9
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ Frullani's theorem could be very usefull here $\endgroup$ – tired Apr 26 '17 at 6:26
  • 1
    $\begingroup$ If you (OP) should follow the approach in the question, you should introduce $I(a)$ without the logarithm in the integral. Then, the logarithm appears when differentiating. As it stands now, it is wrong. $\endgroup$ – mickep Apr 26 '17 at 7:22
  • $\begingroup$ I am really curious: what is your source for the (sometimes rather involved) integrals you propose? $\endgroup$ – Jack D'Aurizio Apr 26 '17 at 13:09
  • $\begingroup$ First of all, I am thankful for your great effort in answering most of the questions. I took most ideas from this site and other like wolfram and Wikipedia and I just play around with it using CAS plus some algebra and thoughts.** Nothing so special how I got these formulas. ** Nearly every formula I gave I gave the answer too. I don't gave formulas without closed form because the integrals could be unsolvable. I will try my best and link the sources next time. $\endgroup$ – gymbvghjkgkjkhgfkl Apr 26 '17 at 19:56
8
+100
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

$\endgroup$
  • $\begingroup$ An interesting technique! $\endgroup$ – gymbvghjkgkjkhgfkl Apr 27 '17 at 5:58
  • $\begingroup$ @Latte' Thanks. Just basic Trigonometry. $\endgroup$ – Felix Marin Apr 27 '17 at 5:59
  • $\begingroup$ That is power(understanding), solving big problem with basic formula.(+1) $\endgroup$ – gymbvghjkgkjkhgfkl Apr 27 '17 at 6:04
  • $\begingroup$ Concise and complete (+1). $\endgroup$ – Mark Viola Apr 27 '17 at 14:46
  • 1
    $\begingroup$ (+1) This hurts so bad, I did not notice such splendid cancellation :D $\endgroup$ – Jack D'Aurizio Apr 27 '17 at 14:55
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Jack, how does one arrive at the RHS of $(1)$? It's 1:16 a.m. and I'm too tired to proceed. $\endgroup$ – Mark Viola Apr 27 '17 at 6:17
  • $\begingroup$ @Dr.MV: Laplace transform of $1-\cos x$, inverse Laplace transform of $\frac{1}{x^\alpha}$. $\endgroup$ – Jack D'Aurizio Apr 27 '17 at 6:24
  • 1
    $\begingroup$ Yes, I thought so. But doing calculations in my head, I couldn't get a reciprocal $\Gamma$ term. Alas, I found a piece of paper and pen, and performed the integration around the branch cut. Then, applying the reflection principal recovered the result herein. Thanks Jack and (+1). $\endgroup$ – Mark Viola Apr 27 '17 at 14:45
  • $\begingroup$ @JackD'Aurizio: I am surprised that nobody mentioned Frullani's integral. $\endgroup$ – Lucian Sep 1 '17 at 19:07
8
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Omg , did you just forget to write $dx$ for each integral :O . $\endgroup$ – Zaid Alyafeai Apr 26 '17 at 9:07
  • 1
    $\begingroup$ How do you justify differentiation under the integral sign? $\endgroup$ – JanG Apr 26 '17 at 11:24
  • $\begingroup$ @JanG that is a very good (and non-trivial) question. since i am very busy today i try to answer it tomorrow $\endgroup$ – tired Apr 26 '17 at 17:30
  • 1
    $\begingroup$ $@$tired Now I think that I know how to justify the differentiation. Use Cauchy condition for uniform convergence and integration by parts to prove that the integral involved in $I'(b)$ converges uniformly. $\endgroup$ – JanG Apr 27 '17 at 6:58
  • $\begingroup$ Feynman's Trick gets a (+1) $\endgroup$ – Mark Viola Apr 27 '17 at 14:45
7
$\begingroup$

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 $I = -\log(2).$

$\endgroup$
  • $\begingroup$ Oh, just using integration by parts. Nicely done! $\endgroup$ – mickep Apr 26 '17 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.