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π ln 2
−π ln 2
0
\[- \frac{\pi}{2}\ln 2\]
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Solution
π ln 2
\[\int_0^\infty \log \left( x + \frac{1}{x} \right) \frac{1}{1 + x^2}dx\]
Substitute x = tan θ
⇒ dx = sec2 θ dθ.
when,
x = 0 ⇒ θ = 0
\[x = \infty \Rightarrow \theta = \frac{\pi}{2}\]
\[ \int_0^\frac{\pi}{2} \left( \tan \theta + \frac{1}{\tan \theta} \right)\frac{1}{1 + \tan^2 \theta} \times \sec^2 \theta d\theta\]
\[ \int_0^\frac{\pi}{2} \log \left( \frac{\tan^2 \theta + 1}{\tan\theta} \right) \frac{1}{1 + \tan^2 \theta} \times \sec^2 \theta d\theta\]
\[ \Rightarrow \int_0^\frac{\pi}{2} \log \left( \frac{\sec^2 \theta}{\tan \theta} \right)\frac{1}{\sec^2 \theta} \times \sec^2 \theta d\theta ................\left[ \because 1 + \tan^2 \theta = \sec^2 \theta \right]\]
\[ \Rightarrow \int_0^\frac{\pi}{2} \log \left( \frac{\sec^2 \theta}{\tan \theta} \right)d\theta\]
\[ \Rightarrow \int_0^\frac{\pi}{2} \log \left( \frac{1}{\sin \theta . \cos \theta} \right)d\theta\]
\[ \Rightarrow - \int_0^\frac{\pi}{2} \log \left( \sin \theta . \cos \theta \right)d\theta\]
\[ \Rightarrow - \int_0^\frac{\pi}{2} \left[ \log \sin \theta + \log \cos \theta \right]d\theta\]
\[ \Rightarrow - \int_0^\frac{\pi}{2} \log \sin \theta d\theta - \int_0^\frac{\pi}{2} \log \cos \theta d\theta\]
Let us consider,
\[\int_0^\frac{\pi}{2} \log \sin \theta d\theta = I .................(1)\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \log \left( \sin \left( \frac{\pi}{2} - \theta \right) \right)d\theta\]
\[ = \int_0^\frac{\pi}{2} \log \cos \theta d\theta ..................(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_0^\frac{\pi}{2} \log \sin \theta d\theta + \int_0^\frac{\pi}{2} \log \cos \theta d\theta\]
\[ = \int_0^\frac{\pi}{2} \log \left( \sin \theta . \cos \theta \right)d\theta\]
\[ = \int_0^\frac{\pi}{2} \log \left( \sin 2\theta \right)d\theta - \int_0^\frac{\pi}{2} \log 2d\theta\]
\[\text{Let us consider } 2\theta = t\]
\[2d\theta = dt\]
\[2I = \frac{1}{2} \int_0^\pi \log \left( \sin t \right)dt - \frac{\pi}{2}\log 2\]
\[2I = \frac{2}{2} \int_0^\frac{\pi}{2} \log \left( \sin t \right)dt - \frac{\pi}{2}\log 2 ................\left[ \because \sin \theta \text{ is positive in both } 1^{st} \text{ and }2^{nd} \text{ quadrants} \right]\]
\[2I = I - \frac{\pi}{2}\log 2\]
\[2I - I = - \frac{\pi}{2}\log 2\]
\[I = - \frac{\pi}{2}\log 2, where I = \int_0^\frac{\pi}{2} \log \sin \theta d\theta\]
\[Now, \]
\[ - \int_0^\frac{\pi}{2} \log\left( \sin \theta \right)d\theta - \int_0^\frac{\pi}{2} \log \cos \theta d\theta\]
\[ - 2 \int_0^\frac{\pi}{2} \log \sin \theta d\theta = - 2 \times I\]
\[ = - 2 \times - \frac{\pi}{2}\log 2 .............\left[ \because \text{where} I = - \frac{\pi}{2}\log2 \right]\]
\[ = \pi \log 2\]
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