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Question Number 138283 by mnjuly1970 last updated on 11/Apr/21

\dots \dots a d v a n c e d \dots \dots \dots . . c a l c u l u s \dots \dots

p r o v e t h a t ::

ϕ = \int_{0}^{1} \frac{l n (1 + x^{2}) . a r c t a n (x)}{x^{2}} d x =

p r o o f :::

ϕ \underset{⟨ s u b s t i t u t i o n ⟩}{\overset{x = t a n (θ)}{=}} \int_{0}^{\frac{π}{4}} \frac{l n (1 + {t a n}^{2} (θ)) . θ}{{t a n}^{2} (θ)} (1 + {t a n}^{2} (θ)) d θ

\overset{⟨ s i m p l i f i c a t i o n ⟩}{=} \int_{0}^{\frac{π}{4}} \frac{θ . l n (\frac{1}{{c o s}^{2} (θ)})}{{s i n}^{2} (θ)} d θ

= - 2 \int_{0}^{\frac{π}{4}} \frac{θ . l n (c o s (θ)}{{s i n}^{2} (θ)} d θ

\overset{i . b . p}{=} 2 {[{(c o t (θ) . θ . l n (c o s (θ))]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} (c o t (θ) . [l n (c o s (θ)) - θ . t a n (θ)] d θ

= 2 . \frac{π}{4} . l n (\frac{\sqrt{2}}{2}) - 2 \int_{0}^{\frac{π}{4}} c o t (θ) . l n (c o s (θ)) d θ + 2 \int_{0}^{\frac{π}{4}} θ d θ

= \frac{- π}{4} l n (2) - Φ + \frac{π^{2}}{16}

Φ = \int_{0}^{\frac{π}{4}} \frac{c o s (θ)}{s i n (θ)} . l n (1 - {s i n}^{2} (θ)) d θ

\overset{s i n (θ) = y}{=} \int_{0}^{\frac{\sqrt{2}}{2}} \frac{l n (1 - y^{2})}{y} d y = - \int_{0}^{\frac{\sqrt{2}}{2}} \underset{n = 1}{\sum^{\infty}} \frac{y^{2 n - 1}}{n} d y

= - Σ {[\frac{y^{2 n}}{2 n^{2}}]}_{0}^{\frac{\sqrt{2}}{2}} = \frac{- 1}{2} {l i}_{2} (\frac{1}{2})

= \frac{- 1}{2} {\frac{π^{2}}{12} - \frac{1}{2} {l n}^{2} (2)} = \frac{- π^{2}}{24} + \frac{1}{4} {l n}^{2} (2) \dots

∴ ϕ = \frac{- π}{4} l n (2) + \frac{π^{2}}{24} + \frac{1}{4} {l n}^{2} (2) + \frac{π^{2}}{16}

\dots \dots \dots ϕ = \frac{5 π^{2}}{48} - \frac{12 π}{48} l n (2) + \frac{12}{48} {l n}^{2} (2) \dots . .

\dots \dots . . ϕ = \frac{1}{48} {5 π^{2} - 12 π l n (2) + 12 {l n}^{2} (2)}

Commented by Dwaipayan Shikari last updated on 11/Apr/21

N i c e s i r!

Commented by mnjuly1970 last updated on 11/Apr/21

t h a n k s a l o t m r p s a a n \dots . .