A limit

Let \mathcal{H}_n denote the n -th harmonic number. Evaluate the limit

\displaystyle \lim_{n\rightarrow +\infty} \left [ e^{\mathcal{H}_n-\gamma} - n \right ]


We begin by the simple observation that

\displaystyle \gamma = \lim_{n \rightarrow +\infty} \left ( \mathcal{H}_n - \ln n \right )

and of course we have the series represantation of the \gamma constant, namely this:

\displaystyle \gamma = \sum_{n=1}^{\infty} \left [ \frac{1}{n} - \ln \left ( 1 + \frac{1}{n} \right ) \right ]


    \begin{align*} \mathcal{H}_n-\gamma&=\sum_{k=1}^n\ln\left(1+\frac{1}{k} \right)+ \\ & \; \quad \quad \quad +\sum_{k=n+1}^{\infty}\left[\ln\left(1+\frac{1}{k} \right)-\frac{1}{k} \right]\\ &=\ln(n+1)+\sum_{k=n+1}^{\infty}\left[\ln\left(1+\frac{1}{k} \right)-\frac{1}{k} \right]\\ &=\ln(n+1)-\frac{1}{2n}+\mathcal{O}(n^{-2}) \end{align*}


    \begin{align*} e^{H_n-\gamma}-n&=e^{\ln(n+1)-\frac{1}{2n}+\mathcal{O}(n^{-2})}-n\\ &=(n+1)e^{-\frac{1}{2n}+\mathcal{O}(n^{-2})}-n\\ &=(n+1)\left(1-\frac{1}{2n}+\mathcal{O}(n^{-2})\right)-n\\ &=n+\frac{1}{2}+\mathcal{O}(n^{-1})-n \end{align*}

We conclude that the desired limit is just \frac{1}{2}.

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A differential equation

Let f:\mathbb{R} \rightarrow \mathbb{R} be a differential equation such that


Find an explicit formula of f.


We consider the function g(x)=f(x)f(-x) which is clearly constant because

    \begin{align*} g'(x) &= f(x) f(-x) \\ &= f'(x) f(-x) - f(x) f'(-x)\\ &= \cancel{f^2(x) f^2(-x) - f^2(x) f^2(-x)}\\ &= 0 \end{align*}

Making use of the initial condition we get that g(x)=1. Thus f(x)f(-x)=1. This also means that f has no roots in the domain given. Hence the initial condition gives us f'(x)=f(x) and the function follows to be f(x)=e^x.

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Does the trigonometric series converge?

Examine if the series

\displaystyle \sum_{n=1}^{\infty} \sin \left[ \pi \left( 2+\sqrt{3} \right)^n \right]



The key lies in the fact that the number

\alpha = \left ( 2 + \sqrt{3} \right )^n +\left ( 2-\sqrt{3} \right )^n

is an integer. Let  x \in \mathbb{R} and let d(x) be the distance from x to the nearest integer. That is

d(x) = \left\{\begin{matrix} x & , & x \in \left [ 0, \frac{1}{2} \right ] \\\\ 1-x& , & x \in \left [ \frac{1}{2}, 1 \right ] \end{matrix}\right.

It is immediate that d can be expanded periodically with period T=1. Since \alpha is an integer we can get that

\displaystyle d\left [ \left ( 2 + \sqrt{3} \right )^n \right ] = \left ( 2-\sqrt{3} \right )^n = \frac{1}{\left ( 2+\sqrt{3} \right )^n}


    \begin{align*} \left |\sum_{n=1}^{\infty} \sin \left [ \pi \left ( 2+\sqrt{3} \right )^n \right ] \right | &\leq \sum_{n=1}^{\infty} \left | \sin \left [ \pi \left ( 2+\sqrt{3} \right )^n \right ] \right | \\ &=\sum_{n=1}^{\infty} \sin \left [ \pi \; d\left ( 2+\sqrt{3} \right )^n \right ] \\ &\leq \pi \sum_{n=1}^{\infty} \frac{1}{\left ( 2+\sqrt{3} \right )^n} \end{align*}

and the conclusion follows.

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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


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


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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