Integral Thing 2

$\begin{aligned} I & = \int_0^\infty \left({\color{#3D99F6}\frac \pi 2 - \arctan x}\right)^2 dx & \small \color{#3D99F6} \text{See Note 1.} \\ & = \int_0^\infty \left({\color{#3D99F6}\arctan \frac 1x}\right)^2 dx & \small \color{#3D99F6} \text{Let }u=\arctan \frac 1x \text{ See Note 2.} \\ & = \int_0^\frac \pi 2 u^2 \csc^2 u \ du & \small \color{#3D99F6} \text{By integration by parts} \\ & = {\color{#D61F06}-u^2\cot u \bigg|_0^\frac \pi 2} + 2 \int_0^\frac \pi 2 u \cot u \ du & \small \color{#D61F06} \text{See Note 3 and 4.} \\ & = {\color{#D61F06}0} + {\color{#D61F06}2u \ln(\sin u) \bigg|_0^\frac \pi 2} - 2 \int_0^\frac \pi 2 \ln(\sin u) \ du \\ & = {\color{#D61F06}0} -\color{#3D99F6} 2 \int_0^\frac \pi 2 \ln(\sin u) \ du & \small \color{#3D99F6} \implies \int_0^\frac \pi 2 \ln(\sin u) \ du = - \frac I2 \\ & = \color{#3D99F6} \int_0^\frac \pi 2 \left(\ln(\sin u) + \ln(\cos u)\right) du & \small \color{#3D99F6} \text{By identity: }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = - \int_0^\frac \pi 2 \ln \left(\frac {\sin 2u}2\right) du \\ & = \int_0^\frac \pi 2 \ln 2 \ du - \int_0^\frac \pi 2 \ln (\sin 2u)\ du & \small \color{#3D99F6} \text{Let }v =2u \implies dv = 2du \\ & = u \ln 2 \bigg|_0^\frac \pi 2 - \frac 12 \int_0^\pi \ln (\sin v)\ dv & \small \color{#3D99F6} \text{Since }\sin \theta \text{ is symmetrical about }\frac \pi 2 \\ & = \frac {\pi \ln 2}2 - \color{#3D99F6} \int_0^\frac \pi 2 \ln (\sin v)\ dv & \small \color{#3D99F6} \text{Note that } \int_0^\frac \pi 2 \ln(\sin v) \ dv = - \frac I2 \\ & = \frac {\pi \ln 2}2 + \color{#3D99F6} \frac I2 \\ \implies \frac I2 & = \frac {\pi \ln 2}2 \\ I & = \pi \ln 2 \end{aligned}$

$\implies a+b = 1+2 = \boxed{3}$

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Note 1: Let $\theta = \arctan x$ , then $\tan \left(\dfrac \pi 2 - \arctan x \right) = \tan \left(\dfrac \pi 2 - \theta \right) = \cot \theta = \dfrac 1{\tan \theta} = \dfrac 1x$ . This implies that $\dfrac \pi 2 - \arctan x = \arctan \dfrac 1x$ .

Note 2: $u = \arctan \dfrac 1x$ , $\implies \tan u = \dfrac 1x$ , $\implies \sec^2 u \ du = - \dfrac 1{x^2} dx$ , $\implies dx = - \dfrac {1+\tan^2 u}{\tan^2 u} du = -\csc^2 u$ , $\implies \displaystyle \int_0^\infty \left(\arctan \frac 1x\right)^2 dx = \int_\frac \pi 2^0 -u^2\csc^2 u \ du = \int^\frac \pi 2_0 u^2\csc^2 u \ du$

Note 3: $\displaystyle u^2\cot u \bigg|_0^\frac \pi 2 = \frac {\pi^2 \cot \frac \pi 2}4 - \lim_{u \to 0} \cancel{\frac u{\sin u}}^1\times u\cos u = 0 - 1(0)(1) = 0$

Note 4:

$\begin{aligned} u \ln(\sin u) \bigg|_0^\frac \pi 2 & = \frac \pi 2 \ln\left(\sin \frac \pi 2\right) - \color{#3D99F6} \lim_{u \to 0} \frac {\ln(\sin u)}{\frac 1u} & \small \color{#3D99F6} \text{A } \infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = 0 - \color{#3D99F6} \lim_{u \to 0} \frac {\frac {\cos u}{\sin u}}{-\frac 1{u^2}} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \lim_{u \to 0} \cancel{\frac u{\sin u}}^1 \times u\cos u = 0 \end{aligned}$