find f(x) = ∫_0 ^1 arctan(1+xt)dt with x real


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Question Number 67020 by mathmax by abdo last updated on 21/Aug/19

$f i n d f (x) = \int_{0}^{1} a r c t a n (1 + x t) d t w i t h x r e a l$
Commented by mathmax by abdo last updated on 22/Aug/19
$f(x)=∫_0 ^1 arctan(1+xt)dt by parts f(x) =t arctan(1+xt)]_0 ^1 −∫_0 ^1 t (x/(1+(1+xt)^2 ))dt =arctan(1+x)−x ∫_0 ^1 (t/(1+(1+xt)^2 ))dt but ∫_0 ^1 (t/(1+(1+xt)^2 ))dt =∫_0 ^1 ((tdt)/(1+1+2xt +x^2 t^2 )) =∫_0 ^1 ((tdt)/(x^2 t^2 +2xt +2)) let decompose F(t)=(t/(x^2 t^2 +2xt+2)) Δ^′ =x^2 −2x^2 =−x^2 <0 if x≠0 so F(t) =(t/(x^2 (t^2 +((2t)/x)+(2/x^2 )))) =(t/(x^2 (t^2 +((2t)/x) +(1/x^2 ) +(2/x^2 )−(1/x^2 )))) =(t/(x^2 {(t+(1/x))^2 +(1/x^2 )})) we use the changement t+(1/x) =(1/(∣x∣))u ⇒xt+1=s(x)u ∫_(s(x)) ^((x+1)s(x)) (t/(1+(1+tx)^2 ))dt =∫_(s(x)) ^((x+1)s(x)) ((((1/(∣x∣))u−(1/x)))/(x^2 ×(1/x^2 )(1+u^2 )))du =(1/(∣x∣))∫_(s(x)) ^((x+1)s(x)) (u/(1+u^2 ))−(1/x) ∫_(s(x)) ^((x+1)s(x)) (du/(1+u^2 )) =(1/(2∣x∣))[ln(1+u^2 )]_(s(x)) ^((x+1)s(x)) −(1/x)[arctanu]_(s(x)) ^((x+1)s(x)) (s^2 (x)=1) =(1/(2∣x∣)){ln(1+(x+1)^2 )−ln2}−(1/x){arctan(x+1)s(x)−arctan(s(x))} ⇒f(x)= arctan(1+x)−((s(x))/2){ln(x^2 +2x+2)−ln(2)} +arctan(x+1)s(x)+x arctan(s(x)) with s(x) =1 if x>0 and s(x)=−1 if x<0$
$f (x) = \int_{0}^{1} a r c t a n (1 + x t) d t b y p a r t s$ ${f (x) = t a r c t a n (1 + x t)]}_{0}^{1} - \int_{0}^{1} t \frac{x}{1 + {(1 + x t)}^{2}} d t$ $= a r c t a n (1 + x) - x \int_{0}^{1} \frac{t}{1 + {(1 + x t)}^{2}} d t b u t$ $\int_{0}^{1} \frac{t}{1 + {(1 + x t)}^{2}} d t = \int_{0}^{1} \frac{t d t}{1 + 1 + 2 x t + x^{2} t^{2}} = \int_{0}^{1} \frac{t d t}{x^{2} t^{2} + 2 x t + 2}$ $l e t d e c o m p o s e F (t) = \frac{t}{x^{2} t^{2} + 2 x t + 2}$ $Δ^{^{'}} = x^{2} - 2 x^{2} = - x^{2} < 0 i f x \neq 0 s o$ $F (t) = \frac{t}{x^{2} (t^{2} + \frac{2 t}{x} + \frac{2}{x^{2}})} = \frac{t}{x^{2} (t^{2} + \frac{2 t}{x} + \frac{1}{x^{2}} + \frac{2}{x^{2}} - \frac{1}{x^{2}})}$ $= \frac{t}{x^{2} {{(t + \frac{1}{x})}^{2} + \frac{1}{x^{2}}}} w e u s e t h e c h a n g e m e n t t + \frac{1}{x} = \frac{1}{∣ x ∣} u \Rightarrow x t + 1 = s (x) u$ $\int_{s (x)}^{(x + 1) s (x)} \frac{t}{1 + {(1 + t x)}^{2}} d t = \int_{s (x)}^{(x + 1) s (x)} \frac{(\frac{1}{∣ x ∣} u - \frac{1}{x})}{x^{2} \times \frac{1}{x^{2}} (1 + u^{2})} d u$ $= \frac{1}{∣ x ∣} \int_{s (x)}^{(x + 1) s (x)} \frac{u}{1 + u^{2}} - \frac{1}{x} \int_{s (x)}^{(x + 1) s (x)} \frac{d u}{1 + u^{2}}$ $= \frac{1}{2 ∣ x ∣} {[l n (1 + u^{2})]}_{s (x)}^{(x + 1) s (x)} - \frac{1}{x} {[a r c t a n u]}_{s (x)}^{(x + 1) s (x)} (s^{2} (x) = 1)$ $= \frac{1}{2 ∣ x ∣} {l n (1 + {(x + 1)}^{2}) - l n 2} - \frac{1}{x} {a r c t a n (x + 1) s (x) - a r c t a n (s (x))}$ $\Rightarrow f (x) = a r c t a n (1 + x) - \frac{s (x)}{2} {l n (x^{2} + 2 x + 2) - l n (2)}$ $+ a r c t a n (x + 1) s (x) + x a r c t a n (s (x))$ $w i t h s (x) = 1 i f x > 0 a n d s (x) = - 1 i f x < 0$
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