Question Number 63395 by mathmax by abdo last updated on 03/Jul/19
$${let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left(\mathrm{1}+{tx}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:{with}\:\:\mid{t}\mid<\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:\:{form}\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by Prithwish 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.......](https://www.tinkutara.com/question/Q63398.png)
$$\mathrm{f}\left(\mathrm{t}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{tx}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{dt} \\ $$$$\mathrm{f}'\left(\mathrm{t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\delta}{\delta\mathrm{t}}\left\{\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{tx}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\right\}\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{tx}\right)}\:\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\mathrm{B}} \left\{\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{t}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:−\frac{\mathrm{t}}{\mathrm{1}+\mathrm{tx}}\:\right\}\mathrm{dx},\:\mathrm{B}\rightarrow\infty \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\left[\:\mathrm{ln}\left(\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{1}+\mathrm{tx}}\right)\:+\:\mathrm{t}\left(\:\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{B}} \:\:\mathrm{B}\rightarrow\infty \\ $$$$\:\:\:\mathrm{Now},\:\mathrm{lim}_{\mathrm{B}\rightarrow\infty} \:\mathrm{ln}\:\left(\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{1}+\mathrm{tx}}\right)\:\Rightarrow\:\mathrm{lim}_{\mathrm{B}\rightarrow\infty} \mathrm{ln}\left(\frac{\sqrt{\mathrm{1}+\mathrm{B}^{\mathrm{2}} }}{\mathrm{1}+\mathrm{Bt}}\right) \\ $$$$\mathrm{put}\:\mathrm{B}=\:\frac{\mathrm{1}}{\mathrm{n}}\:\Rightarrow\:\mathrm{lim}_{\mathrm{n}\rightarrow\mathrm{0}} \:\mathrm{ln}\left(\frac{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{n}+\mathrm{t}}\right)\Rightarrow\:−\mathrm{ln}\left(\mathrm{t}\right) \\ $$$$\:\therefore\:\:\mathrm{f}\left(\mathrm{t}\right)\:=\:−\int\frac{\mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt}+\:\frac{\pi}{\mathrm{2}}\:\int\frac{\mathrm{tdt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}\:} } \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{continue}……. \\ $$$$ \\ $$
Commented by mathmax 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...](https://www.tinkutara.com/question/Q63482.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}+{tx}\right)}{dx}\:\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{{x}}{\left({tx}+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{tx}+\mathrm{1}}\:+\frac{{bx}+{c}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}={lim}_{{x}\rightarrow−\frac{\mathrm{1}}{{t}}} \:\:\:\left({tx}+\mathrm{1}\right){F}\left({x}\right)=\frac{−\mathrm{1}}{{t}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}\right)}\:=\frac{−{t}^{\mathrm{2}} }{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=−\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}=\frac{{a}}{{t}}\:+{b}\:\Rightarrow{b}=−\frac{{a}}{{t}}\:=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\Rightarrow{F}\left({x}\right)=−\frac{{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({tx}+\mathrm{1}\right)}\:+\frac{\frac{{x}}{\mathrm{1}+{t}^{\mathrm{2}} }+{c}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{0}\:=−\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+{c}\:\Rightarrow{c}\:=\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{−{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({tx}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\frac{{x}+{t}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${f}^{'} \left({t}\right)\:=\frac{−{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{{tx}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:+\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\left\{\:\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−{ln}\mid{tx}\:+\mathrm{1}\mid\right]_{\mathrm{0}} ^{+\infty} \right\}\:+\frac{\pi{t}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\left[{ln}\mid\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{tx}+\mathrm{1}}\mid\right]_{\mathrm{0}} ^{+\infty} \:+\frac{\pi{t}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{ln}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)\:+\frac{\pi{t}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$=−\frac{{ln}\mid{t}\mid}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\pi{t}}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=−\int_{\mathrm{0}} ^{{t}} \:\:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:+\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{t}} \:\:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:+{C} \\ $$$$=−\int_{\mathrm{0}} ^{{t}} \:\:\:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:+\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+{C} \\ $$$${C}={f}\left(\mathrm{0}\right)\:=\mathrm{0}\:\Rightarrow{f}\left({t}\right)\:=\:\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)−\int_{\mathrm{0}} ^{{t}} \:\:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:\:\:\left({we}\:{suppose}\:{t}\geqslant\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:={f}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${changement}\:\:{x}\:={tan}\theta\:{give}\: \\ $$$$\:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{ln}\left({tan}\theta\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({tan}\theta\right){d}\theta \\ $$$$=\left[\theta\:{ln}\left({tan}\theta\right)\:\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\theta\:\frac{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{{tan}\theta}\:{d}\theta\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\theta\:\:\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta\:\frac{{sin}\theta}{{cos}\theta}}{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\frac{\theta}{{cos}\theta\:{sin}\theta}\:{d}\theta\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{\theta}{{sin}\left(\mathrm{2}\theta\right)}{d}\theta\:=_{\mathrm{2}\theta\:={u}} \:\:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{u}}{\mathrm{2}{sin}\left({u}\right)}\:\frac{{du}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{u}}{{sinu}}\:{du}\:=_{{tan}\left(\frac{{u}}{\mathrm{2}}\right)=\alpha} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{arctan}\left(\alpha\right)}{\frac{\mathrm{2}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }}\:\frac{\mathrm{2}{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{arctan}\left(\alpha\right)}{\alpha}\:{d}\alpha\:\:\:\:\:…{be}\:{continued}… \\ $$
Commented by mathmax by abdo last updated on 04/Jul/19
![let try another way we have ∫_0 ^∞ ((ln(1+x))/(1+x^2 ))dx =(π/4)ln(2)−∫_0 ^1 ((ln(x))/(1+x^2 ))dx ∫_0 ^1 ((ln(x))/(1+x^2 ))dx =∫_0 ^1 ln(x)(Σ_(n=0) ^∞ (−1)^n x^(2n) )dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(2n) ln(x)dx by[parts ∫_0 ^1 x^(2n) ln(x)dx =[(1/(2n+1))x^(2n+1) ln(x)]_0 ^1 −∫_0 ^1 (1/(2n+1))x^(2n+1) (dx/x) =−(1/(2n+1)) ∫_0 ^1 x^(2n) dx =−(1/((2n+1)^2 )) ⇒∫_0 ^1 ((ln(x))/(1+x^2 ))dx =−Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) rest to find the value of this serie by fourier series.....](https://www.tinkutara.com/question/Q63483.png)
$${let}\:{try}\:{another}\:{way}\:{we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} \right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}{n}} {ln}\left({x}\right){dx}\:\:{by}\left[{parts}\right. \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{\mathrm{2}{n}} {ln}\left({x}\right){dx}\:=\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} \:\frac{{dx}}{{x}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}{n}} {dx}\:=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${rest}\:{to}\:{find}\:{the}\:{value}\:{of}\:{this}\:{serie}\:\:\:{by}\:{fourier}\:{series}….. \\ $$