find-f-x-0-1-arctan-xt-dt-x-from-R-

Question Number 41135 by math khazana by abdo last updated on 02/Aug/18

f i n d f (x) = \int_{0}^{1} a r c t a n (x t) d t x f r o m R

Answered by math khazana by abdo last updated on 03/Aug/18

w e h a v e f^{^{'}} (x) = \int_{0}^{1} \frac{t}{1 + x^{2} t^{2}} d t

=_{x t = u} \int_{0}^{x} \frac{u}{x (1 + u^{2})} \frac{d u}{x} = \frac{1}{x^{2}} \int_{0}^{x} \frac{u d u}{1 + u^{2}}

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

f (x) = \int_{0}^{x} \frac{l n (1 + t^{2})}{2 t^{2}} d t + c

c = f (0) = 0 \Rightarrow f (x) = \int_{0}^{x} \frac{l n (1 + t^{2})}{2 t^{2}} d t

b y p a r t s 2 f (x) = {[- \frac{1}{t} l n (1 + t^{2})]}_{0}^{x} + \int_{0}^{x} \frac{1}{t} \frac{2 t}{1 + t^{2}} d t

= - \frac{1}{x} l n (1 + x^{2}) + 2 a r c t a n x \Rightarrow

f (x) = a r c t a n (x) - \frac{1}{2 x} l n (1 + x^{2}) .

Commented by math khazana by abdo last updated on 03/Aug/18

l e t p r o v e {l i m}_{t \to 0} \frac{l n (1 + t^{2})}{t} = 0 w e h a v e

l n (1 + t^{2}) \sim t^{2} \Rightarrow \frac{l n (1 + t^{2})}{t} \sim t (t \to 0) r e s u l t i s p r o v e d