How may we show that $\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^2}\cdot\ln(x)\,\mathrm dx=-\ln(2)?$

Question

Simple closed form given by this complicated integral

$$\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^2}\cdot\ln(x)\,\mathrm dx=-\ln(2)\tag1$$

Making an attempt:

Splitting $(1)$ results in the integral to be diverges.

$$I(a)=\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^a}\cdot\ln(x)\,\mathrm dx\tag2$$

$$I'(a)=\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^a}\,\mathrm dx\tag3$$

$$I'(2)=\int_{0}^{\infty}{{\cos\left({2x\over \pi}\right)-\cos^2\left({2x\over \pi}\right)}\over x^2}\,\mathrm dx\tag4$$

$u={2x\over \pi}$

$$I'(2)={2\over \pi}\int_{0}^{\infty}{{\cos\left({u}\right)-\cos^2\left({u}\right)}\over u^2}\,\mathrm du=I_1-I_2\tag5$$ Recall from table of integral, I was thinking of using $(6)$ for $I_1$ but it is only valid for $0\le p\le 1$ $$\int_{0}^{\infty}{\cos x\over x^p}\,\mathrm dx={\pi\over 2\Gamma(p)\cos(p\pi/2)}\tag6$$

How may we prove (1)?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\infty} {\cos\pars{2x/\pi} - \cos^{2}\pars{2x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x \\[5mm] = &\ \int_{0}^{\infty} {1 - \cos^{2}\pars{2x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x - \int_{0}^{\infty} {1 - \cos\pars{2x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x \\[5mm] = &\ \int_{0}^{\infty} {\sin^{2}\pars{2x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x - \int_{0}^{\infty} {2\sin^{2}\pars{x/\pi} \over x^{2}}\,\ln\pars{x}\,\dd x \\[5mm] = &\ {2 \over \pi}\int_{0}^{\infty} {\sin^{2}\pars{x} \over x^{2}}\,\ln\pars{\pi x \over 2}\,\dd x - {2 \over \pi}\int_{0}^{\infty} {\sin^{2}\pars{x} \over x^{2}}\,\ln\pars{\pi x}\,\dd x \\[5mm] = &\ {2 \over \pi}\int_{0}^{\infty} {\sin^{2}\pars{x} \over x^{2}}\,\bracks{\ln\pars{1 \over 2}}\,\dd x = \bbx{\ds{-\ln\pars{2}}} \end{align}

Answer 2

Define

$$ I(b)=\int_0^{\infty}\frac{\cos(b x)-\cos^2(b x)}{x^2}\log(x) $$

so the integral in question is $I(2/\pi)$. Now

$$ -I'(b)=\int_0^{\infty}\frac{\sin(b x)(1-2\cos(b x))}{x}\log(x)\underbrace{=}_{xb\rightarrow y}\\I'(1)-\log(b)\int_0^{\infty}\frac{\sin(x)(1-2\cos(x))}{x}=I'(1) $$

or

$$ I(b)=-bI'(1)+c $$

now $I'(1)$ can be calculated pretty straightfowardy from using $2\cos(x)\sin(x)=\sin(2x)$ .

$$ I'(1)=\log(2)\int_0^{\infty}\frac{\sin(x)}{x}=\log(2)\frac{\pi}{2} $$

Afterwards the only thing left is to fix $c$ which can be done by observing that $\lim_{b\rightarrow 0}I(b)=0$ which follows from the Taylorexpansion of $\cos(bx)-\cos^2(bx)$. This means

$$ I(b)=-b\frac{\pi}{2}\log(2) $$

from which it follows that $I(2/\pi)=-\log(2)$ as proposed

Answer 3

I will propose an alternative (but extremely similar to tired's) approach.
By the Laplace transform, for any $k>0$ and any $\alpha\in(1,3)$ we have

$$ \int_{0}^{+\infty}\frac{1-\cos(kx)}{x^\alpha}\,dx = k^{\alpha-1}\int_{0}^{+\infty}\frac{1-\cos(x)}{x^{\alpha}}\,dx =\frac{k^{\alpha-1}}{\Gamma(\alpha)}\int_{0}^{+\infty}\frac{s^{\alpha-2}}{1+s^2}\,ds\tag{1}$$ and the last integral can be computed through the Beta function. In particular we get:

$$\int_{0}^{+\infty}\frac{1-\cos(kx)}{x^\alpha}\,dx = \frac{\pi\,k^{\alpha-1}}{2\cos\left(\frac{\pi}{2}(a-2)\right)\Gamma(\alpha)} \tag{2}$$ and by differentiating both sides with respect to $\alpha$ we simply get: $$ g(k)=\int_{0}^{+\infty}\frac{1-\cos(kx)}{x^2}\log(x)\,dx = \frac{k\pi}{2}\left(1-\gamma-\log k\right) \tag{3}$$ so the original integral equals:

$$ \int_{0}^{+\infty}\frac{\cos\left(\frac{2x}{\pi}\right)-\cos^2\left(\frac{2x}{\pi}\right)}{x^2}\log(x)\,dx = \frac{1}{2}\,g\left(\frac{4}{\pi}\right)-g\left(\frac{2}{\pi}\right)=\color{red}{-\log 2}\tag{4} $$ as wanted.

Answer 4

I will propose an answer based on integration by parts.

Put $b = \frac{2}{\pi}$. We get \begin{gather*} I = \int_{0}^{\infty}\dfrac{\cos(bx)-\cos^2(bx)}{x^2}\log(x)\, dx = \left[-\dfrac{\cos(bx)-\cos^2(bx)}{x}\log(x)\right]_{0}^{\infty}+\\[2ex]b\int_{0}^{\infty}\dfrac{2\sin(bx)\cos(bx)-\sin(bx)}{x}\log(x)\, dx + \int_{0}^{\infty}\dfrac{\cos(bx)-\cos^2(bx)}{x^2}\, dx = 0+bI_1+I_2\tag{1} \end{gather*} where \begin{gather*} I_1 =\int_{0}^{\infty}\dfrac{2\sin(bx)\cos(bx)-\sin(bx)}{x}\log(x)\, dx = \int_{0}^{\infty}\dfrac{\sin(2bx)}{x}\log(x)\, dx -\\[2ex] \int_{0}^{\infty}\dfrac{\sin(bx)}{x}\log(x)\, dx = \int_{0}^{\infty}\dfrac{\sin(bx)}{x}\log\left(\dfrac{x}{2}\right)\, dx -\int_{0}^{\infty}\dfrac{\sin(bx)}{x}\log(x)\, dx =\\[2ex] -\log(2)\int_{0}^{\infty}\dfrac{\sin(bx)}{x}\, dx = -\log(2)\int_{0}^{\infty}\dfrac{\sin(x)}{x}\, dx = -\log(2)\dfrac{\pi}{2}= -\log(2)\dfrac{1}{b} \end{gather*} and \begin{gather*} I_2 = \int_{0}^{\infty}\dfrac{\cos(bx)-\cos^2(bx)}{x^2}\, dx = \left[-\dfrac{\cos(bx)-\cos^2(bx)}{x}\right]_{0}^{\infty} +\\[2ex] b\int_{0}^{\infty}\dfrac{2\sin(bx)\cos(bx)-\sin(bx)}{x}\, dx = 0. \end{gather*} Consequently $I = -\log(2).$