let f(t) =∫_0 ^∞ ((ln(1+tx))/(1+x^2 ))dx with ∣t∣<1 1) determine a explicit form of f(t) 2) find the value of ∫


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Question Number 63395 by mathmax by abdo last updated on 03/Jul/19

$l e t f (t) = \int_{0}^{\infty} \frac{l n (1 + t x)}{1 + x^{2}} d x w i t h ∣ t ∣< 1$ $1) d e t e r m i n e a e x p l i c i t f o r m o f f (t)$ $2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{l n (1 + x)}{1 + x^{2}} d x$
Commented byPrithwish sen last updated on 03/Jul/19
$f(t)=∫_0 ^∞ ((ln(1+tx))/(1+x^2 )) dt f′(t) = ∫_0 ^∞ (δ/(δt)){((ln(1+tx))/(1+x^2 )) }dx = ∫_0 ^∞ (x/((1+x^2 )(1+tx))) dx = (1/(1+t^2 )) ∫_0 ^B {(x/(1+x^2 )) +(t/(1+x^2 )) −(t/(1+tx)) }dx, B→∞ = (1/(1+t^2 )) [ ln(((√(1+x^2 ))/(1+tx))) + t( tan^(−1) x)]_0 ^B B→∞ Now, lim_(B→∞) ln (((√(1+x^2 ))/(1+tx))) ⇒ lim_(B→∞) ln(((√(1+B^2 ))/(1+Bt))) put B= (1/n) ⇒ lim_(n→0) ln(((√(n^2 +1))/(n+t)))⇒ −ln(t) ∴ f(t) = −∫((ln(t))/(1+t^2 )) dt+ (π/2) ∫((tdt)/(1+t^(2 ) )) to be continue.......$
$f (t) = \int_{0}^{\infty} \frac{\ln (1 + tx)}{1 + x^{2}} dt$ $f^{'} (t) = \int_{0}^{\infty} \frac{δ}{δ t} {\frac{\ln (1 + tx)}{1 + x^{2}}} dx$ $= \int_{0}^{\infty} \frac{x}{(1 + x^{2}) (1 + tx)} dx$ $= \frac{1}{1 + t^{2}} \int_{0}^{B} {\frac{x}{1 + x^{2}} + \frac{t}{1 + x^{2}} - \frac{t}{1 + tx}} dx, B \to \infty$ $= \frac{1}{1 + t^{2}} {[\ln (\frac{\sqrt{1 + x^{2}}}{1 + tx}) + t (\tan^{- 1} x)]}_{0}^{B} B \to \infty$ $Now, \lim_{B \to \infty} \ln (\frac{\sqrt{1 + x^{2}}}{1 + tx}) \Rightarrow \lim_{B \to \infty} \ln (\frac{\sqrt{1 + B^{2}}}{1 + Bt})$ $put B = \frac{1}{n} \Rightarrow \lim_{n \to 0} \ln (\frac{\sqrt{n^{2} + 1}}{n + t}) \Rightarrow - \ln (t)$ $∴ f (t) = - \int \frac{\ln (t)}{1 + t^{2}} dt + \frac{π}{2} \int \frac{tdt}{1 + t^{2}}$ $to be continue . . . . . . .$
Commented bymathmax by abdo last updated on 04/Jul/19
$1) we have f^′ (t) =∫_0 ^∞ (x/((x^2 +1)(1+tx)))dx let decompose F(x)=(x/((tx+1)(x^2 +1))) F(x)=(a/(tx+1)) +((bx+c)/(x^2 +1)) a=lim_(x→−(1/t)) (tx+1)F(x)=((−1)/(t((1/t^2 )+1))) =((−t^2 )/(t(1+t^2 ))) =−(t/(1+t^2 )) lim_(x→+∞) xF(x)=0=(a/t) +b ⇒b=−(a/t) =(1/(1+t^2 )) ⇒F(x)=−(t/((1+t^2 )(tx+1))) +(((x/(1+t^2 ))+c)/(x^2 +1)) F(0)=0 =−(t/(1+t^2 )) +c ⇒c =(t/(1+t^2 )) ⇒ F(x)=((−t)/((1+t^2 )(tx+1))) +(1/(1+t^2 )) ((x+t)/(x^2 +1)) ⇒ f^′ (t) =((−t)/(1+t^2 )) ∫_0 ^∞ (dx/(tx+1)) +(1/(2(1+t^2 ))) ∫_0 ^∞ ((2x)/(x^2 +1))dx +(t/(1+t^2 ))∫_0 ^∞ (dx/(1+x^2 )) =(1/(1+t^2 )){ [(1/2)ln(x^2 +1)−ln∣tx +1∣]_0 ^(+∞) } +((πt)/(2(1+t^2 ))) =(1/(1+t^2 ))[ln∣((√(x^2 +1))/(tx+1))∣]_0 ^(+∞) +((πt)/(2(1+t^2 ))) =(1/(1+t^2 ))ln((1/t^2 )) +((πt)/(2(1+t^2 ))) =−((ln∣t∣)/(1+t^2 )) +((πt)/(2(1+t^2 ))) ⇒ f(t) =−∫_0 ^t ((lnx)/(1+x^2 ))dx +(π/2) ∫_0 ^t (x/(1+x^2 )) dx +C =−∫_0 ^t ((lnx)/(1+x^2 ))dx +(π/4)ln(1+t^2 ) +C C=f(0) =0 ⇒f(t) = (π/4)ln(1+t^2 )−∫_0 ^t ((lnx)/(1+x^2 ))dx (we suppose t≥0) 2) we have ∫_0 ^∞ ((ln(1+x))/(1+x^2 ))dx =f(1) =(π/4)ln(2)−∫_0 ^1 ((ln(x))/(1+x^2 ))dx changement x =tanθ give ((ln(x))/(1+x^2 ))dx = ∫_0 ^(π/4) ((ln(tanθ))/(1+tan^2 θ))(1+tan^2 θ)dθ = ∫_0 ^(π/4) ln(tanθ)dθ =[θ ln(tanθ) ]_0 ^(π/4) −∫_0 ^(π/4) θ ((1+tan^2 θ)/(tanθ)) dθ =−∫_0 ^(π/4) θ (1/(cos^2 θ ((sinθ)/(cosθ))))dθ =∫_0 ^(π/4) (θ/(cosθ sinθ)) dθ = 2 ∫_0 ^(π/4) (θ/(sin(2θ)))dθ =_(2θ =u) 2 ∫_0 ^(π/2) (u/(2sin(u))) (du/2) =(1/2) ∫_0 ^(π/2) (u/(sinu)) du =_(tan((u/2))=α) (1/2) ∫_0 ^1 ((2arctan(α))/((2α)/(1+α^2 ))) ((2dα)/(1+α^2 )) =(1/2) ∫_0 ^1 ((arctan(α))/α) dα ...be continued...$
$1) w e h a v e f^{^{'}} (t) = \int_{0}^{\infty} \frac{x}{(x^{2} + 1) (1 + t x)} d x l e t d e c o m p o s e F (x) = \frac{x}{(t x + 1) (x^{2} + 1)}$ $F (x) = \frac{a}{t x + 1} + \frac{b x + c}{x^{2} + 1}$ $a = {l i m}_{x \to - \frac{1}{t}} (t x + 1) F (x) = \frac{- 1}{t (\frac{1}{t^{2}} + 1)} = \frac{- t^{2}}{t (1 + t^{2})} = - \frac{t}{1 + t^{2}}$ ${l i m}_{x \to + \infty} x F (x) = 0 = \frac{a}{t} + b \Rightarrow b = - \frac{a}{t} = \frac{1}{1 + t^{2}} \Rightarrow F (x) = - \frac{t}{(1 + t^{2}) (t x + 1)} + \frac{\frac{x}{1 + t^{2}} + c}{x^{2} + 1}$ $F (0) = 0 = - \frac{t}{1 + t^{2}} + c \Rightarrow c = \frac{t}{1 + t^{2}} \Rightarrow$ $F (x) = \frac{- t}{(1 + t^{2}) (t x + 1)} + \frac{1}{1 + t^{2}} \frac{x + t}{x^{2} + 1} \Rightarrow$ $f^{^{'}} (t) = \frac{- t}{1 + t^{2}} \int_{0}^{\infty} \frac{d x}{t x + 1} + \frac{1}{2 (1 + t^{2})} \int_{0}^{\infty} \frac{2 x}{x^{2} + 1} d x + \frac{t}{1 + t^{2}} \int_{0}^{\infty} \frac{d x}{1 + x^{2}}$ $= \frac{1}{1 + t^{2}} {{[\frac{1}{2} l n (x^{2} + 1) - l n ∣ t x + 1 ∣]}_{0}^{+ \infty}} + \frac{π t}{2 (1 + t^{2})}$ $= \frac{1}{1 + t^{2}} {[l n ∣ \frac{\sqrt{x^{2} + 1}}{t x + 1} ∣]}_{0}^{+ \infty} + \frac{π t}{2 (1 + t^{2})} = \frac{1}{1 + t^{2}} l n (\frac{1}{t^{2}}) + \frac{π t}{2 (1 + t^{2})}$ $= - \frac{l n ∣ t ∣}{1 + t^{2}} + \frac{π t}{2 (1 + t^{2})} \Rightarrow$ $f (t) = - \int_{0}^{t} \frac{l n x}{1 + x^{2}} d x + \frac{π}{2} \int_{0}^{t} \frac{x}{1 + x^{2}} d x + C$ $= - \int_{0}^{t} \frac{l n x}{1 + x^{2}} d x + \frac{π}{4} l n (1 + t^{2}) + C$ $C = f (0) = 0 \Rightarrow f (t) = \frac{π}{4} l n (1 + t^{2}) - \int_{0}^{t} \frac{l n x}{1 + x^{2}} d x (w e s u p p o s e t ⩾ 0)$ $2) w e h a v e \int_{0}^{\infty} \frac{l n (1 + x)}{1 + x^{2}} d x = f (1) = \frac{π}{4} l n (2) - \int_{0}^{1} \frac{l n (x)}{1 + x^{2}} d x$ $c h a n g e m e n t x = t a n θ g i v e$ $\frac{l n (x)}{1 + x^{2}} d x = \int_{0}^{\frac{π}{4}} \frac{l n (t a n θ)}{1 + {t a n}^{2} θ} (1 + {t a n}^{2} θ) d θ = \int_{0}^{\frac{π}{4}} l n (t a n θ) d θ$ $= {[θ l n (t a n θ)]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} θ \frac{1 + {t a n}^{2} θ}{t a n θ} d θ = - \int_{0}^{\frac{π}{4}} θ \frac{1}{{c o s}^{2} θ \frac{s i n θ}{c o s θ}} d θ$ $= \int_{0}^{\frac{π}{4}} \frac{θ}{c o s θ s i n θ} d θ = 2 \int_{0}^{\frac{π}{4}} \frac{θ}{s i n (2 θ)} d θ =_{2 θ = u} 2 \int_{0}^{\frac{π}{2}} \frac{u}{2 s i n (u)} \frac{d u}{2}$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{u}{s i n u} d u =_{t a n (\frac{u}{2}) = α} \frac{1}{2} \int_{0}^{1} \frac{2 a r c t a n (α)}{\frac{2 α}{1 + α^{2}}} \frac{2 d α}{1 + α^{2}}$ $= \frac{1}{2} \int_{0}^{1} \frac{a r c t a n (α)}{α} d α . . . b e c o n t i n u e d . . .$
Commented bymathmax by abdo last updated on 04/Jul/19

$l e t t r y a n o t h e r w a y w e h a v e \int_{0}^{\infty} \frac{l n (1 + x)}{1 + x^{2}} d x = \frac{π}{4} l n (2) - \int_{0}^{1} \frac{l n (x)}{1 + x^{2}} d x$ $\int_{0}^{1} \frac{l n (x)}{1 + x^{2}} d x = \int_{0}^{1} l n (x) (\sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n}) d x$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{2 n} l n (x) d x b y [p a r t s$ $\int_{0}^{1} x^{2 n} l n (x) d x = {[\frac{1}{2 n + 1} x^{2 n + 1} l n (x)]}_{0}^{1} - \int_{0}^{1} \frac{1}{2 n + 1} x^{2 n + 1} \frac{d x}{x}$ $= - \frac{1}{2 n + 1} \int_{0}^{1} x^{2 n} d x = - \frac{1}{{(2 n + 1)}^{2}} \Rightarrow \int_{0}^{1} \frac{l n (x)}{1 + x^{2}} d x = - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}}$ $r e s t t o f i n d t h e v a l u e o f t h i s s e r i e b y f o u r i e r s e r i e s . . . . .$
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