6

Observe this integral

Where $\phi={\sqrt{5}+1\over 2}$

and $\gamma=0.5772156...$ is Euler's Constant

$$\int_{0}^{\infty}{\ln(x)\sin(x)\cos\left(x\over \sqrt{\phi}\right)\over x}\mathrm dx={1\over 2}\pi \left(\ln\phi-\gamma\right)\tag1$$

$$I(a)=\int_{0}^{\infty}{\ln(x)\sin(x)\cos\left(ax\right)\over x}\mathrm dx\tag2$$

$$I^{'}(a)=-\int_{0}^{\infty}{\ln(x)\sin(x)\sin\left(ax\right)}\mathrm dx\tag3$$

Using

$2\sin A\sin B=\cos(A-B)-\cos(A+B)$

$\sin x\sin(ax)=\cos[(1-a)x]-\cos[(1+a)x]$

$$I^{'}(a)=-\int_{0}^{\infty}\cos[(1-a)x]\ln x \mathrm dx+\int_{0}^{\infty}\cos[(1+a)x]\ln x\mathrm dx\tag4$$

$$\int \cos[(1-a)x]\ln x \mathrm dx={\sin[(1-a)x]\ln x\over 1-a}-\int {\sin[(1-a)x]\over x}\tag 5$$

$$\int \cos[(1-a)x]\ln x \mathrm dx={\sin[(1+a)x]\ln x\over 1+a}-\int {\sin[(1+a)x]\over x}\tag 6$$

$(5)$ and $(6)$ showed divgergent integrals

This approach I have tried is not working, what other method can we use to verify $(1)?$

  • 2
    i somehow know this kind of questions...do you know a guy called 'bui'? – tired May 21 '17 at 23:27
  • This looks like something that could be calculated with Frullani's Theorem easily. I'll take a stab at it soon – Brevan Ellefsen May 22 '17 at 00:40
  • The presence of $\phi$ is really irrelevant here; we generally have that $$\int_{0}^{\infty}{\ln(x)\sin(x)\cos\left(x\over \sqrt{\phi}\right)\over x}\mathrm dx = -\frac{1}{4} \pi \left(\log \left(1-\frac{1}{n^2}\right)+2 \gamma \right)$$ – Brevan Ellefsen May 22 '17 at 02:01
  • @tired: there are many questions on integrals with similar pattern and even I think most of it comes from same source. Anyway this does not matter much as long as we get good questions and more importantly great answers like the one here by user Felix Marin. – Paramanand Singh May 22 '17 at 05:18
  • Through Integration by Parts, we find that $$\int_0^\infty \frac{\log(x)\sin(x)\cos\left(\frac{x}{n}\right)}{x} dx =L-\frac{1}{2}\int_0^\infty \frac{\operatorname{Si}\left(\left[1+\frac{1}{n}\right]x\right)-\operatorname{Si}\left(\left[\frac{1}{n}-1\right]x\right)}{x} dx$$ which is so close to being a Frullani Integral. Unfortunately, both $L$ and the rightmost integral blow up, and the easiest way to show that their singularities cancel seems to me, thus far at least, to evaluate the integral by a different method – Brevan Ellefsen May 22 '17 at 06:18
  • Welcome back ... – Zaid Alyafeai May 22 '17 at 20:26
  • @ZaidAlyafeai are you welcoming back this post? – Brevan Ellefsen May 22 '17 at 21:06
  • @BrevanEllefsen, I am welcoming back the person who posted it. – Zaid Alyafeai May 22 '17 at 21:11

2 Answers2

5

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\infty} {\ln\pars{x}\sin\pars{x}\cos\pars{x\over \root{\phi}} \over x}\,\dd x \\[5mm] = &\ {1 \over 2}\int_{0}^{\infty} {\ln\pars{x}\sin\pars{\bracks{1 + \phi^{-1/2}}x}\over x}\,\dd x + {1 \over 2}\int_{0}^{\infty} {\ln\pars{x}\sin\pars{\bracks{1 - \phi^{-1/2}}x}\over x}\,\dd x \\[5mm] = &\ {1 \over 2}\int_{0}^{\infty} {\ln\pars{x/\bracks{1 + \phi^{-1/2}}}\sin\pars{x}\over x}\,\dd x + {1 \over 2}\int_{0}^{\infty} {\ln\pars{x/\bracks{1 - \phi^{-1/2}}}\sin\pars{x}\over x}\,\dd x \\[5mm] = &\ -\,{1 \over 2}\,\ln\pars{1 - \phi^{-1}} \int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x + \int_{0}^{\infty}\ln\pars{x}\,{\sin\pars{x} \over x}\,\dd x \\[5mm] = &\ -\,{\pi \over 4}\,\ln\pars{1 - \phi^{-1}} + \bbox[#ffd,10px]{\ds{% \int_{0}^{\infty}\ln\pars{x}\,{\sin\pars{x} \over x}\,\dd x}} \end{align}

\begin{align} &\bbox[#ffd,10px]{\ds{% \int_{0}^{\infty}\ln\pars{x}\,{\sin\pars{x} \over x}\,\dd x}} = \left.\partiald{}{\mu}\Im\int_{0}^{\infty}x^{\mu - 1}\expo{\ic x} \,\dd x\,\right\vert_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ \left.\partiald{}{\mu}\Im\int_{0}^{\infty\ic}x^{\mu - 1}\expo{\ic\pars{1 - \mu}\pi/2}\expo{-x} \pars{-\ic}\,\dd x\,\right\vert_{\ \mu\ =\ 0^{+}} \\[5mm] = &\ \partiald{}{\mu}\bracks{\sin\pars{\mu\,{\pi \over 2}} \Gamma\pars{\mu}}_{\ \mu\ =\ 0^{+}} = \left.{1 \over 2}\,\pi\,\partiald{\Gamma\pars{\mu + 1}}{\mu} \right\vert_{\ \mu\ =\ 0^{+}} = -\,{1 \over 2}\,\pi\gamma \end{align}
$$ \bbx{\ds{\int_{0}^{\infty} {\ln\pars{x}\sin\pars{x}\cos\pars{x\over \root{\phi}} \over x}\,\dd x = {\pi \over 2}\bracks{-\,{1 \over 2}\,\ln\pars{1 - \phi^{-1}} - \gamma}}} $$

Note that $\ds{-\ln\pars{1 - \phi^{-1}} = \ln\pars{3 + \root{5} \over 2}}$.

Since $\ds{\phi^{2} - \phi - 1 = 0 \implies -\,{1 \over 2}\,\ln\pars{1 - \phi^{-1}} = \ln\pars{\phi}}$, an alternative expression is $$ \bbx{\ds{\int_{0}^{\infty} {\ln\pars{x}\sin\pars{x}\cos\pars{x\over \root{\phi}} \over x}\,\dd x = \color{#f00}{+}\,{1 \over 2}\,\pi\bracks{\ln\pars{\phi} - \gamma}}} $$

Felix Marin
  • 89,464
2

Hint: let $$I(a)=\int_{0}^{\infty}x^a\sin(x)\cos\left(x\over \sqrt{\phi}\right)dx$$ and then use Mellon transform to calculate it. Finally evaluate $I'(-1)$ which will be the answer.

xpaul
  • 44,000