Work along an oriented field

Prove that the work

\displaystyle \mathcal{W}=- \oint \limits_{\gamma} \frac{(x, y, z)}{\left ( x^2+y^2+z^2 \right )^{3/2}} \cdot \, {\rm d}(x, y, z)

produced along a \mathcal{C}^1 oriented curve \gamma of \mathbb{R}^3 \setminus \{(0, 0, 0) \} depends only on the distances of starting and ending point of \gamma about the origin.

Solution

Let us consider the function  f:\mathbb{R}^3 \setminus \{0 \} \rightarrow \mathbb{R} such that

\displaystyle f\left ( x,y,z \right ) = - \frac{1}{\sqrt{x^2+y^2+z^2}}

which is continuously differentiable in \mathbb{R}^3 \setminus \{0\} and it holds that:

\displaystyle {\rm grad }f\left ( x, y, z \right ) = \frac{(x, y, z)}{\left ( x^2+y^2+z^2 \right )^{3/2}}

Hence the vector field \displaystyle \frac{(x,y,z)}{( x^2+y^2+z^2)^{3/2}} is a conservative one. This , in return,  means that \mathcal{W} is actually independent of the road we choose , meaning that it only depends on distances of starting and ending point of \gamma about the origin.

This was an exam’s question somewhere in Greece. The answer to this question was migrated from the Greek team of mathimatikoi.org forum.

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Curves and line integrals

Let \gamma be defined as

\gamma(t) = e^{-t} (\cos t, \sin t )\quad , \; t \geq 0

(i) Sketch the graph of \gamma.

(ii) Evaluate the line integrals:

\begin{matrix} & \displaystyle ({\rm i})\; \oint \limits_{\gamma}\left ( x^2 +y^2 \right )\, {\rm d}s & & ({\rm ii}) \displaystyle \oint \limits_{\gamma} (-y, x)\cdot {\rm d}(x, y) \end{matrix}

Solution

(i) The graph of the curve \gamma  is depicted below:

Rendered by QuickLaTeX.com

(ii)

   (i) For the first integral we have successively:

    \begin{align*} \oint \limits_{\gamma} \left(x^2+y^2\right)\,\mathrm{d}s&=\int_{0}^{\infty}\left(x^2(t)+y^2(t)\right)\,\||\gamma^\prime(t)||\,\mathrm{d}t\\ &=\int_{0}^{\infty}e^{-2\,t}\,\sqrt{2}\,e^{-t}\,\mathrm{d}t\\ &=\sqrt{2}\,\int_{0}^{\infty}e^{-3\,t}\,\mathrm{d}t\\ &=\left[-\dfrac{\sqrt{2}}{3}\,e^{-3\,t}\right]_{0}^{\infty}\\ &=\dfrac{\sqrt{2}}{3} \end{align*}

 (ii) For the second integral we have successively:

    \begin{align*} \oint \limits_{\gamma} \left(-y,x\right)\cdot d(x,y)&=\int_{0}^{\infty}\left(-y(t),x(t)\right)\cdot \gamma^\prime(t)\,\mathrm{d}t\\ &=\int_{0}^{\infty}-e^{-t}\cdot (-e^{-t})\,\mathrm{d}t\\ &=\int_{0}^{\infty}e^{-2\,t}\,\mathrm{d}t\\ &=\left[-\dfrac{e^{-2\,t}}{2}\right]_{0}^{\infty}\\ &=\dfrac{1}{2} \end{align*}

This exercise was an exam’s question somewhere in Greece.

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A dilogarithm and log Γ integral

Let {\rm Li}_2 denote the dilogarithm function. Prove that

\displaystyle \zeta(3) = 2 \bigintsss_{0}^{1} \bigg( {\rm Li}_2 \left ( e^{-2 \pi i x} \right ) + {\rm Li}_2 \left ( e^{2\pi i x} \right ) \bigg) \log \Gamma (x) \, {\rm d}x

Solution

Take a loot at the JoM Forum.
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Not Lebesgue integrable function

Let  x \in \mathbb{R} . Given the series

(1)   \begin{equation*} \sum_{n=2}^{\infty} \frac{\sin nx}{\ln n} \end{equation*}

(i) Prove that (1) converges forall  x \in \mathbb{R} .

(ii) Prove that (1) is not a Fourier series of a Lebesgue integrable function.

(i) Let \kappa \in \mathbb{Z}.We note that for x=\kappa \pi the series , trivially, converges since \sin \kappa \pi =0 . For all other x we are using Dirichlet’s test. It is well known that  \sum \limits_{m=1}^{n} \sin mx is bounded and we also note that \frac{1}{\ln n} \searrow 0. The result follows.

(ii) Suppose that such f exists. Integrating (we can do that since Fourier series is integrated term by term) we take a continuous and of bounded variance function. The Fourier series of this function should converge at 0. This is not the case here since \displaystyle \sum \limits_{n=2}^{\infty} \frac{1}{n \ln n} is known to diverge. Thus, such function does not exist.

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Not a uniformly convergent polynomial sequence

Prove that there does not exist a sequence \{ p_n(z)\}_{n \in \mathbb{N}} of complex polynomials such that p_n(z) \rightarrow \frac{1}{z} uniformly on \mathcal{C}_R=\{ z \in \mathbb{C} \mid \left| z \right| = R\} .

Solution

 If such sequence existed then the convergence on the compact set  \left| z \right| = r would be uniform.

However,

    \begin{align*} 2 \pi i &= \oint \limits_{\mathcal{C}_R} \frac{{\rm d}z}{z} \\ &= \oint \limits_{\mathcal{C}_R} \lim_{n \rightarrow +\infty} p_n (z) \, {\rm d}z \\ &= \lim_{n \rightarrow +\infty} \oint \limits_{\mathcal{C}_R} p_n (z) \, {\rm d}z \\ &= \lim_{n \rightarrow +\infty} 0 = 0 \end{align*}

which is an obscurity.

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