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Harmonic function

Consider the points O(0, 0) and  A(1, 0). Let \Gamma(x, y) be a point of the plane such that y>0. Set \varphi(x, y) to be the angle that is defined by O\Gamma and A \Gamma. ( the one that is less than \pi.) Prove that the function \varphi(x, y) is harmonic.

Solution [by Demetres Skouteris]

The complex function \log \left( 1 - \frac{1}{z} \right) has a holomorphic branch in the half plane \mathfrak{Im}(z)>0 and its imaginary part is the desired angle. Hence, the function of the angle is harmonic.

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A convergent series

Let f be holomorphic on the open unit disk \mathbb{D} and suppose that the integral \displaystyle \iint \limits_{\mathbb{D}} \left| f(z) \right|^2 \, {\rm d}(x, y) converges. If the Taylor expansion of f is of the form \sum \limits_{n=0}^{\infty} a_n z^n then prove that the sum

\displaystyle \mathcal{S}= \sum_{n=0}^{\infty} \frac{|a_n|^2}{n+1}

converges.

Solution

We evaluate the integral using the standard orthogonality results for e^{in \theta}. Thus:

\begin{aligned} \iint \limits_{\mathbb{D}}|f(z)|^2 \, {\rm d}z&=\int_0^1 r\int_0^{2\pi}\left|\sum_{n=0}^{\infty}a_nr^ne^{in\theta}\right|^2\,{\rm d}\theta\,{\rm d}r\\ &=\int_0^1r\int_0^{2\pi}\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_n\overline{a_m}r^{n+m}e^{i(n-m)\theta}\,{\rm d}\theta\,{\rm d}r\\ &=2\pi\int_0^1r\sum_{n=0}^{\infty}|a_n|^2r^{2n}\,{\rm d}r\\ &=\pi\sum_{n=0}^{\infty}\frac{|a_n|^2}{n+1} \end{aligned}

and thus the series converges.

Note: The set of functions satisfying this is a Hilbert space of functions, but it is not the same as the Hardy space \mathbb{H}^2.

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An integral with trigonometric and rational function

Evaluate the integral

\displaystyle \int_0^{\infty} \frac{x^2-4}{x^2+4} \frac{\sin 2x}{x} \, {\rm d}x

Solution

Firstly , we begin by noticing the following

 \displaystyle \int_{0}^{\infty} \frac{(x^2-4) \sin 2x}{\left ( x^2+4 \right )x} \, {\rm d}x = \frac{1}{2} \int_{-\infty}^{\infty} \frac{(x^2-4) \sin 2x}{\left ( x^2+4 \right )x} \, {\rm d}x \quad (1)

Let us now consider the function \displaystyle f(z)=\frac{(z^2-4) e^{2iz}}{z(z^2+4)} as well as the contour

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The function has three simple poles of which only z=2i is included within the contour. The residue at z=2i turns out to be e^{-4}. Thus

\displaystyle \oint \limits_{\gamma} f(z) \, {\rm d}z = 2 \pi i {\rm Res}\left ( f ;z =2i \right ) = 2 \pi i e^{-4}

The contribution of the large circle as R \rightarrow +\infty is 0 whereas the contribution of the small circle as \epsilon \rightarrow 0 is \pi i . This can be seen by parametrising the small circle (  \epsilon e^{it} \; , \; t \in [0, \pi] ). Hence:

\begin{aligned} \frac{2 \pi i }{e^4} = \int_{-\infty}^{\infty} f(z) \, {\rm d}z +i \pi &\Leftrightarrow \left ( \frac{2}{e^4} -1 \right ) \pi i = \int_{-\infty}^{\infty} \frac{\left ( z^2-4 \right ) e^{2iz}}{z \left ( z^4+4 \right )} \, {\rm d}z \\ &\Leftrightarrow \left ( \frac{2}{e^4} -1 \right ) \pi = \int_{-\infty}^{\infty} \frac{\left ( x^2-4 \right ) \sin 2x}{x(x^2+4)} \, {\rm d}x \end{aligned}

Using (1) we get that

\displaystyle \int_{0}^{\infty} \frac{(x^2-4) \sin 2x}{x(x^2+4)} \, {\rm d}x = \pi e^{-4} - \frac{\pi}{2}

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A logarithmic integral

Prove that

\displaystyle \int_{0}^{\infty} \frac{\sqrt{x} \log x}{\left ( 1+x \right )^2} \, {\rm d}x = \pi

Solution

We begin by making the substitution u=\sqrt{x} thus:

\displaystyle \int_{0}^{\infty} \frac{\sqrt{x} \log x}{(1+x)^2} \, {\rm d}x \overset{u=\sqrt{x}}{=\! =\! =\! =\!} \int_{0}^{\infty} \frac{4u^2 \log u}{\left ( 1+u^2 \right )^2} \, {\rm d}u

Now let us consider the complex function \displaystyle f(z)=\frac{z^2 \log^2 z}{(1+z^2)^2} where the principal arguement of z lies within the interval (-\pi, \pi] as well as the contour below

It is clear that f has two poles of order 2 at z=2i and z=-2i. The residue at z=i is equal to  \frac{\pi}{4} + \frac{i \pi^2}{16} whereas the residue at z=-i is equal to  \frac{\pi}{4} - \frac{i \pi^2}{16} . Thus

\displaystyle \oint \limits_{\gamma} f(z) \, {\rm d}z = \oint \limits_{\mathcal{C}_R} f(z) \, {\rm d}z + \oint \limits_{\mathcal{C}_\epsilon} f(z) \, {\rm d}z + \int_{-R}^{-\epsilon} f(z) \, {\rm d}z + \int_{-\epsilon}^{-R} f(z) \, {\rm d}z

Sending \epsilon \rightarrow 0 and R \rightarrow + \infty the contribution of both the large and the small circle is 0. Hence:

\begin{aligned} i \pi^2 &= \int_{-\infty}^{0} \frac{x^2 \left ( \log \left | x \right | + i \pi \right )^2}{\left ( 1+x^2 \right )^2} \, {\rm d}x + \int_{0}^{-\infty} \frac{x^2 \left ( \log \left | x \right | - i \pi \right )^2}{\left ( 1+x^2 \right )^2} \, {\rm d}x \\ &\!\!\!\!\! \overset{y=-x}{=\! =\! =\! =\! =\!} \int_{0}^{\infty} \frac{x^2 \left ( \log x + i \pi \right )^2}{\left ( 1+x^2 \right )^2} - \int_{0}^{\infty} \frac{x^2 \left ( \log x - i \pi \right )^2}{\left ( 1+x^2 \right )^2} \, {\rm d}x \\ &=\bigintss_{0}^{\infty} \frac{x^2 \bigg( \left ( \log x + i \pi \right )^2 - \left ( \log x - i \pi \right )^2 \bigg)}{\left ( 1+x^2 \right )^2} \, {\rm d}x \\ &= 4 i \pi \int_{0}^{\infty} \frac{x^2 \log x}{\left ( 1+x^2 \right )^2} \, {\rm d}x \end{aligned}

Thus the conclusion follows.

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A contour integral

Let a, b \in \mathbb{C} such that |b|<1. Prove that:

\displaystyle \frac{1}{2\pi} \oint \limits_{\left | z \right |=1} \left | \frac{z-a}{z-b} \right |^2 \left | {\rm d} z\right | = \frac{\left | a-b \right |^2}{1- \left | b \right |^2} + 1

Solution

We have successively:

\begin{aligned} \frac{1}{2\pi}\oint \limits_{|z|=1}\left | \frac{z-a}{z-b} \right |^2 \left | {\rm d} z\right | &=\frac{1}{2\pi} \oint \limits_{|z|=1} \frac{\left ( z-a \right )\left ( \bar{z}-\bar{a} \right )}{\left ( z-b \right )\left ( \bar{z}-\bar{b} \right )}\frac{{\rm d}z}{iz}\\ &= \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\left ( z-a \right )\left ( \frac{1}{z}-\bar{a} \right )}{\left ( z-b \right )\left ( \frac{1}{z}-\bar{b} \right )}\frac{{\rm d}z}{z}\\ &= \frac{1}{2\pi i}\oint \limits_{|z|=1}\frac{\left ( z-a \right )\left ( 1-\bar{a}z \right )}{b\left ( 1-\bar{b}z \right )}\left ( \frac{1}{z-b}- \frac{1}{z} \right )\, {\rm d}z\\ &=\frac{\left ( z-a \right )\left ( 1-\bar{a}z \right )}{b\left ( 1-\bar{b}z \right )}\bigg|_{z=b}- \frac{\left ( z-a \right )\left ( 1-\bar{a}z \right )}{b\left ( 1-\bar{b}z \right )} \bigg|_{z=0} \\ &= \frac{\left ( a-\bar{b} \right )\left ( \bar{a}-b \right )}{1-b\bar{b}}+1 \\&= \frac{\left | a-b \right |^2}{1-|b|^2}+1 \end{aligned}

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