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Proposed:

$$\int_{0}^{1}\cos(\ln x)\cdot{\mathrm dx\over 1+x^2}={\pi\over 4}\cdot{1\over \cosh\left({\pi\over 2}\right)}\tag1$$

My try:

$x=\tan u\implies dx=\sec^2 u du$, then $(1)$ becomes

$$\int_{0}^{\pi/4}\cos(\ln\tan u)\mathrm du\tag2$$

Recall of $$\int_{0}^{\pi/2}\ln\sin x\mathrm dx=\int_{0}^{\pi/2}\ln\cos x\mathrm dx=-{\pi\over 2}\ln(2)\tag3$$

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  • $\begingroup$ Line $(2)$ to $(3)$, $$\cos(\ln\tan u)=\cos(\ln\sin u - \ln\cos u)\neq\cos(\ln\sin u)-\cos\ln\cos u)$$ $\endgroup$ – John Doe May 10 '17 at 4:54
  • $\begingroup$ Oh yeah thank you, just woke up! What was I thinking! $\endgroup$ – gymbvghjkgkjkhgfkl May 10 '17 at 4:56
  • $\begingroup$ I suppose that the antiderivative can be expressed in terms of hypergeometric functions and that this integral would come from polygamma functions. What should be also interesting is to be able to show that $$\int_{0}^{\infty}\frac{\cos(\ln x)}{1+x^2}\,dx=\pi \sinh \left(\frac{\pi }{2}\right) \text{csch}(\pi )$$ I hope that some one would come with a simpler solution. $\endgroup$ – Claude Leibovici May 10 '17 at 6:11
  • $\begingroup$ is it possible to solve the equation with complex numbers? $$\cos(\log(x)) = \frac{x^{-i}}{2} + \frac{x^î}{2}$$ $\endgroup$ – Darío A. Gutiérrez May 10 '17 at 6:30
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    $\begingroup$ Related question $\endgroup$ – A.Γ. May 10 '17 at 14:49
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First note that

$$2 \int^1_0 \frac{\cos(\log x) }{x^2+1}\,dx = \int^\infty_0 \frac{\cos(\log x)}{x^2+1}\,dx$$

Integrate the following function

$$f(z) = \frac{e^{i\log(z)}}{z^2+1}$$

Around a semi-circle in the upper half place. Where we avoid the branch point at $z=0$ by a semi-circle. We assume that the branch cut is taken on the negative imaginary axis.

enter image description here

$$\int^{-r}_{-R}f(z) \,dz +\int^{R}_{r}f(z) \,dz+ \int_{C_R}f(z)\,dz+\int_{c_r}f(z)\,dz =2\pi i \mathrm{Res}(f,i)$$

The integral on semi-circles

\begin{align}\left|\int_{C_R} \frac{e^{i\log(z)}}{z^2+1}\right| &\leq R \int^{\pi}_0 \left|\frac{e^{i \log(R e^{it})}}{R^2e^{2it}+1}\right| dt\\ &\leq R \int^{\pi}_0 \frac{e^{i \log R-t}}{|R^2-1|} dt\\ &\leq \frac{R}{R^2-1} \int^{\pi}_0 e^{-t} dt\\ &\leq \frac{R(1-e^{-\pi})}{R^2-1} dt \sim_{\infty} 0 \end{align}

Similarily

$$\lim_{r \to 0}\int_{C_r} \frac{e^{i\log(z)}}{z^2+1} = 0$$

Hence we have as $ R \to \infty , r \to 0 $ $$\int_{-\infty}^0 \frac{e^{i\log|x|-\pi}}{x^2+1}\,dx +\int^\infty_0 \frac{e^{i\log(x)}}{x^2+1}\,dx =2\pi i \mathrm{Res}(f,i)$$

$$(1+e^{-\pi})\int^\infty_0 \frac{e^{i\log(x)}}{x^2+1}\,dx =2\pi i \mathrm{Res}(f,i)$$

Note that

$$\mathrm{Res}(f,i) = \frac{e^{i\log(i)}}{2i} = \frac{e^{-\pi/2}}{2i}$$

Hence

$$\int^\infty_0 \frac{e^{i\log(x)}}{x^2+1}\,dx =\pi \frac{e^{-\pi/2}}{1+e^{-\pi}} = \frac{\pi}{e^{\pi/2}+e^{-\pi/2}} = \frac{\pi}{2}\mathrm{sech}\left( \frac{\pi}{2}\right)$$

Which implies that

$$\int^\infty_0 \frac{\cos(\log x)}{x^2+1}\,dx = \frac{\pi}{2}\mathrm{sech}\left( \frac{\pi}{2}\right)$$

Hence we have our result

$$\int^1_0 \frac{\cos(\log x)}{x^2+1}\,dx = \frac{\pi}{4}\mathrm{sech}\left( \frac{\pi}{2}\right)$$


The Tikz code for the contour drawing

\begin{tikzpicture}[decoration={markings,
   mark=at position 0.5cm with {\arrow[line width=1pt]{>}},
   mark=at position 2cm with {\arrow[line width=1pt]{>}},
   mark=at position 3.85cm with {\arrow[line width=1pt]{>}},
   mark=at position 7cm with {\arrow[line width=1pt]{>}}
   }
   ]
   % The axes
   \draw[help lines,->] (-3,0) -- (3,0) coordinate (xaxis);
   \draw[help lines,->] (0,-1) -- (0,3) coordinate (yaxis);
   % The path
   \path[draw,line width=0.8pt,postaction=decorate] (-2,0) node[below] {$\minus 
   R$} -- (-0.5,0) node[below] {$\minus r$} -- (-0.5,0) arc (180:0:0.5) -- 
   (0.5,0) node[below] {$r$} -- (2,0) node[below] {$R$} arc (0:180:2);
   % The labels
   \node[below] at (xaxis) {$x$};
   \node[left] at (yaxis) {$y$};
   \node at (0.5,0.7) {$c_{r}$};
   \node at (0.5,2.3) {$C_{R}$};
   \node at (-0.2,1) {$i$};
   \draw[red,fill=red] (0,0) circle (.3ex);
   \draw[green,fill=green] (0,1) circle (.3ex);
   \end{tikzpicture}
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Combining the substituion $\ln x =-t$, the fact that $\cos t$ is an even function and $\sin t$ an odd one we get \begin{equation*} I = \int_{0}^{1}\cos(\ln x)\dfrac{1}{1+x^2}\, dx = \int_{0}^{\infty}\cos t\dfrac{e^{-t}}{1+e^{-2t}}\, dt = \dfrac{1}{4}\int_{-\infty}^{\infty} e^{-i\omega t}\dfrac{1}{\cosh t}\, dt. \end{equation*} But it well known that the Fourier transform of $\dfrac{1}{\cosh t}$ is $\dfrac{\pi}{\cosh\left(\frac{\pi \omega}{2}\right)}$, where the Fourier transform is given by \begin{equation*} \hat{f}(\omega) = \int_{-\infty}^{\infty}e^{-i\omega t}f(t)\, dt. \end{equation*} Consequently \begin{equation*} I = \dfrac{\pi}{4\cosh\left(\frac{\pi}{2}\right)}. \end{equation*}

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  • $\begingroup$ The fourier is shown using contour integration. (+1) nevertheless. $\endgroup$ – Zaid Alyafeai May 10 '17 at 17:41
  • $\begingroup$ $@$Zaid Alyafeai Thanks. You are right. Integration along a rectangle with corners in $\pm R$ and $\pm R+i\pi$ will work. $\endgroup$ – JanG May 11 '17 at 13:25
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Another approach, Note that

$$2 \int^1_0 \frac{\cos(\log x) }{x^2+1}\,dx = \int^\infty_0 \frac{\cos(\log x)}{x^2+1}\,dx$$

Now we use that

$$x^i= e^{i\log(x)} = \cos(\log x)+i \sin(\log x)$$

Consider the integral

$$\int^\infty_0 \frac{x^i}{x^2+1}\,dx$$

Let $x^2 = y $

$$\frac{1}{2}\int^\infty_0 \frac{y^{(i-1)/2}}{y+1}\,dy$$

Note that

$$\int^\infty_0 \frac{t^{x-1}}{(t+1)^{x+y}}\,dt = B(x,y)$$

We conclude that

$$\frac{1}{2}\int^\infty_0 \frac{y^{(i-1)/2}}{y+1}\,dy= \Gamma\left( \frac{i+1}{2}\right)\Gamma\left(1-\frac{i+1}{2} \right) = \pi \csc\left(\frac{i+1}{2}\pi \right) = \frac{\pi}{2}\mathrm{sech}\left(\frac{\pi}{2} \right)$$

Hence

$$ \int^1_0 \frac{\cos(\log x) }{x^2+1}\,dx = \frac{1}{2} \Re \int^\infty_0 \frac{x^i}{x^2+1}\,dx =\frac{\pi}{4}\mathrm{sech}\left(\frac{\pi}{2} \right) $$

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}\cos\pars{\ln\pars{x}}\,{\dd x \over 1 + x^{2}} & = \Re\int_{0}^{1}\expo{\ic\ln\pars{x}}\,{\dd x \over 1 + x^{2}} = \Re\int_{0}^{1}{x^{\ic} \over 1 + x^{2}}\,\dd x = \Re\int_{0}^{1}{x^{\ic} - x^{\ic + 2} \over 1 - x^{4}}\,\dd x \\[5mm] & = {1 \over 4}\,\Re\int_{0}^{1}{x^{\ic/4 -3/4} - x^{\ic/4 - 1/4} \over 1 - x}\,\dd x \\[5mm] & = {1 \over 4}\,\Re\pars{% \int_{0}^{1}{1 - x^{\ic/4 - 1/4} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{\ic/4 - 3/4} \over 1 - x}\,\dd x} \\[5mm] & = {1 \over 4}\,\Re\pars{H_{\ic/4 - 1/4} - H_{\ic/4 - 3/4}}\qquad \pars{~H_{z}:\ Harmonic\ Number~} \\[5mm] & = {1 \over 4}\ \overbrace{\Re\pars{H_{\ic/4 - 1/4} - H_{-\ic/4 - 1/4}}}^{\ds{=\ 0}}\ -\ {1 \over 4}\ \underbrace{\Re\pars{\pi\cot\pars{\pi\bracks{{3 \over 4} - {\ic \over 4}}}}} _{\ds{\substack{\ds{=\ -\,{\pi \over \cosh\pars{\pi/2}}\\ {\tiny after\ a\ tedious\ algebraic\ task}}}}} \\[5mm] & = \bbx{{1 \over 4}\,\pi\,\mrm{sech}\pars{\pi \over 2}} \end{align}

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  • $\begingroup$ Congrats on +50k reputation. $\endgroup$ – Zaid Alyafeai May 19 '17 at 11:22
  • $\begingroup$ @ZaidAlyafeai Thanks. The last step is somehow tricky. $\endgroup$ – Felix Marin May 19 '17 at 18:44
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Let $$ f(z)=\frac{2z+1}{(2z+1)^2+1} $$ and hence $f(z)$ has two poles $z_{1,2}=-\frac12\pm\frac12 i$. Using $t=-\ln x$, then one has \begin{eqnarray} &&\int_{0}^{1}\cos(\ln x)\cdot{\mathrm dx\over 1+x^2}\\ &=&\int_{0}^{\infty}\cos(t)\cdot{e^{-t}\mathrm dt\over 1+e^{-2t}}\\ &=&\int_{0}^{\infty}\sum_{n=0}^\infty (-1)^ne^{-(2n+1)t}\cos(t)\mathrm dt\\ &=&\sum_{n=0}^\infty (-1)^n\frac{2n+1}{(2n+1)^2+1}\\ &=&\frac12\sum_{n=-\infty}^\infty (-1)^n\frac{2n+1}{(2n+1)^2+1}\\ &=&-\frac12\bigg[\text{Res}(f(z)\csc(\pi z),z_1)+\text{Res}(f(z)\csc(\pi z),z_2)\bigg]\\ &=&{\pi\over 4}\cdot{1\over \cosh\left({\pi\over 2}\right)}. \end{eqnarray} Here $$ \sum_{n=-\infty}^{\infty}(-1)^nf(n)=-\sum_{k=1}^n \text{Res}(f(z)\csc(\pi z),a_k) $$ where $a_1,a_2,\cdots, a_n$ are poles of $f$.

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  • $\begingroup$ Nice approach... $\endgroup$ – Zaid Alyafeai May 10 '17 at 14:50
  • $\begingroup$ @ZaidAlyafeai, thanks. Your approach is nice too. $\endgroup$ – xpaul May 10 '17 at 14:56
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    $\begingroup$ Could you explain why it is legitimate to use the last formula? Because what I have seen there is an assumption on $f$ that $|f(z)|\le\frac{M}{|z|^k}$, $k>1$, for large $|z|$, to get the integral vanish, but this $f$ has $k=1$. $\endgroup$ – A.Γ. May 10 '17 at 19:18
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    $\begingroup$ In your link they say "If we now demand that both $F(z)$ and $G(z)$ vanish on a rectangular contour enclosing all the poles..." This is exactly what I am wondering about - how do we know that $G(z)=\pi f(z)\csc(\pi z)$ vanishes on the large rectangular contour? $\endgroup$ – A.Γ. May 10 '17 at 20:46
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    $\begingroup$ @mickep, sorry about that. $\endgroup$ – xpaul May 11 '17 at 13:21

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