Multiple integral

Let \langle \cdot, \cdot \rangle denote the usual inner product of \mathbb{R}^m. Evaluate the integral

    \[\mathcal{M} = \int \limits_{\mathbb{R}^m} \exp \left( - ( \langle x, \mathcal{S} ^{-1} x \rangle )^a \right) \, {\rm d}x\]

where \mathcal{S} is a positive symmetric m \times m matrix and a>0.

Solution

Since \mathcal{S} is a positive symmetric matrix , so is \mathcal{S}^{-1}. For a positive symmetric matrix \mathcal{A} there exists an \mathcal{R} positive symmetric matrix such that \mathcal{A} = \mathcal{R}^2. Applying this to \mathcal{S}^{-1} our integral becomes

    \[\mathcal{M} = \int \limits_{\mathbb{R}^m} \exp \left ( - \left \| \mathcal{R} x \right \|^{2a} \right ) \, {\rm d}x\]

where \left \| \cdot \right \| is the Euclidean norm. Applying a change of variables we have that

    \[\mathcal{M} = \det \left ( \mathcal{R}^{-1} \right ) \int \limits_{\mathbb{R}^m} e^{-\left \| y \right \|^{2a}} \, {\rm d}y\]

Since \det \left ( \mathcal{R}^{-1} \right ) = \sqrt{\det \left ( \mathcal{S} \right )} then by converting to polar coordinates we have that

    \begin{align*} \mathcal{M} &=  \omega_m \sqrt{\det \left ( \mathcal{S} \right )} \int_{0}^{\infty} r^{m-1} e^{-r^{2a}} \, {\rm d}r \\ &= \frac{\omega_m}{m} \sqrt{\det \left ( \mathcal{S} \right )}\Gamma \left ( \frac{m}{2a} + 1 \right ) \end{align*}

Here \omega_m denotes the surface area measure of the unit sphere and it is known to be

    \[\omega_m = \frac{2 \pi^{m/2}}{\Gamma \left ( \frac{m}{2} \right )}\]

hence

    \[\mathcal{M}=\frac{\sqrt{\det (\mathcal{S}) }\pi^{m/2}\Gamma\left(\frac{m}{2a}\right)}{2^{a-1}\Gamma\left(\frac{m}{2}\right)}\]

where \Gamma denotes the Gamma Euler function for which it holds that

    \[\Gamma(x+1) = x \Gamma(x) \quad \text{forall} \quad x>0\]

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Limit of a trigonometric integral

Let A \subseteq \mathbb{R} be a measurable of finite length set. Evaluate the limit:

    \[\ell = \lim_{\lambda \rightarrow +\infty} \int \limits_{A} | \sin \lambda x | \, {\rm d}x\]

Solution

Let us begin with the Fourier series of |\sin x | which is of the form:

    \[|\sin x | = \frac{2}{\pi} + \sum_{n=1}^{\infty} a_n \cos nx\]

Hence

(1)   \begin{equation*} |\sin \lambda x | = \frac{2}{\pi} + \sum_{n=1}^{\infty} a_n \cos n \lambda x  \end{equation*}

Integrating ( 1 ) we get that

    \begin{align*} \ell &= \lim_{\lambda \rightarrow +\infty} \int \limits_{A} \left | \sin \lambda x \right | \, {\rm d}x \\ &=\lim_{\lambda \rightarrow +\infty} \int \limits_{A} \left ( \frac{2}{\pi} + \sum_{n=1}^{\infty} a_n \cos n \lambda x \right ) \, {\rm d}x \\ &= \lim_{\lambda \rightarrow +\infty} \int \limits_{A} \frac{2}{\pi} + \lim_{\lambda \rightarrow +\infty} \sum_{n=1}^{\infty} a_n \int \limits_{A} \cos n \lambda x \, {\rm d}x \\ &= \frac{2 \left | A \right |}{\pi} \end{align*}

since it follows from Riemann – Lebesgue lemma that

    \[\lim_{\lambda \rightarrow +\infty} \int \limits_{A} \cos n \lambda x \, {\rm d}x = 0\]

 

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A generating function involving harmonic number of even index

Let \mathcal{H}_n denote the n-th harmonic number. Prove that forall |x|<1 it holds that

    \[\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_{2n} x^{2n+1}}{2n+1} = \frac{\arctan x \log (1+x^2)}{2}\]

Solution

Well we are stating two lemmata.

Lemma 1: For all |x|<1 it holds that

    \[\sum_{n=1}^{\infty} \mathcal{H}_n x^n = -\frac{\log(1-x)}{1-x}\]

Proof: Pretty straight forward calculations show that

    \begin{align*} \frac{\log \left ( 1-x \right )}{1-x} &= \left ( -\sum_{n=1}^{\infty} \frac{x^n}{n} \right )\left ( \sum_{n=0}^{\infty} x^n \right ) \\ &= -\sum_{n=1}^{\infty} \mathcal{H}_n x^n \end{align*}

and Lemma 1 is proved. \blacksquare

Lemma 2: For all |x|<1 it holds that

    \[\sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} = - \frac{1}{2} \left [ \frac{\log \left ( 1-x \right )}{1-x} + \frac{\log\left ( 1+x \right )}{1+x} \right ]\]

Proof: We begin by lemma 1 and successively we have

    \begin{align*} -\frac{\log \left ( 1-x \right )}{1-x} &= \sum_{n=1}^{\infty} \mathcal{H}_{n} x^n \\ &= \sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} + \\ &\quad \quad + \sum_{n=0}^{\infty} \mathcal{H}_{2n+1} x^{2n+1} \\ &=\sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} + x + \\ & \quad \quad +\sum_{n=1}^{\infty} \left ( \mathcal{H}_{2n} + \frac{1}{2n+1} \right ) x^{2n+1} \\ &=\sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} + x + \\ &\quad \quad +x \sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} + \sum_{n=1}^{\infty} \frac{x^{2n+1}}{2n+1} \\ &= \sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} + x + x \sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} + \\ & \quad \quad +\frac{1}{2} \left [ \log (1+x) - \log \left ( 1-x \right ) \right ] \end{align*}

and Lemma 2 follows. \blacksquare

Now, mapping x \mapsto ix back at Lemma 2 we have that

\displaystyle\sum_{n=1}^{\infty} (-1)^n \mathcal{H}_{2n} x^{2n} = -\frac{1}{2} \left [ \frac{\log(1-ix)}{1-ix} + \frac{\log(1+ix)}{1+ix} \right ]

Integrating we have that

\begin{aligned} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_{2n} x^{2n+1} }{2n+1} &= \frac{i}{4} \left [ \log^2 (1-ix) -\log^2 (1+ix) \right ] \\ &=\frac{i}{4} \bigg [ \log^2 \left ( 1+x^2 \right )-i \arctan x \log \left ( 1+x^2 \right ) \\ & \quad \quad - \log^2 \left ( 1+x^2 \right ) - i \arctan x \log \left ( 1+x^2 \right )\bigg] \\ &= \frac{\arctan x \log \left ( 1+x^2 \right )}{2} \end{aligned}

 

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Constant function

Let f:\mathbb{R} \rightarrow \mathbb{R} be a continuous function such that

    \[f(x) = f \left(x + \frac{1}{\sqrt{n}} \right)\]

where n=1, 2, \dots. Prove that f is constant.

Solution

Well it does hold

    \[f(x) = f \left(x + \frac{k}{\sqrt{n}} \right)\]

where k \in \mathbb{Z} and n =1, 2, \dots. The set \{ \frac{k}{\sqrt{n}} \; , \; k \in \mathbb{Z} , \; n =1, 2, \dots \} is dense in \mathbb{R} and since f is continous and characterised by its values in a dense set we conclude that f is constant since

    \[f(0) = f \left( \frac{k}{\sqrt{n}} \right)\]

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On a known pattern

Let x \in \left( - \frac{\pi}{2} , \frac{\pi}{2} \right) and consider the function

    \[f(x) = a_1 \tan x + a_2 \tan \frac{x}{2} + \cdots +a_n \tan \frac{x}{n}\]

where a_1, a_2, \dots, a_n \in \mathbb{R} and n \in \mathbb{N}. If \left| f(x) \right| \leq \left| \tan x  \right| for all x \in \left( - \frac{\pi}{2} , \frac{\pi}{2} \right) then prove that

    \[\left| a_1 + \frac{a_2}{2} + \cdots + \frac{a_n}{n} \right| \leq 1\]

Solution

The function f is differentiable in \left( - \frac{\pi}{2} , \frac{\pi}{2} \right) and the derivative is given by

    \[f'(x) = \frac{a_1}{\cos^2 x} +\frac{a_2}{2 \cos^2 \frac{x}{2}} + \cdots + \frac{a_n}{n \cos^2 \frac{x}{n}}\]

Hence we want to prove that \left| f'(0) \right| \leq 1. We note that forall x \in \left ( - \frac{\pi}{2} , 0 \right ) \cup \left ( 0, \frac{\pi}{2} \right ) it holds that

    \[\left | \frac{f(x)}{x} \right |\leq \left | \frac{\tan x}{x} \right |\]

Thus

    \begin{align*} \left | f'(0) \right | &=\left | \lim_{x\rightarrow 0} \frac{f(x) - f(0)}{x-0} \right | \\ &= \lim_{x\rightarrow 0} \left | \frac{f(x)}{x} \right |\\ &\leq \lim_{x\rightarrow 0} \left | \frac{\tan x}{x} \right | \\ &=1 \end{align*}

and the conclusion follows.

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