Question Number 53284 by maxmathsup by imad last updated on 20/Jan/19
$${find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({xt}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:\:{with}\:{x}\:{real}\:. \\ $$
Commented by prof Abdo imad last updated on 21/Jan/19
![we have f^′ (x)=∫_0 ^∞ (t/((1+x^2 t^2 )(1+t^2 )))dt =_(xt =u) ∫_0 ^∞ (u/(x(1+u^2 )(1+(u^2 /x^2 )))) (du/x) =∫_0 ^∞ (u/((u^2 +1))(u^2 +x^2 )))du let decompose F(u)=(u/((u^2 +1)(u^2 +x^2 ))) ⇒ F(u)=((au+b)/(u^2 +1)) +((cu +d)/(u^2 +x^2 )) F(−u)=−F(u) ⇒((−au +b)/(u^2 +1)) +((−cu +d)/(u^2 +x^2 )) =((−au−b)/(u^2 +1)) +((−cu−d)/(u^2 +x^2 )) ⇒b=d=0 ⇒ F(u) =((au)/(u^2 +1)) +((cu)/(u^2 +x^2 )) lim_(u→+∞) u F(u)=0 =a+c ⇒c=−a ⇒ F(u) =((au)/(u^2 +1)) −((au)/(u^2 +x^2 )) F(1)= (1/(2(1+x^2 ))) =(a/2)−(a/(1+x^2 ))=((a +ax^2 −2a)/(2(1+x^2 ))) ⇒ (x^2 −1)a =1 ⇒a =(1/(x^2 −1)) (we suppose x≠+^− 1) ⇒ F(u) =(1/(x^2 −1)){ (u/(u^2 +1)) −(u/(u^2 +x^2 ))} ⇒ f^′ (x) =(1/(x^2 −1))∫_0 ^∞ ((u/(u^2 +1)) −(u/(u^2 +x^2 )))du =(1/(2(x^2 −1)))[ln∣((u^(2 ) +1)/(u^2 +x^2 ))∣]_0 ^(+∞) =(1/(2(x^2 −1)))(2ln∣x∣) =((ln∣x∣)/(x^2 −1)) let suppose x>1 ⇒ f^′ (x) =((ln(x))/(x^2 −1)) ⇒ f(x)=∫_1 ^x ((ln(t))/(t^2 −1)) dt +λ λ =f(1) =∫_0 ^∞ ((arctan(t))/(1+t^2 ))dt by parts λ =[arctan^2 (t)]_0 ^∞ −∫_0 ^∞ ((arctan(t))/(1+t^2 ))dt =(π^2 /4) −λ ⇒2λ =(π^2 /4) ⇒λ=(π^2 /8) ⇒ f(x)=∫_1 ^x ((ln(t))/(t^2 −1))dt +(π^2 /8)](https://www.tinkutara.com/question/Q53429.png)
$${we}\:{have}\:{f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}}{\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=_{{xt}\:={u}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}}{{x}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{{u}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)}\:\frac{{du}}{{x}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{u}}{\left.\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\right)\left({u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)}{du}\:{let}\:{decompose} \\ $$$${F}\left({u}\right)=\frac{{u}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${F}\left({u}\right)=\frac{{au}+{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} } \\ $$$${F}\left(−{u}\right)=−{F}\left({u}\right)\:\Rightarrow\frac{−{au}\:+{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{cu}\:+{d}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} } \\ $$$$=\frac{−{au}−{b}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{cu}−{d}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\:\Rightarrow{b}={d}=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{{au}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{cu}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} } \\ $$$${lim}_{{u}\rightarrow+\infty} {u}\:{F}\left({u}\right)=\mathrm{0}\:={a}+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{{au}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{au}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=\frac{{a}}{\mathrm{2}}−\frac{{a}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{{a}\:+{ax}^{\mathrm{2}} −\mathrm{2}{a}}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{1}\right){a}\:=\mathrm{1}\:\:\Rightarrow{a}\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\:\left({we}\:{suppose}\:{x}\neq\overset{−} {+}\mathrm{1}\right)\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\left\{\:\frac{{u}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{u}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\right\}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \:\left(\frac{{u}}{{u}^{\mathrm{2}} +\mathrm{1}}\:−\frac{{u}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\left[{ln}\mid\frac{{u}^{\mathrm{2}\:} +\mathrm{1}}{{u}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }\mid\right]_{\mathrm{0}} ^{+\infty} =\frac{\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\left(\mathrm{2}{ln}\mid{x}\mid\right) \\ $$$$=\frac{{ln}\mid{x}\mid}{{x}^{\mathrm{2}} −\mathrm{1}}\:\:{let}\:{suppose}\:{x}>\mathrm{1}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow\:{f}\left({x}\right)=\int_{\mathrm{1}} ^{{x}} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} −\mathrm{1}}\:{dt}\:+\lambda \\ $$$$\lambda\:={f}\left(\mathrm{1}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{by}\:{parts} \\ $$$$\lambda\:=\left[{arctan}^{\mathrm{2}} \left({t}\right)\right]_{\mathrm{0}} ^{\infty} \:−\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:−\lambda\:\Rightarrow\mathrm{2}\lambda\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow\lambda=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\int_{\mathrm{1}} ^{{x}} \:\:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$
Commented by prof Abdo imad last updated on 21/Jan/19
![let detetmine ∫_1 ^x ((ln(t))/(t^2 −1))dt chsng.t =(1/u) give ∫_1 ^x ((ln(t))/(t^2 −1))dt =∫_1 ^(1/x) ((−ln(u))/(((1/u^2 )−1))) (−(du/u^2 )) =∫_1 ^(1/x) ((ln(u))/(u^2 −1)) du =∫_(1/x) ^1 ((ln(u))/(1−u^2 ))du =(1/2)∫_(1/x) ^1 ln(u){ (1/(1−u)) +(1/(1+u))}du =(1/2) ∫_(1/x) ^1 ((ln(u))/(1−u))du +(1/2) ∫_(1/x) ^1 ((ln(u))/(1+u))du ∫_(1/x) ^1 ((ln(u))/(1−u))du =∫_(1/x) ^1 ln(u){Σ_(n=0) ^∞ u^n )du =Σ_(n=0) ^∞ ∫_(1/x) ^1 u^n ln(u)du =Σ_(n=0) ^∞ A_n by parts A_n =[(1/(n+1))u^(n+1) ln(u)]_(1/x) ^1 −∫_(1/x) ^1 (u^n /(n+1))du =((ln(x))/((n+1)x^(n+1) )) −(1/((n+1)))[(1/(n+1)) u^(n+1) ]_(1/x) ^1 =((ln(x))/((n+1)x^(n+1) )) −(1/((n+1)^2 )){1−(1/x^(n+1) )} ⇒ ∫_(1/x) ^1 ((ln(u))/(1−u)) du =ln(x)Σ_(n=0) ^∞ (1/((n+1)x^(n+1) )) −Σ_(n=0) ^∞ (1/((n+1)^2 )) +Σ_(n=0) ^∞ (1/((n+1)^2 x^(n+1) )) Σ_(n=0) ^∞ (1/((n+1)^2 )) =ξ(2) =(π^2 /6) let find Σ_(n=0) ^∞ (1/((n+1)x^(n+1) )) be continued...](https://www.tinkutara.com/question/Q53432.png)
$${let}\:{detetmine}\:\int_{\mathrm{1}} ^{{x}} \:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\:{chsng}.{t}\:=\frac{\mathrm{1}}{{u}}\:{give} \\ $$$$\int_{\mathrm{1}} ^{{x}} \:\:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\:=\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{{x}}} \:\frac{−{ln}\left({u}\right)}{\left(\frac{\mathrm{1}}{{u}^{\mathrm{2}} }−\mathrm{1}\right)}\:\left(−\frac{{du}}{{u}^{\mathrm{2}} }\right) \\ $$$$=\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{{x}}} \:\:\frac{{ln}\left({u}\right)}{{u}^{\mathrm{2}} −\mathrm{1}}\:{du}\:=\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:\:\frac{{ln}\left({u}\right)}{\mathrm{1}−{u}^{\mathrm{2}} }{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:{ln}\left({u}\right)\left\{\:\frac{\mathrm{1}}{\mathrm{1}−{u}}\:+\frac{\mathrm{1}}{\mathrm{1}+{u}}\right\}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:\:\frac{{ln}\left({u}\right)}{\mathrm{1}−{u}}{du}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:\:\frac{{ln}\left({u}\right)}{\mathrm{1}+{u}}{du} \\ $$$$\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:\frac{{ln}\left({u}\right)}{\mathrm{1}−{u}}{du}\:=\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} {ln}\left({u}\right)\left\{\sum_{{n}=\mathrm{0}} ^{\infty} {u}^{{n}} \right){du} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:{u}^{{n}} \:{ln}\left({u}\right){du}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \\ $$$${by}\:{parts}\:{A}_{{n}} =\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}{u}^{{n}+\mathrm{1}} {ln}\left({u}\right)\right]_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:−\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:\frac{{u}^{{n}} }{{n}+\mathrm{1}}{du} \\ $$$$=\frac{{ln}\left({x}\right)}{\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} }\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{u}^{{n}+\mathrm{1}} \right]_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \\ $$$$=\frac{{ln}\left({x}\right)}{\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} }\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\left\{\mathrm{1}−\frac{\mathrm{1}}{{x}^{{n}+\mathrm{1}} }\right\}\:\Rightarrow \\ $$$$\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:\frac{{ln}\left({u}\right)}{\mathrm{1}−{u}}\:{du}\:={ln}\left({x}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} } \\ $$$$−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} {x}^{{n}+\mathrm{1}} } \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\xi\left(\mathrm{2}\right)\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${let}\:{find}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} }\:\:{be}\:{continued}… \\ $$