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Multiple logarithmic integral

Prove that

    \[\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{\mathrm{d}(x, y,z)}{ \ln x + \ln y + \ln z} = - \frac{1}{2}\]

Solution

First of all we observe that the integral \bigintsss_0^1 \frac{\mathrm{d}x}{\ln x} as well as the integral \bigintsss_0^1 \bigintsss_0^1 \frac{\mathrm{d}(x, y)}{\ln x + \ln y} diverge whereas the proposed integral converges which is an interesting fact. Now,

    \begin{align*} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{\mathrm{d}(x, y, z)}{\ln x + \ln y + \ln z} &= \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{\mathrm{d}(x, y, z)}{\ln xyz} \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{u=xyz \; , \; v=y \; , \; w=z}{=\! =\! =\! =\! =\! =\! =\! =\! =\!=\!=\!} \iiint \limits_{\mathbb{D}} \frac{\mathrm{d}(v, w, u)}{v w \ln u} \\ &=\int_{0}^{1} \int_{u}^{1} \int_{u/w}^{1} \frac{\mathrm{d}(v, w, u)}{v w \ln u} \\ &=-\frac{1}{2} \end{align*}

since \mathbb{D} is determined by the inequalities

    \[\frac{u}{w}\leq v\leq 1 \quad , \quad u\leq w\leq 1\quad ,\quad 0\leq u\leq 1\]

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Conservative field

(i) Let f \in \mathcal{C}^2(\mathbb{R}) such that {\rm div(grad \;f)}=0 . Let \mathbb{D} \subseteq \mathbb{R}^2 be a \mathcal{C}^1 normal area. Prove that

    \[\oint \limits_{\partial \mathbb{D}} \left ( \frac{\partial f}{\partial y} , -\frac{\partial f}{\partial x} \right ) \cdot {\rm d}(x, y) =0\]

(ii) Examine if \bar{f}(x, y)=(2x \cos y , -x^2 \sin y) is a conservative field. If so, fiend a force of it.

Solution

(i) Using Green’s theorem we have that

    \begin{align*} \oint \limits_{\partial \mathbb{D}} \left ( \frac{\partial f}{\partial y} , -\frac{\partial f}{\partial x} \right ) \cdot {\rm d}(x, y) &= \iint \limits_{\mathbb{D}} \left ( -\frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2} \right )\, {\rm d}(x, y) \\ &= \iint \limits_{\mathbb{D}} \left ( -\nabla^2 f \right ) \, {\rm d}(x, y)\\ &= 0 \end{align*}

(ii) Yes, it is. We note that

    \[\nabla \left ( x^2 \cos y \right ) = 2x\cos y - x^2 \sin y\]

and of course a force is g(x, y)=x^2\cos y.

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Multiple integral on a zero measured set

Let A \subseteq \mathbb{R}^n be a Jordan measurable set of zero measure. Prove that \bigintsss \limits_{A} 1 \, {\rm d} \bar{x}= 0.

Solution

Since \mu(A)=0 there exists a sequence U_n of closed rectangles of \mathbb{R}^n such that A \subseteq \bigcup \limits_{n} U_n and \forall \epsilon>0 it is \sum \limits_{n} \mathcal{V} \left( U_n \right)< \epsilon where \mathcal{V}(U_n) is the volume of the rectangle U_n. Then foreach \epsilon>0

    \begin{align*} 0 &\leq \int \limits_{A} 1 \, {\rm d} \bar{x} \\ &\leq \int \limits_{\bigcup \limits_{n} U_n} 1 \, {\rm d} \bar{x} \\ &\leq \sum_{n} \int \limits_{U_n} 1\, {\rm d}\bar{x} \\ &= \sum_{n} \mathcal{V} \left ( U_n \right )\\ &< \epsilon \end{align*}

 

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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)   \begin{equation*} \mathcal{A} = \sum_{k=0}^{n} \frac{(x_{k+1} + x_k)(y_{k+1}-y_{k})}{2} \end{equation*}

where n is the number of vertices, (x_k, y_k) is the k-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  (x_{n+1}, y_{n+1}) = (x_0, y_0).

Solution

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

(2)   \begin{equation*} \oint \limits_\mathcal{C} P(x, y)\; {\rm d}x + Q(x, y)\; {\rm d} y = \iint \limits_{\mathbb{D}} \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\; {\rm d} A \end{equation*}

Since the area of \mathbb{D} is equal to \iint \limits_{\mathbb{D}} \; {\rm d}A we can use Green’s Theorem to calculate area  by choosing P(x, y) =0 and Q(x, y) = x that satisfy

    \[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}=1\]

Thus letting \mathcal{A} be the area of region \mathbb{D} we have that

(3)   \begin{equation*} \mathcal{A} = \oint \limits_{\mathcal{C}} x \, {\rm d}y \end{equation*}

Now, consider the polygon below, bordered by the piecewise-smooth curve C = C_0\cup C_1\cup\cdots\cup C_{n-1}\cup C_n where C_k starts at the point (x_k, y_k) 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)   \begin{equation*} \mathcal{A} = \oint \limits_{\mathcal{C}} x {\rm d} y  = \int \limits_{C_0} x\; {\rm d} y + \cdots + \int \limits_{C_n} x\; {\rm d} y \end{equation*}

To compute the k-th line line integral above, parametrise the segment from (x_k, y_k) to (x_{k+1} , y_{k+1}). Hence

(5)   \begin{equation*} C_k: \bigg((x_{k+1} - x_k)t + x_k,\; (y_{k+1} - y_k)t + y_k \bigg) \; , \;  0\leq t\leq 1 \end{equation*}

Substituting this parametrisation into the integral, we find:

    \begin{align*} \int_{C_k} x\; {\rm d} y &=\int_0^1 \left((x_{k+1} - x_k)t + x_k\right)\left(y_{k+1} - y_k\right) \; {\rm d}t \\ &= \frac{(x_{k+1} + x_k)(y_{k+1}-y_{k})}{2} \end{align*}

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

    \[\mathcal{A} =\sum_{k=0}^{n} \frac{(x_{k+1} + x_k)(y_{k+1}-y_{k})}{2}\]

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 (0, 0) and proceeding counter-clockwise we apply our formula:

\begin{aligned} \mathcal{A} &= \sum_{k=0}^{n} \frac{(x_{k+1} + x_k)(y_{k+1}-y_{k})}{2} \\ &= \frac{(1.5 + 0)(0 - 0)}{2} + \frac{(2.5 + 1.5)(-1 - 0)}{2} + \frac{(3.5 + 2.5)(0 - (-1))}{2} + \frac{(5+3.5)(0-0)}{2} \\ &\quad + \frac{(5+5)(2-0)}{2} + \frac{(3.5+5)(2-2)}{2} + \frac{(2.5+3.5)(3-2)}{2}\\ &\quad +\frac{(1.5+2.5)(2-3)}{2} + \frac{(0+1.5)(2-2)}{2} + \frac{(0+0)(0-2)}{2}\\ &= 0 + (-2) + 3 + 0 + 10 + 0 + 3 + (-2) + 0 + 0\\ &= 12 \end{aligned}

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

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An everywhere zero function

Let f:\mathbb{R}^3 \rightarrow \mathbb{R} be a continuous function such that \bigintsss_{\mathbb{R}^3} \left| f(x) \right| \, {\rm d}x is real. If for every plane \mathcal{P} of \mathbb{R}^3 it holds that \bigintsss_{\mathcal{P}} f(x) \, {\rm d}s =0 then prove that f(x)=0 forall x \in \mathbb{R}^3.

Solution

Since \bigintsss_{\mathbb{R}^3} \left| f(x) \right| \, {\rm d}x is real, we can define the Fourier transform of f as

    \[\hat{f}(a) = \int \limits_{\mathbb{R}^3} f(x) e^{-2\pi i ax} \, {\rm d}x\]

If a is fixed and t \in \mathbb{R} then the planes defined as

    \[\mathcal{P} =\left \{ x \in \mathbb{R}^3 \bigg| ax = t \right \}\]

are parallel and cover \mathbb{R}^3. Integrating over the planes and then over t we have that

    \begin{align*} \hat{f}(a) = \int \limits_{\mathbb{R}^3} f(x) e^{-2\pi i ax} \, {\rm d}x &= \int \limits_{\mathbb{R}} \int \limits_{xa =t} f(x) e^{-2\pi i t} \, {\rm d} \sigma {\rm d}t \\ &= \int \limits_{\mathbb{R}} e^{-2 \pi i t } \int \limits_{ax =t} f(x) \, {\rm d} \sigma {\rm d}t \\ &= 0 \end{align*}

Hence forall a \in \mathbb{R}^3 we have that \hat{f}(a)=0. Since our function is continuous and its Fourier transform is 0 it follows that f=0 forall x \in \mathbb{R}^3.

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