let f(x)=∫_0 ^(+∞) ((arctan(xt))/(1+t^2 ))dt with x≥0 1) calculate f^′ (x) then a simple form of f(x) 2) calculate ∫_0 ^(+∞) ((arctant)/(1+t^2 ))dt 3) calculate ∫


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 38310 by maxmathsup by imad last updated on 23/Jun/18

$l e t f (x) = \int_{0}^{+ \infty} \frac{a r c t a n (x t)}{1 + t^{2}} d t w i t h x ⩾ 0$ $1) c a l c u l a t e f^{^{'}} (x) t h e n a s i m p l e f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{+ \infty} \frac{a r c t a n t}{1 + t^{2}} d t$ $3) c a l c u l a t e \int_{0}^{+ \infty} \frac{a r c t a n (2 t)}{1 + t^{2}} d t$
Commented by prof Abdo imad last updated on 24/Jun/18
$1)we have f^′ (x)=∫_0 ^∞ (t/((1+t^2 )(1+x^2 t^2 )))dt let decompose F(t)= (t/((1+t^2 )(1+x^2 t^2 ))) F(t)= ((at +b)/(t^2 +1)) +((ct +d)/(x^2 t^2 +1)) F(−t)=−F(t)⇒((−at +b)/(t^2 +1)) +((−ct +d)/(x^2 t^2 +1)) =((−at−b)/(t^2 +1)) +((−ct−d)/(x^2 t^2 +1)) ⇒n=d=0 ⇒ F(t)= ((at)/(t^2 +1)) +((ct)/(x^2 t^(2 ) +1)) lim_(t→+∞) t F(t)=0=a +(c/x^2 ) ⇒ax^2 +c=0 ⇒ c=−ax^2 ⇒F(t)= ((at)/(t^2 +1)) −x^2 ((at)/(x^2 t^2 +1)) F(1)= (1/(2(1+x^2 ))) = (a/2) −((ax^2 )/(x^2 +1)) ⇒ (1/2) = ((1+x^2 )/2)a −ax^2 =((1+x^2 −2x^2 )/2)a =((1−x^2 )/2)a ⇒ (1−x^2 )a=1 ⇒a= (1/(1−x^2 )) if x^2 ≠1 ⇒ F(x)= (1/(1−x^2 )) (t/(t^2 +1)) −(x^2 /(1−x^2 )) (t/(x^2 t^2 +1)) =(1/(1−x^2 )){ (t/(t^2 +1)) − ((x^2 t)/(x^2 t^2 +1))} .$
$1) w e h a v e f^{^{'}} (x) = \int_{0}^{\infty} \frac{t}{(1 + t^{2}) (1 + x^{2} t^{2})} d t$ $l e t d e c o m p o s e F (t) = \frac{t}{(1 + t^{2}) (1 + x^{2} t^{2})}$ $F (t) = \frac{a t + b}{t^{2} + 1} + \frac{c t + d}{x^{2} t^{2} + 1}$ $F (- t) = - F (t) \Rightarrow \frac{- a t + b}{t^{2} + 1} + \frac{- c t + d}{x^{2} t^{2} + 1}$ $= \frac{- a t - b}{t^{2} + 1} + \frac{- c t - d}{x^{2} t^{2} + 1} \Rightarrow n = d = 0 \Rightarrow$ $F (t) = \frac{a t}{t^{2} + 1} + \frac{c t}{x^{2} t^{2} + 1}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + \frac{c}{x^{2}} \Rightarrow {a x}^{2} + c = 0 \Rightarrow$ $c = - {a x}^{2} \Rightarrow F (t) = \frac{a t}{t^{2} + 1} - x^{2} \frac{a t}{x^{2} t^{2} + 1}$ $F (1) = \frac{1}{2 (1 + x^{2})} = \frac{a}{2} - \frac{{a x}^{2}}{x^{2} + 1} \Rightarrow$ $\frac{1}{2} = \frac{1 + x^{2}}{2} a - {a x}^{2} = \frac{1 + x^{2} - 2 x^{2}}{2} a = \frac{1 - x^{2}}{2} a \Rightarrow$ $(1 - x^{2}) a = 1 \Rightarrow a = \frac{1}{1 - x^{2}} i f x^{2} \neq 1 \Rightarrow$ $F (x) = \frac{1}{1 - x^{2}} \frac{t}{t^{2} + 1} - \frac{x^{2}}{1 - x^{2}} \frac{t}{x^{2} t^{2} + 1}$ $= \frac{1}{1 - x^{2}} {\frac{t}{t^{2} + 1} - \frac{x^{2} t}{x^{2} t^{2} + 1}} .$
Commented by abdo.msup.com last updated on 24/Jun/18

$f^{,} (x) = \int_{0}^{\infty} F (t) d t \Rightarrow (1 - x^{2}) f^{^{'}} (x)$ $= \int_{0}^{\infty} \frac{t}{1 + t^{2}} d t - \frac{1}{2} \int_{0}^{\infty} \frac{2 x^{2} t}{x^{2} t^{2} + 1} d t$ $= {[\frac{1}{2} l n ∣ \frac{1 + t^{2}}{x^{2} t^{2} + 1} ∣]}_{0}^{+ \infty}$ $= \frac{1}{2} l n (\frac{1}{x^{2}}) = - l n ∣ x ∣ \Rightarrow f^{^{'}} (x) = - \frac{l n ∣ x ∣}{1 - x^{2}}$
Commented by abdo.msup.com last updated on 24/Jun/18

$f o r x > 0 w e g e t f^{^{'}} (x) = - \frac{l n (x)}{1 - x^{2}} \Rightarrow$ $f (x) = - \int_{1}^{x} \frac{l n (t)}{1 - t^{2}} d t + c$ $c = f (1) = \int_{0}^{\infty} \frac{a r c t a n t}{1 + t^{2}} d t c h a n g e m e n t$ $t = t a n θ g i v e$ $f (1) = \int_{0}^{\frac{π}{2}} \frac{θ}{1 + {t a n}^{2} θ} (1 + {t a n}^{2} θ) d θ$ $= \int_{0}^{\frac{π}{2}} θ d θ = {[\frac{θ^{2}}{2}]}_{0}^{\frac{π}{2}} = \frac{π^{2}}{8} \Rightarrow$ $f (x) = \frac{π^{2}}{8} - \int_{1}^{x} \frac{l n (t)}{1 - t^{2}} d t .$
Commented by prof Abdo imad last updated on 24/Jun/18
$2) ∫_0 ^∞ ((arctan(t))/(1+t^2 ))dt =(π^2 /8) 3) ∫_0 ^∞ ((arctan(2t))/(1+t^2 ))dt =f(2) =(π^2 /8) −∫_1 ^2 ((ln(t))/(1−t^2 ))dt changement t=1+x give ∫_1 ^2 ((ln(t))/(1−t^2 ))dt = ∫_0 ^1 ((ln(1+x))/(1−(1+x)^2 ))dx =∫_0 ^1 ((ln(1+x))/(1−1−x^2 −2x))dx = −∫_0 ^1 ((ln(1+x))/(x(x+2)))dx =−∫_0 ^1 ln(1+x)((1/x) −(1/(x+2)))dx =−(1/2) ∫_0 ^1 ( ((ln(1+x))/x) −((ln(1+x))/(x+2)))dx =(1/2) ∫_0 ^1 ((ln(1+x))/(x+2))dx −(1/2) ∫_0 ^1 ((ln(1+x))/x)dx but ln^′ (1+x) = (1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n ⇒ ln(1+x)=Σ_(n=0) ^∞ (((−1)^n )/(n+1))x^(n+1) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n ⇒∫_0 ^1 ((ln(1+x))/x)dx=∫_0 ^1 (Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^(n−1) ) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n^ )∫_0 ^1 x^(n−1) dx =−Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =−(−(π^2 /(12)))=(π^2 /(12)) by parts ∫_0 ^1 ((ln(1+x))/(x+2))dx =[ln(1+x)ln(2+x)]_0 ^1 −∫_0 ^1 ((ln(x+2))/(x+1))dx =ln(2)ln(3)−∫_0 ^1 ((ln(x+2))/(x+1))dx ∫_0 ^1 ((ln(x+2))/(x+1))dx=∫_0 ^1 (Σ_(n=0) ^∞ (−1)^n x^n )ln(x+2)dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^n ln(x+2)dx by parts A_n = ∫_0 ^1 x^n ln(x+2)dx =[(1/(n+1))x^(n+1) ln(x+2)]_0 ^1 −∫_0 ^1 (x^(n+1) /((n+1)(x+2)))dx =(1/(n+1)){ln(3) −2^(n+1) ln(2)}−(1/(n+1)) ∫_0 ^1 (x^(n+1) /(x+2))dx ...be continued...$
$2) \int_{0}^{\infty} \frac{a r c t a n (t)}{1 + t^{2}} d t = \frac{π^{2}}{8}$ $3) \int_{0}^{\infty} \frac{a r c t a n (2 t)}{1 + t^{2}} d t = f (2)$ $= \frac{π^{2}}{8} - \int_{1}^{2} \frac{l n (t)}{1 - t^{2}} d t c h a n g e m e n t$ $t = 1 + x g i v e$ $\int_{1}^{2} \frac{l n (t)}{1 - t^{2}} d t = \int_{0}^{1} \frac{l n (1 + x)}{1 - {(1 + x)}^{2}} d x$ $= \int_{0}^{1} \frac{l n (1 + x)}{1 - 1 - x^{2} - 2 x} d x$ $= - \int_{0}^{1} \frac{l n (1 + x)}{x (x + 2)} d x$ $= - \int_{0}^{1} l n (1 + x) (\frac{1}{x} - \frac{1}{x + 2}) d x$ $= - \frac{1}{2} \int_{0}^{1} (\frac{l n (1 + x)}{x} - \frac{l n (1 + x)}{x + 2}) d x$ $= \frac{1}{2} \int_{0}^{1} \frac{l n (1 + x)}{x + 2} d x - \frac{1}{2} \int_{0}^{1} \frac{l n (1 + x)}{x} d x b u t$ ${l n}^{^{'}} (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} \Rightarrow$ $l n (1 + x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} x^{n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n}$ $\Rightarrow \int_{0}^{1} \frac{l n (1 + x)}{x} d x = \int_{0}^{1} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n - 1})$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{}} \int_{0}^{1} x^{n - 1} d x$ $= - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = - (- \frac{π^{2}}{12}) = \frac{π^{2}}{12}$ $b y p a r t s \int_{0}^{1} \frac{l n (1 + x)}{x + 2} d x$ $= {[l n (1 + x) l n (2 + x)]}_{0}^{1} - \int_{0}^{1} \frac{l n (x + 2)}{x + 1} d x$ $= l n (2) l n (3) - \int_{0}^{1} \frac{l n (x + 2)}{x + 1} d x$ $\int_{0}^{1} \frac{l n (x + 2)}{x + 1} d x = \int_{0}^{1} (\sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n}) l n (x + 2) d x$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{n} l n (x + 2) d x b y p a r t s$ $A_{n} = \int_{0}^{1} x^{n} l n (x + 2) d x$ $= {[\frac{1}{n + 1} x^{n + 1} l n (x + 2)]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 1}}{(n + 1) (x + 2)} d x$ $= \frac{1}{n + 1} {l n (3) - 2^{n + 1} l n (2)} - \frac{1}{n + 1} \int_{0}^{1} \frac{x^{n + 1}}{x + 2} d x$ $. . . b e c o n t i n u e d . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18

	$2) \int_{0}^{\infty} \frac{{t a n}^{- 1} t}{1 + t^{2}} d t$ $y = {t a n}^{- 1} t t a n y = t d t = {s e c}^{2} y d y$ $\int_{0}^{\frac{Π}{2}} \frac{y \times {s e c}^{2} y d y}{{s e c}^{2} y}$ $= \frac{1}{2} \times {(\frac{Π}{2})}^{2} = \frac{\prod^{2}}{8}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum