calculate-I-0-pi-4-arctanx-1-x-and-J-0-pi-4-arctanx-1-x-dx-

Question Number 44002 by maxmathsup by imad last updated on 19/Sep/18

c a l c u l a t e I = \int_{0}^{\frac{π}{4}} \frac{a r c t a n x}{1 + x} a n d J = \int_{0}^{\frac{π}{4}} \frac{a r c t a n x}{1 - x} d x

Commented by maxmathsup by imad last updated on 24/Sep/18

l e t f (α) = \int_{0}^{\frac{π}{4}} \frac{a r c t a n (α x)}{1 + x} d x \Rightarrow I = f (1) w e h a v e

f^{^{'}} (α) = \int_{0}^{\frac{π}{4}} \frac{x d x}{(1 + x) (1 + α^{2} x^{2})} =_{α x = t} \int_{0}^{\frac{α π}{4}} \frac{t d t}{α (1 + \frac{t}{α}) (1 + t^{2})}

= \int_{0}^{\frac{α π}{4}} \frac{t d t}{(t + α) (t^{2} + 1)} l e t d e c o m p o s e F (t) = \frac{t}{(t + α) (t^{2} + 1)}

F (t) = \frac{a}{t + α} + \frac{b t + c}{t^{2} + 1} w e h a v e a = {l i m}_{t \to - α} (t + α) F (t) = \frac{- α}{1 + α^{2}}

{l i m}_{t \to + \infty} t F (t) = 0 = a + b \Rightarrow b = \frac{α}{1 + α^{2}} \Rightarrow F (t) = - \frac{α}{(1 + α^{2}) (t + α)} + \frac{\frac{α}{1 + α^{2}} t + c}{1 + t^{2}}

F (0) = 0 = - \frac{α}{(1 + α^{2}) α} + c \Rightarrow c = \frac{1}{1 + α^{2}} \Rightarrow

F (t) = - \frac{α}{(1 + α^{2}) (t + α)} + \frac{1}{1 + α^{2}} \frac{α t + 1}{t^{2} + 1} \Rightarrow

f^{^{'}} (α) = \frac{- α}{1 + α^{2}} \int_{0}^{\frac{α π}{4}} \frac{d t}{t + α} + \frac{1}{1 + α^{2}} \int_{0}^{\frac{α π}{4}} \frac{α t + 1}{t^{2} + 1} d t b u t

\int_{0}^{\frac{α π}{4}} \frac{d t}{t + α} = {[l n ∣ t + α ∣]}_{0}^{\frac{α π}{4}} = l n ∣ \frac{α π}{4} + α ∣ - l n ∣ α ∣ = l n ∣ 1 + \frac{π}{4} ∣

\int_{0}^{\frac{α π}{4}} \frac{α t + 1}{t^{2} + 1} d t = \frac{α}{2} {[l n (t^{2} + 1)]}_{0}^{\frac{α π}{4}} + a r c t a n (\frac{α π}{4})

= \frac{α}{2} l n (1 + \frac{α^{2} π^{2}}{16}) + a r c t a n (\frac{α π}{4}) \Rightarrow

f^{^{'}} (α) = \frac{- α}{1 + α^{2}} l n (1 + \frac{π}{4}) + \frac{α}{2 (1 + α^{2})} l n (1 + \frac{α^{2} π^{2}}{16}) + \frac{1}{1 + α^{2}} a r c t a n (\frac{α π}{4}) \Rightarrow

f (α) = - l n (1 + \frac{π}{4}) \int_{0}^{α} \frac{x}{1 + x^{2}} d x + \frac{1}{2} \int_{0}^{α} \frac{x}{1 + x^{2}} l n (1 + \frac{x^{2} π^{2}}{16}) d x + \int_{0}^{α} \frac{1}{1 + x^{2}} a r c t a n (\frac{π x}{4}) d x

+ c (c = f (0) = 0) \Rightarrow

f (1) = - l n (1 + \frac{π}{4}) \int_{0}^{1} \frac{x}{1 + x^{2}} d x + \frac{1}{2} \int_{0}^{1} \frac{x}{1 + x^{2}} l n (1 + \frac{x^{2} π^{2}}{16}) d x

+ \int_{0}^{1} \frac{a r c t a n (\frac{π x}{4})}{1 + x^{2}} d x \dots . b e c o n t i n u e d \dots