Question

\[\int\limits_0^\infty \log\left( x + \frac{1}{x} \right) \frac{1}{1 + x^2} dx =\] 

विकल्प

• π ln 2

• −π ln 2

• 0

• \[- \frac{\pi}{2}\ln 2\]

Answer 1

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, 

\[\int_0^\frac{\pi}{2} \log \sin \theta d\theta = I .................(1)\]

\[ \Rightarrow I = \int_0^\frac{\pi}{2} \log \left( \sin \left( \frac{\pi}{2} - \theta \right) \right)d\theta\]

\[ = \int_0^\frac{\pi}{2} \log \cos \theta d\theta ..................(2)\]

\[\text{Adding (1) and (2)}\]

\[2I = \int_0^\frac{\pi}{2} \log \sin \theta d\theta + \int_0^\frac{\pi}{2} \log \cos \theta d\theta\]

\[ = \int_0^\frac{\pi}{2} \log \left( \sin \theta . \cos \theta \right)d\theta\]

\[ = \int_0^\frac{\pi}{2} \log \left( \sin 2\theta \right)d\theta - \int_0^\frac{\pi}{2} \log 2d\theta\]

\[\text{Let us consider } 2\theta = t\]

\[2d\theta = dt\]

\[2I = \frac{1}{2} \int_0^\pi \log \left( \sin t \right)dt - \frac{\pi}{2}\log 2\]

\[2I = \frac{2}{2} \int_0^\frac{\pi}{2} \log \left( \sin t \right)dt - \frac{\pi}{2}\log 2 ................\left[ \because \sin \theta \text{ is positive in both } 1^{st} \text{ and }2^{nd} \text{ quadrants} \right]\]

\[2I = I - \frac{\pi}{2}\log 2\]

\[2I - I = - \frac{\pi}{2}\log 2\]

\[I = - \frac{\pi}{2}\log 2, where I = \int_0^\frac{\pi}{2} \log \sin \theta d\theta\]

\[Now, \]

\[ - \int_0^\frac{\pi}{2} \log\left( \sin \theta \right)d\theta - \int_0^\frac{\pi}{2} \log \cos \theta d\theta\]

\[ - 2 \int_0^\frac{\pi}{2} \log \sin \theta d\theta = - 2 \times I\]

\[ = - 2 \times - \frac{\pi}{2}\log 2 .............\left[ \because \text{where} I = - \frac{\pi}{2}\log2 \right]\]

\[ = \pi \log 2\]