let f(x) =∫_0 ^∞ ((ln(1+xt^2 ))/(2+t^2 )) dt determine a explicit form of f(x) 2)calculate ∫


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Question Number 59278 by Mr X pcx last updated on 07/May/19

$l e t f (x) = \int_{0}^{\infty} \frac{l n (1 + {x t}^{2})}{2 + t^{2}} d t$ $d e t e r m i n e a e x p l i c i t f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{\infty} \frac{l n (1 + 3 x^{2})}{2 + t^{2}} d t$
Commented by maxmathsup by imad last updated on 13/May/19
$1) we have f^′ (x) =∫_0 ^∞ (t^2 /((2+t^2 )(1+xt^2 ))) dt let decompose F(t) =(t^2 /((t^2 +2)(xt^2 +1))) case 1 x>0 F(t) =((at +b)/(t^2 +2)) +((ct +d)/(xt^2 +1)) we have F(−t)=F(t) ⇒ ((−at +b)/(t^2 +2)) +((−ct +d)/(xt^2 +1)) = F(t) ⇒a=c=0 ⇒F(t) =(b/(t^2 +2)) +(d/(xt^2 +1)) lim_(t→+∞) t^2 F(t) =(1/x) =b +(d/x) ⇒1 =bx +d ⇒d =1−bx F(0) =0 =(b/2) +d ⇒b+2d =0 ⇒b =−2d ⇒d =1−(−2d)x =1+2dx ⇒ (1−2x)d =1 ⇒d =(1/(1−2x)) (we suppose x≠(1/2)) ⇒b =((−2)/(1−2x)) =(2/(2x−1)) ⇒ F(t) =(2/((2x−1)(t^2 +2))) −(1/((2x−1)(xt^2 +1))) ⇒f^′ (x) =∫_0 ^∞ F(t)dt =(2/(2x−1)) ∫_0 ^∞ (dt/(t^2 +2)) −(1/(2x−1))∫_0 ^∞ (dt/(xt^2 +1)) but ∫_0 ^∞ (dt/(t^2 +2)) =_(t =(√2)u) ∫_0 ^∞ (((√2)du)/(2(1+u^2 ))) =(1/(√2)) ∫_0 ^∞ (du/(1+u^2 )) =(π/(2(√2))) ∫_0 ^∞ (dt/(xt^2 +1)) =_((√x)t =u) ∫_0 ^∞ (du/((√x)(1+u^2 ))) =(π/(2(√x))) ⇒ f^′ (x) = (2/(2x−1)) (π/(2(√2))) −(1/(2x−1)) (π/(2(√x))) ⇒f(x) =(π/(√2)) ∫ (dx/(2x−1)) −(π/2) ∫ (dx/((√x)(2x−1))) +c =(π/(2(√2)))ln∣2x−1∣ −(π/2) ∫ (dx/((√x)(2x−1))) +c ∫ (dx/((√x)(2x−1))) =_((√x)=u) ∫ ((2udu)/(u(2u^2 −1))) =2 ∫ (du/(2u^2 −1)) =2 ∫ (du/(((√2)u−1)((√2)u +1))) = ∫ ((1/((√2)u−1)) −(1/((√2)u +1)))du =(1/(√2))ln∣(((√2)u−1)/((√2)u +1))∣ =(1/(√2))ln∣(((√(2x))−1)/((√(2x)) +1))∣ ⇒ f(x) =(π/(2(√2)))ln∣2x−1∣ −(π/(2(√2)))ln∣(((√(2x))−1)/((√(2x)) +1))∣ +c f(0) =0 =c ⇒ f(x) =(π/(2(√2))){ln∣(√(2x))−1∣ +ln∣(√(2x))+1∣−ln∣(√(2x))−1∣ +ln∣(√(2x))+1∣} =(π/(2(√2))) (2ln∣(√(2x)) +1∣ ⇒ f(x) =(π/(√2))ln∣(√(2x))+1∣ with x >0 rest to stydy the case x<0$
$1) w e h a v e f^{^{'}} (x) = \int_{0}^{\infty} \frac{t^{2}}{(2 + t^{2}) (1 + {x t}^{2})} d t$ $l e t d e c o m p o s e F (t) = \frac{t^{2}}{(t^{2} + 2) ({x t}^{2} + 1)} c a s e 1 x > 0$ $F (t) = \frac{a t + b}{t^{2} + 2} + \frac{c t + d}{{x t}^{2} + 1} w e h a v e F (- t) = F (t) \Rightarrow$ $\frac{- a t + b}{t^{2} + 2} + \frac{- c t + d}{{x t}^{2} + 1} = F (t) \Rightarrow a = c = 0 \Rightarrow F (t) = \frac{b}{t^{2} + 2} + \frac{d}{{x t}^{2} + 1}$ ${l i m}_{t \to + \infty} t^{2} F (t) = \frac{1}{x} = b + \frac{d}{x} \Rightarrow 1 = b x + d \Rightarrow d = 1 - b x$ $F (0) = 0 = \frac{b}{2} + d \Rightarrow b + 2 d = 0 \Rightarrow b = - 2 d \Rightarrow d = 1 - (- 2 d) x = 1 + 2 d x \Rightarrow$ $(1 - 2 x) d = 1 \Rightarrow d = \frac{1}{1 - 2 x} (w e s u p p o s e x \neq \frac{1}{2}) \Rightarrow b = \frac{- 2}{1 - 2 x} = \frac{2}{2 x - 1} \Rightarrow$ $F (t) = \frac{2}{(2 x - 1) (t^{2} + 2)} - \frac{1}{(2 x - 1) ({x t}^{2} + 1)} \Rightarrow f^{^{'}} (x) = \int_{0}^{\infty} F (t) d t$ $= \frac{2}{2 x - 1} \int_{0}^{\infty} \frac{d t}{t^{2} + 2} - \frac{1}{2 x - 1} \int_{0}^{\infty} \frac{d t}{{x t}^{2} + 1} b u t$ $\int_{0}^{\infty} \frac{d t}{t^{2} + 2} =_{t = \sqrt{2} u} \int_{0}^{\infty} \frac{\sqrt{2} d u}{2 (1 + u^{2})} = \frac{1}{\sqrt{2}} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} = \frac{π}{2 \sqrt{2}}$ $\int_{0}^{\infty} \frac{d t}{{x t}^{2} + 1} =_{\sqrt{x} t = u} \int_{0}^{\infty} \frac{d u}{\sqrt{x} (1 + u^{2})} = \frac{π}{2 \sqrt{x}} \Rightarrow$ $f^{^{'}} (x) = \frac{2}{2 x - 1} \frac{π}{2 \sqrt{2}} - \frac{1}{2 x - 1} \frac{π}{2 \sqrt{x}} \Rightarrow f (x) = \frac{π}{\sqrt{2}} \int \frac{d x}{2 x - 1} - \frac{π}{2} \int \frac{d x}{\sqrt{x} (2 x - 1)} + c$ $= \frac{π}{2 \sqrt{2}} l n ∣ 2 x - 1 ∣ - \frac{π}{2} \int \frac{d x}{\sqrt{x} (2 x - 1)} + c$ $\int \frac{d x}{\sqrt{x} (2 x - 1)} =_{\sqrt{x} = u} \int \frac{2 u d u}{u (2 u^{2} - 1)} = 2 \int \frac{d u}{2 u^{2} - 1}$ $= 2 \int \frac{d u}{(\sqrt{2} u - 1) (\sqrt{2} u + 1)} = \int (\frac{1}{\sqrt{2} u - 1} - \frac{1}{\sqrt{2} u + 1}) d u$ $= \frac{1}{\sqrt{2}} l n ∣ \frac{\sqrt{2} u - 1}{\sqrt{2} u + 1} ∣ = \frac{1}{\sqrt{2}} l n ∣ \frac{\sqrt{2 x} - 1}{\sqrt{2 x} + 1} ∣ \Rightarrow$ $f (x) = \frac{π}{2 \sqrt{2}} l n ∣ 2 x - 1 ∣ - \frac{π}{2 \sqrt{2}} l n ∣ \frac{\sqrt{2 x} - 1}{\sqrt{2 x} + 1} ∣ + c$ $f (0) = 0 = c \Rightarrow f (x) = \frac{π}{2 \sqrt{2}} {l n ∣ \sqrt{2 x} - 1 ∣ + l n ∣ \sqrt{2 x} + 1 ∣ - l n ∣ \sqrt{2 x} - 1 ∣ + l n ∣ \sqrt{2 x} + 1 ∣}$ $= \frac{π}{2 \sqrt{2}} (2 l n ∣ \sqrt{2 x} + 1 ∣ \Rightarrow f (x) = \frac{π}{\sqrt{2}} l n ∣ \sqrt{2 x} + 1 ∣ w i t h x > 0 r e s t t o s t y d y t h e c a s e x < 0$
Commented by maxmathsup by imad last updated on 13/May/19

$d u o t o x > 0 w e h a v e f (x) = \frac{π}{\sqrt{2}} l n (\sqrt{2 x} + 1)$
Commented by maxmathsup by imad last updated on 13/May/19

$2) \int_{0}^{\infty} \frac{l n (1 + 3 t^{2})}{2 + t^{2}} d t = f (3) = \frac{π}{\sqrt{2}} l n (1 + \sqrt{6}) .$
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