Integrate it #1

$\begin{aligned} I_1 & = \int_0^\infty \frac {\ln(1+x^2)}{1+x^2} dx & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = \sec^2 \theta \ d\theta \\ & = \int_0^\frac \pi 2 \ln(\sec^2 \theta) \ d\theta \\ & = - 2 \int_0^\frac \pi 2 \ln(\cos \theta) \ d\theta & \small \color{#3D99F6} \text{By identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = - \int_0^\frac \pi 2 \left(\ln(\cos \theta) + \ln (\sin \theta) \right) \ d\theta \\ & = - \int_0^\frac \pi 2 \ln(\sin \theta \cos \theta) \ d\theta \\ & = - \int_0^\frac \pi 2 \ln \left(\frac {\sin 2 \theta}2 \right) \ d\theta \\ & = - {\color{#3D99F6} \int_0^\frac \pi 2 \ln (\sin 2 \theta) \ d\theta} + \int_0^\frac \pi 2 \ln 2 \ d\theta & \small \color{#3D99F6} \text{Let }\phi = 2\theta \implies d \phi = 2\ d\theta \\ & = - {\color{#3D99F6} \frac 12 \int_0^\pi \ln (\sin \phi) \ d\phi} + \frac {\pi \ln 2}2 & \small \color{#3D99F6} \text{Note that }\sin x \text{ is symmetrical about }\frac \pi 2 \\ & = - {\color{#3D99F6} \int_0^{\color{#D61F06}\frac \pi 2} \ln (\sin \phi) \ d\phi} + \frac {\pi \ln 2}2 & \small \color{#3D99F6} \text{Note that } \int_0^\frac \pi 2 \ln(\sin \theta) \ d\theta = \int_0^\frac \pi 2 \ln(\cos \theta) \ d\theta = - \frac {I_1}2 \\ & = {\color{#3D99F6} \frac {I_1}2} + \frac {\pi \ln 2}2 \\ & = \pi \ln 2 \end{aligned}$

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$\begin{aligned} I_2 & = \int_0^1 \frac {\ln(1+x)}{1+x^2} dx & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = \sec^2 \theta \ d\theta \\ & = \int_0^\frac \pi 4 \ln(1 + \tan \theta) \ d\theta & \small \color{#3D99F6} \text{By identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\frac \pi 4 \left( \ln(1 + \tan \theta) + \ln \left(1 + \tan \left(\frac \pi 4 - \theta\right) \right) \right) d\theta \\ & = \frac 12 \int_0^\frac \pi 4 \left( \ln(1 + \tan \theta) + \ln \left(1 + \frac {1-\tan \theta}{1+\tan \theta} \right) \right) d\theta \\ & = \frac 12 \int_0^\frac \pi 4 \left( \ln(1 + \tan \theta) + \ln \left(\frac 2{1+\tan \theta} \right) \right) d\theta \\ & = \frac 12 \int_0^\frac \pi 4 \left( \ln(1 + \tan \theta) + \ln 2 - \ln (1+\tan \theta) \right) d\theta \\ & = \frac 12 \int_0^\frac \pi 4 \ln 2 \ d\theta \\ & = \frac {\pi \ln 2}8 = \frac 18 I_1 \end{aligned}$

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$\implies \lambda = \boxed{8}$

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

First calculate $I_{2}$ :

Let $\tan^{-1}x=t \implies \dfrac{dx}{1+x^2}=dt$

$I_{2}=\displaystyle\int_{0}^{\frac{π}{4}} \log(1+ \tan t)dt$

$I_{2}$ can be easily calculated by using the following fact:

$\displaystyle\int_{a}^{b} f(x)dx= \displaystyle\int_{a}^{b} f(a+b-x)dx$

$I_{2}=\displaystyle\int_{0}^{\frac{π}{4}} \log(1+ \tan(\frac{π}{4}-t))dt$

$I_{2}=\displaystyle\int_{0}^{\frac{π}{4}} \log(1+ \dfrac{1-\tan t}{1+\tan t}))dt$

$I_{2}=\displaystyle\int_{0}^{\frac{π}{4}} \log(2)dt - I_{2} \implies I_{2}= \dfrac{π}{8}log2$

For $I_{1}$ , first divide it into two intervals :

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

For second interval let $\dfrac{1}{x}=t \implies dx=-\dfrac{1}{t^2}dt$

$I_{1}=\displaystyle\int_{0}^{1} \dfrac{\log(1+x^2)}{1+x^2}dx +\displaystyle\int_{0}^{1} \dfrac{\log(1+\dfrac{1}{t^2})}{t^2(1+\dfrac{1}{t^2})}dt$

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

$I_{1}=2\displaystyle\int_{0}^{1} \dfrac{\log(1+x^2)}{1+x^2}dx -2\displaystyle\int_{0}^{1}\dfrac{logx}{(1+x^2)}dx$

Using the substitution: $\tan^{-1}x=t$ , we get :

$I_{1}=π\log2$

$I_{1}$ can be calculated by using following fact:

$\displaystyle\int_{0}^{\frac{π}{2}} \log(\sin x)dx = -\dfrac{π}{2}\log2$

So, $\dfrac{I_{1}}{I_{2}}=\lambda=\boxed{8}$