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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Is f necessarily constant?

Let f:\mathbb{R} \rightarrow \mathbb{R} be a differentiable function such that f'(x)=0 forall  x \in \mathbb{Q}. Does it follow that f is necessarily constant?

Solution

Since f'(x)=0 forall x \in \mathbb{Q} it follows that f(x)=c forall x \in \mathbb{Q}. Let x_0 \in \mathbb{R} \setminus \mathbb{Q} and suppose that f(x_0) \neq c. Since \mathbb{Q} is dense there will exist a sequence \{q_n\}_{n \in \mathbb{N}} of rational numbers that it converges to x_0. Thus:

\displaystyle c \neq f\left ( x_0 \right ) = \lim_{x\rightarrow x_0} f(x)= \lim_{n \rightarrow +\infty} f \left ( q_n \right ) = c

contradicting what we had assumed in the first place. Hence f is constant.

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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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Convergence of series

Examine if the series:

\displaystyle \sum_{n=1}^{\infty}\bigg ( e - \left ( 1+\tfrac{1}{n} \right )^n \bigg )

converges.

Solution

We are using the basic inequality

e^x \geq x +1 \quad \text{forall} \; x \in \mathbb{R}

as well as the Hermite Hadamard inequality. Applying the Hermite – Hadamard inequality we have that

    \begin{align*} n \ln \left ( 1 + \frac{1}{n} \right ) &= n \int_{n}^{n+1} \frac{{\rm d}x}{x} \\ &\leq \frac{n}{2} \left ( \frac{1}{n} + \frac{1}{n+1} \right ) \\ &= 1 - \frac{1}{2n+2} \end{align*}

Exponentiating we get that

    \begin{align*} \left ( 1 + \frac{1}{n} \right )^n & \leq e e^{-1/(2n+2)} \\ &\leq \frac{e}{1+ \frac{1}{2n+2}} \quad \quad (e^x \geq x +1)\\ &= e \left ( 1 -\frac{1}{2n+3} \right ) \end{align*}

and hence

\displaystyle e - \left ( 1 + \frac{1}{n} \right )^n \geq \frac{e}{2n+3}

This leads us to the conclusion that the series diverges.

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