# Question #d811b

Oct 4, 2017

$\pi \cdot \log 2$

#### Explanation:

1) I called initial integral as $I$.

2) I used $x = \tan y$ and $\mathrm{dx} = {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$ transforms.

3) I used $\tan y + \cot y = \frac{2}{\sin 2 y}$ identity.

4) I called second integral as $J$.

5) I used $z = 2 y$ and $\mathrm{dz} = 2 \mathrm{dy}$ transformations in $J$.

6) I used symmetry of the sine function about $\frac{\pi}{2}$:

${\int}_{0}^{\pi} f \left(\sin z\right) \cdot \mathrm{dz}$=${\int}_{0}^{\frac{\pi}{2}} 2 f \left(\sin z\right) \cdot \mathrm{dz}$

7) I used ${\int}_{0}^{\frac{\pi}{2}} f \left(z\right) \cdot \mathrm{dz}$=${\int}_{0}^{\frac{\pi}{2}} f \left(\frac{\pi}{2} - z\right) \cdot \mathrm{dz}$ identity.

8) I collected 2 $J$'s.

9) I used $\log m + \log n = \log \left(m \cdot n\right)$ and $\log \left(\frac{a}{b}\right) = \log a - \log b$ identities.

10) I found value of $J$.

11) I found value of $I$.

$I$=${\int}_{0}^{\infty} \log \left(x + \frac{1}{x}\right) \cdot \frac{\mathrm{dx}}{1 + {x}^{2}}$

After using $x = \tan y$ and $\mathrm{dx} = {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$ transforms,

$I$=${\int}_{0}^{\frac{\pi}{2}} \log \left(\tan y + \cot y\right) \cdot \mathrm{dy}$

=${\int}_{0}^{\frac{\pi}{2}} \log \left(\frac{2}{\sin 2 y}\right) \cdot \mathrm{dy}$

=$\frac{\pi}{2} \cdot \log 2$-${\int}_{0}^{\frac{\pi}{2}} \log \left(\sin 2 y\right) \cdot \mathrm{dy}$

I called new integral as $J$,

$J$=${\int}_{0}^{\frac{\pi}{2}} \log \left(\sin 2 y\right) \cdot \mathrm{dy}$

After using $z = 2 y$ and $\mathrm{dz} = 2 \mathrm{dy}$ transformations,

$J$=${\int}_{0}^{\pi} \frac{1}{2} \cdot \log \left(\sin z\right) \cdot \mathrm{dz}$

After using symmetry of the sine function about $\frac{\pi}{2}$,

$J$=${\int}_{0}^{\frac{\pi}{2}} \frac{1}{2} \cdot 2 \cdot \log \left(\sin z\right) \cdot \mathrm{dz}$

=${\int}_{0}^{\frac{\pi}{2}} \log \left(\sin z\right) \cdot \mathrm{dz}$

=${\int}_{0}^{\frac{\pi}{2}} \log \left[\sin \left(\frac{\pi}{2} - z\right)\right] \cdot \mathrm{dz}$

=${\int}_{0}^{\frac{\pi}{2}} \log \left(\cos z\right) \cdot \mathrm{dz}$

After collecting 2 integrals,

$2 J$=${\int}_{0}^{\frac{\pi}{2}} \log \left(\sin z\right) \cdot \mathrm{dz}$+${\int}_{0}^{\frac{\pi}{2}} \log \left(\cos z\right) \cdot \mathrm{dz}$

=${\int}_{0}^{\frac{\pi}{2}} \log \left(\sin z \cdot \cos z\right) \cdot \mathrm{dz}$

=${\int}_{0}^{\frac{\pi}{2}} \log \left(\frac{\sin 2 z}{2}\right) \cdot \mathrm{dz}$

=${\int}_{0}^{\frac{\pi}{2}} \log \left(\sin 2 z\right) \cdot \mathrm{dz}$-$\frac{\pi}{2} \cdot \log 2$

=$J$-$\frac{\pi}{2} \cdot \log 2$

Hence $J = - \frac{\pi}{2} \cdot \log 2$

Thus,

$I$=${\int}_{0}^{\infty} \log \left(x + \frac{1}{x}\right) \cdot \frac{\mathrm{dx}}{1 + {x}^{2}}$

=$\frac{\pi}{2} \cdot \log 2 - J$

=$\frac{\pi}{2} \cdot \log 2 - \left(- \frac{\pi}{2} \cdot \log 2\right)$

=$\pi \cdot \log 2$