let-I-0-1-ln-1-x-1-x-2-dx-and-J-0-1-2-x-1-x-2-1-xy-dxdy-find-J-by-two-method-and-deduce-the-valueof-I-

Question Number 78707 by abdomathmax last updated on 20/Jan/20

l e t 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

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

Answered by mind is power last updated on 20/Jan/20

J = \int_{0}^{1} \frac{1}{1 + x^{2}} \int_{0}^{1} \frac{x}{1 + xy} dydx

= \int_{0}^{1} \frac{1}{1 + x^{2}} {[\ln (1 + xy)]}_{0}^{1} dx

= \int_{0}^{1} \frac{\ln (1 + x)}{1 + x^{2}} = I

By Fubini

J = \int_{0}^{1} \int_{0}^{1} \frac{x}{(1 + x^{2}) (1 + xy)} dxdy

\frac{x}{(1 + x^{2}) (1 + xy)} = \frac{- y}{1 + y^{2}} . \frac{1}{1 + {xy}^{}} + \frac{cx + d}{1 + x^{2}}

d = \frac{y}{1 + y^{2}}

\frac{1}{1 + y^{2}} = c \Rightarrow \frac{x}{(1 + x^{2}) (1 + xy)} = \frac{1}{1 + y^{2}} (\frac{- y}{1 + xy} + \frac{x + y}{1 + x^{2}})

J = \int_{0}^{1} (\frac{1}{1 + y^{2}} (\int_{0}^{1} {\frac{- y}{1 + xy} + \frac{x + y}{1 + x^{2}}} dx) dy)

= \int_{0}^{1} (\frac{1}{1 + y^{2}} {[_{0}^{1} - \ln (1 + xy) + \frac{\ln (1 + x^{2})}{2} + yarctan (x)] dy

= \int_{0}^{1} {\frac{- \ln (1 + y)}{1 + y^{2}} + \frac{\ln (2)}{2 (1 + y^{2})} + \frac{y}{1 + y^{2}} . \frac{π}{4}} dy

J = - \int_{0}^{1} \frac{\ln (1 + y) dy}{1 + y^{2}} + \frac{\ln (2)}{2} \int_{0}^{1} \frac{dy}{1 + y^{2}} + \frac{π}{4} \int_{0}^{1} \frac{y}{1 + y^{2}} dy = I by first

\Rightarrow - I + \frac{\ln (2)}{2} . \frac{π}{4} + \frac{π}{4} {[\frac{\ln (1 + y^{2})}{2}]}_{0}^{1} = I

\Rightarrow 2 I = \frac{π \ln (2)}{4} \Rightarrow I = \frac{π \ln (2)}{8} = π \ln (2^{\frac{1}{8}})