Question

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

• A. $\pi$ log2
• B. $-\pi$ log2
• C. $-\frac{\pi }{2}$ log2
• D. $\frac{\pi }{2}$ log2

Answer 1

The correct option is A $\pi$ log2

Let $I={\int }_0^{\infty }\frac{\log (1+x^2)}{1+x^2}dx$

Substitute $x=\tan \theta$

Then, $I={\int }_0^{\pi /2}-2\log \left(\cos \theta \right)d\theta$

Let $L={\int }_0^{\pi /2}\log \left(\cos \theta \right)d\theta$ .....(1)

Using $(a+b-x)$ rule,

$L={\int }_0^{\pi /2}\log \left(\sin \theta \right)d\theta$ .....(2)

Adding (1) and (2), we get

$L={\int }_0^{\pi /2}\frac{1}{2}\log \left(\sin \theta \cos \theta \right)d\theta$

$L=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\log \left(\sin 2\theta \right)d\theta -\frac{\pi }{4}\log \left(2\right)$

Substituting $2\theta =\frac{\pi }{2}-t$, we get

$L=\frac{1}{4}{\int }_{-\pi /2}^{\pi /2}\log \left(\cos t\right)dt-\frac{\pi }{4}\log \left(2\right)=\frac{1}{2}{\int }_0^{\pi /2}\log \left(\cos t\right)dt-\frac{\pi }{4}\log \left(2\right)$

$\Rightarrow L=\frac{L}{2}-\frac{\pi }{4}\log \left(2\right)$

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

Since, $I=-2L$

$\Rightarrow I=\pi \log \left(2\right)$

So, option A is correct.