Question

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

• A. $4$
• B. $\pi$
• C. $8$
• D. $2\pi$

Answer 1

Answer: (C)

$8$

Let $I_1={\int }_0^{\infty }\frac{\log (1+x^2)}{1+x^2}dx$ and $I_2={\int }_0^1\frac{\log (1+x)}{1+x^2}dx$
Put $x=\tan \theta$ in both integral
$\Rightarrow dx={\sec }^2\theta d\theta$
$I_1={\int }_0^{\frac{\pi }{2}}\frac{\log \left({\sec }^2\theta \right)}{{\sec }^2\theta }\cdot {\sec }^2\theta d\theta =-{\int }_0^{\frac{\pi }{2}}\log {\cos }^2\theta d\theta ......\left(1\right)$
Also using $\left({\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$
$I_1=-{\int }_0^{\frac{\pi }{2}}\log {\sin }^2\theta d\theta ........\left(2\right)$
Adding (1) and (2) we get $2I_1=-2{\int }_0^{\frac{\pi }{2}}\log \left(\sin \theta \cos \theta \right)d\theta$
$\Rightarrow I_1=-{\int }_0^{\frac{\pi }{2}}\log \frac{\sin 2\theta }{2}d\theta =\log 2{\int }_0^{\frac{\pi }{2}}d\theta -{\int }_0^{\frac{\pi }{2}}\log \left(\sin 2\theta \right)d\theta$
Put $2\theta =x\Rightarrow 2d\theta =dx$
$\Rightarrow I_1=\frac{\pi }{2}\log 2-\frac{1}{2}{\int }_0^{\pi }\log \left(\sin \theta \right)d\theta =\frac{\pi }{2}\log 2-{\int }_0^{\frac{\pi }{2}}\log \left(\sin \theta \right)d\theta =\frac{\pi }{2}\log 2+\frac{I_1}{2}$, using (2)
$\therefore I_1=\pi \log 2$
Now $I_2={\int }_0^{\frac{\pi }{4}}\log (1+\tan \theta )d\theta$
$\Rightarrow I_2={\int }_0^{\frac{\pi }{4}}\log (1+\tan (\frac{\pi }{4}-\theta \left)\right)d\theta ={\int }_0^{\frac{\pi }{4}}\log (1+\frac{1-\tan \theta }{1+\tan \theta })d\theta$
$\Rightarrow I_2={\int }_0^{\frac{\pi }{4}}\log \left(\frac{2}{1+\tan \theta }\right)d\theta =\log 2{\int }_0^{\frac{\pi }{4}}d\theta -{\int }_0^{\frac{\pi }{4}}\log (1+\tan \theta )d\theta$
$\Rightarrow I_2=\frac{\pi }{4}\log 2-I_2\Rightarrow I_2=\frac{\pi }{8}\log 2$
Hence $\lambda =\frac{I_1}{I_2}=8$