find the total area:

which is the desired result.

Application: To demonstrate use of this formula, let us apply this to the shape below. A copy of the image is found below, but this one is marked with the coordinates of the vertices.

Starting with the coordinate and proceeding counter-clockwise we apply our formula:

This post is just a migration from math.overflow.com blog.

## An everywhere zero function

Let be a continuous function such that is real. If for every plane of it holds that then prove that forall .

Solution

Since is real, we can define the Fourier transform of as

If is fixed and then the planes defined as

are parallel and cover . Integrating over the planes and then over we have that

Hence forall we have that . Since our function is continuous and its Fourier transform is it follows that forall .