Mathematics

$$\displaystyle \int_{0}^{\infty}\frac{\log(1+x^{2})}{1+x^{2}}dx=$$


ANSWER

$$\pi$$ log2


SOLUTION
Let $$I = \displaystyle\int_{0}^{\infty }\dfrac{\log(1+x^{2})}{1+x^{2}}dx$$
Substitute  $$x=\tan\theta $$
Then, $$I=\displaystyle\int_{0}^{\pi/2}-2\log(\cos\theta )d\theta $$
Let $$L=\displaystyle\int_{0}^{\pi/2}\log(\cos\theta )d\theta $$       .....(1)
Using $$(a+b-x)$$ rule,
$$L=\displaystyle\int_{0}^{\pi/2}\log(\sin\theta)d\theta$$           .....(2)
Adding (1) and (2), we get
$$L=\displaystyle\int_{0}^{\pi/2}\dfrac{1}{2}\log(\sin\theta \text{ }\cos\theta) d\theta $$
$$L=\dfrac{1}{2}\displaystyle\int_{0}^{\dfrac{\pi }{2}}\log(\sin 2\theta )d\theta -\dfrac{\pi }{4}\log(2)$$
Substituting $$2\theta =\dfrac{\pi }{2}-t$$, we get
$$L=\dfrac{1}{4}\displaystyle \int_{-\pi/2}^{\pi/2}\log(\cos t)dt-\dfrac{\pi}{4}\log(2)=\dfrac{1}{2}\displaystyle\int_{0}^{\pi/2}\log(\cos t)dt-\dfrac{\pi}{4}\log(2)$$
$$\Rightarrow L=\dfrac{L}{2} -\dfrac{\pi }{4}\log(2)$$
$$\Rightarrow L=-\dfrac{\pi }{2}\log(2)$$
Since, $$I=-2L$$
$$\Rightarrow I=\pi \log(2)$$
So, option A is correct.
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Single Correct Hard Published on 17th 09, 2020
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