answer-to-25955-we-introduce-the-parametric-function-F-t-0-ln-1-1-x-2-t-1-x-2-1-dx-after-verifying-that-F-is-derivable-on-0-we-find-F-t-0-1-1-x-2-t-1-dx-F-

Question Number 25960 by abdo imad last updated on 16/Dec/17

a n s w e r t o 25955 . w e i n t r o d u c e t h e p a r a m e t r i c f u n c t i o n

F (t) = \int_{0}^{\infty} l n (1 + (1 + x^{2_{}}) t) {(1 + x^{2})}^{- 1} d x a f t e r v e r i f y i n g t h a t

F i s d e r i v a b l e o n [0 . \propto [w e f i n d \partial F / \partial t = \int_{0}^{\infty} ({(1 + (1 + x^{2}) t)}^{- 1} d x

\partial F / \partial t = 1 / 2 \int_{R} {({t x}^{2} + t + 1)}^{- 1} d x w e p u t f (z) = {({t z}^{2} + z + 1)}^{- 1}

l e t f i n d t h e p o l e s o f f . . {t z}^{2} + z + 1 = 0 < - > z = + - i {((t + 1) t^{- 1})}^{1 / 2}

a n d t h e p o l e s a r e z_{0} = i {((t + 1) t^{- 1})}^{1 / 2} a n d z_{1} = - i {((t + 1) t^{- 1})}^{1 / 2}

a n d f (t) = {(t (t - z_{0}) (t - z_{1}))}^{- 1} b y r e s i d u s t h e o r e m

\int_{R} f (z) d z = 2 i π R (f . z_{0}) = 2 i π {(t (z_{0} - z_{1}))}^{- 1}

= π t^{- 1 / 2} {(t + 1)}^{- 1 / 2} - > \partial F / \partial t = π 2^{- 1} t^{- 1 / 2} {(1 + t)}^{- 1 / 2}

- > F (t) = π 2^{- 1} \int_{0}^{t} x^{- 1 / 2} {(1 + x)}^{- 1 / 2} d x + α

α = F (0) = 0 a n d F (t) = π 2^{- 1} \int_{0}^{t} x^{- 1 / 2} {(1 + x)}^{- 1 / 2} d x

a n d b y t h e c h a n g e m e n t x^{1 / 2} = u w e f i n d

F (t) = π l n (t^{1 / 2} + {(1 + t)}^{1 / 2}) s o \int_{0}^{\infty} l n (2 + x^{2}) {(1 + x^{2})}^{- 1} d x = F (1) = π l n (1 + 2^{1 / 2})