How can we show that $\int_{0}^{1}\cos(\ln x)\cdot{\mathrm dx\over 1+x^2}={\pi\over 4}\cdot{1\over \cosh\left({\pi\over 2}\right)}?$

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.

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

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}

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*}

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

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