No rational function

Prove that there exists no rational function such that

\displaystyle f(n)=1+ \frac{1}{2} + \cdots + \frac{1}{n} \quad \text{forall} \; n \in \mathbb{N}

Solution

Suppose , on the contrary , that such function exists. Since the harmonic series diverges we conclude that the limit of our function in infinity is infinity. This, in return means that the degree of the nominator , call that m is greater that the one of the denominator , call that n. Extracting x^{n-m} in the nominator we get that

\displaystyle R(x)=\frac{P(x)}{Q(x)} = x^{m-n} s(x)

The limit of s at infinity is finite and call that \ell. Hence:

\begin{aligned} \lim_{x \rightarrow +\infty} \left ( \frac{f(x)}{g(x)} - \ln x \right ) &= \lim_{x \rightarrow +\infty} \left [ x^{m-n} s(x) - \ln x \right ] \\ &= \lim_{n \rightarrow +\infty} x^{m-n} \left [ s(x) - \frac{\ln x}{x^{m-n}} \right ]\\ &= +\infty \end{aligned}

and of course this contradicts the fact that

\displaystyle \lim \left ( \mathcal{H}_n - \ln n \right ) = \gamma

where \gamma is the Euler  – Mascheroni constant .

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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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On Ahmed’s integral

Evaluate the integral

\displaystyle \mathcal{J} = \int_{0}^{1} \frac{\arctan \sqrt{2+x^2}}{\left ( 1+x^2 \right ) \sqrt{2+x^2}} \, {\rm d}x

Solution

Let us begin by the identity

\displaystyle \arctan x + \arctan \frac{1}{x} = \frac{\pi}{2} \quad \text{forall} \; x>0

Thus

    \begin{align*} \mathcal{J} &= \int_{0}^{1} \frac{\arctan \sqrt{2+x^2}}{\left ( 1+x^2 \right ) \sqrt{2+x^2}} \, {\rm d}x \\ &= \frac{\pi}{2} \int_{0}^{1} \frac{{\rm d}x}{\left ( 1+x^2 \right ) \sqrt{2+x^2}} - \\ &\quad \quad \quad -\int_{0}^{1} \frac{\arctan \left ( \frac{1}{\sqrt{2+x^2}} \right )}{\left ( 1+x^2 \right )\sqrt{2+x^2}} \, {\rm d}x \\ &=\frac{\pi^2}{12} - \int_{0}^{1} \frac{\arctan \left ( \frac{1}{\sqrt{2+x^2}} \right )}{\left ( 1+x^2 \right )\sqrt{2+x^2}} \, {\rm d}x \end{align*}

Now one of the \arctan‘s definition is

\displaystyle \arctan \frac{1}{a} = \int_{0}^{1} \frac{a}{x^2+a^2} \, {\rm d}x \Leftrightarrow \frac{1}{a} \arctan \frac{1}{a} = \int_{0}^{1} \frac{{\rm d}x}{x^2+a^2}

Hence

\begin{aligned} \int_{0}^{1} \frac{\arctan \left ( \frac{1}{\sqrt{2+x^2}} \right )}{\left ( 1+x^2 \right )\sqrt{2+x^2}} \, {\rm d}x &= \int_{0}^{1} \int_{0}^{1} \frac{{\rm d}(x, y)}{\left ( 1+x^2 \right )\left ( 2+x^2+y^2 \right )} \\ &=\int_{0}^{1} \int_{0}^{1} \frac{1}{y^2+1} \bigg( \frac{1}{1+x^2} - \\ & \quad \quad \quad - \frac{1}{2+x^2+y^2} \bigg )\, {\rm d}(x, y) \\\\ &=\int_{0}^{1} \int_{0}^{1} \frac{{\rm d}(x, y)}{\left ( 1+x^2 \right ) \left ( 1+y^2 \right )} - \\ &\quad \quad \quad -\int_{0}^{1} \int_{0}^{1} \frac{{\rm d}(x, y)}{2+x^2+y^2} \\\\ &= \frac{1}{2} \int_{0}^{1} \int_{0}^{1} \frac{{\rm d}(x, y)}{\left ( 1+x^2 \right )\left ( 1+y^2 \right )} \\ &= \frac{1}{2}\left ( \int_{0}^{1} \frac{{\rm d}x}{1+x^2} \right )^2 \\ & = \frac{\pi^2}{32} \end{aligned}

Finally,

    \begin{align*} \mathcal{J} &= \frac{\pi^2}{12} - \frac{\pi^2}{32} \\ &= \frac{5 \pi^2}{96} \end{align*}

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Integral with trigonometric and exponential

Let a\in \mathbb{R}. Prove that

\displaystyle \int_{-\infty}^{\infty} \frac{\cos ax}{e^x+e^{-x}} \, {\rm d}x

Solution

Let us start things off by recalling the famous identity

\displaystyle \pi {\rm sech} \;\pi a = 4 \sum_{n=0}^{\infty} \frac{(-1)^n \left ( 2n+1 \right )}{4a^2 +\left ( 2n+1 \right )^2}

which is a simple application of residues.

Thus making use of parity we have that

\begin{aligned} \int_{-\infty}^{\infty} \frac{\cos ax}{e^x + e^{-x}} \, {\rm d}x &= 2\int_{0}^{\infty} \frac{\cos ax}{e^x + e^{-x}} \, {\rm d}x \\ &= 2 \int_{0}^{\infty} \cos ax \sum_{n=0}^{\infty} (-1)^n e^{-(2n+1)x} \, {\rm d}x\\ &= 2\sum_{n=0}^{\infty} (-1)^n \int_{0}^{\infty} e^{-(2n+1) x} \cos ax \, {\rm d}x \\ &= 2 \sum_{n=0}^{\infty} \frac{(-1)^n \left ( 2n+1 \right )}{a^2 + \left ( 2n+1 \right )^2} \\ &= \frac{\pi {\rm sech} \left ( \frac{\pi a}{2} \right )}{2} \end{aligned}

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Non existence of sequence of continuous functions

Prove that there does not exist a sequence of continuous functions f_n:[0, 1] \rightarrow \mathbb{R} such that converges pointwise, to the function \chi_{\mathbb{Q}} , where \chi_{\mathbb{Q}} is the characteristic polynomial of the rationals in [0, 1].

Solution

The indicator of the rationals is no other function than

\chi_{\mathbb{Q}}= \left\{\begin{matrix} 1& ,& x \in \mathbb{Q}\\ 0& , & \text{elsewhere} \end{matrix}\right.

It is known that pointwise limits of continuous functions have a meagre set of points of continuity. However, this function is discontinuous everywhere and thus we cannot expect a sequence of continuous functions to converge pointwise to it.

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