let-give-I-0-1-ln-1-x-1-x-2-dx-and-J-0-1-2-x-1-x-2-1-xy-dxdy-calculate-J-by-two-methods-then-find-the-value-of-I-

Question Number 28200 by abdo imad last updated on 21/Jan/18

l e t g i v e I = \int_{0}^{1} \frac{l n (1 + x)}{1 + x^{2}} d x a n d J = \int \int_{{[0, 1]}^{2}} \frac{x}{(1 + x^{2}) (1 + x y)} d x d y

c a l c u l a t e J b y t w o m e t h o d s t h e n f i n d t h e v a l u e o f I .

Commented by abdo imad last updated on 22/Jan/18

w e h a v e J = \int_{0}^{1} (\int_{0}^{1} \frac{x}{1 + x y} d y) \frac{d x}{1 + x^{2}} b u t

\int_{0}^{1} \frac{x}{1 + x y} d y = {[l n (1 + x y)]}_{y = 0}^{y = 1} = l n (1 + x) s o

J = \int_{0}^{1} \frac{l n (1 + x)}{1 + x^{2}} d x f r o m a n o t h e r s i d e

J = \int_{0}^{1} (\int_{0}^{1} \frac{x}{(1 + x^{2}) (1 + x y)} d x) d y l e t d e c o m p o s e

F (x) = \frac{x}{(1 + x^{2}) (1 + x y)} = \frac{α}{1 + x y} + \frac{a x + b}{1 + x^{2}}

α = {l i m}_{x \to \frac{- 1}{y}} (1 + x y) F (x) = \frac{- 1}{y (1 + \frac{1}{y^{2}})} = \frac{- 1}{y + \frac{1}{y}} = \frac{- y}{1 + y^{2}}

{l i m}_{x \to + \propto} x F (x) = 0 = \frac{α}{y} + a \Rightarrow a = - \frac{α}{y} = \frac{1}{1 + y^{2}}

F (x) = \frac{- y}{(1 + y^{2}) (1 + x y)} + \frac{\frac{1}{1 + y^{2}} x + b}{1 + x^{2}}

F (0) = 0 = \frac{- y}{1 + y^{2}} + b \Rightarrow b = \frac{y}{1 + y^{2}} a n d

F (x) = \frac{- y}{(1 + y^{2}) (1 + x y)} + \frac{1}{1 + y^{2} (} \frac{x + y}{1 + x^{2})}

\int_{0}^{1} F (x) d x = \frac{- 1}{1 + y^{2}} \int_{0}^{1} \frac{y d x}{1 + x y} + \frac{1}{2 (1 + y^{2})} \int_{0}^{1} \frac{2 x}{1 + x^{2}} d x + \frac{y}{1 + y^{2}} \int_{0}^{1} \frac{d x}{1 + x^{2}}

= \frac{- 1}{1 + y^{2}} {[l n (1 + x y)]}_{x = 0}^{x = 1} + \frac{1}{2 (1 + y^{2})} {[l n (1 + x^{2})]}_{0}^{1}

+ \frac{π}{4} \frac{y}{1 + y^{2}}

- \frac{l n (1 + y)}{1 + y^{2}} + \frac{l n 2}{2 (1 + y^{2})} + \frac{π y}{4 (1 + y^{2})}

J = - \int_{0}^{1} \frac{l n (1 + y)}{1 + y^{2}} d y + \frac{l n 2}{2} \int_{0}^{1} \frac{d y}{1 + y^{2}} + \frac{π}{8} \int_{0}^{1} \frac{2 y d y}{1 + y^{2}}

2 J = \frac{π}{8} l n 2 + \frac{π}{8} l n 2 = \frac{π}{4} l n 2 \Rightarrow J = \frac{π}{8} l n 2 .