Question Number 64635 by mathmax by abdo last updated on 19/Jul/19
$$\left.\mathrm{1}\right){calculate}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left(\alpha{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{with}\:\alpha\:{real} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 21/Jul/19
![1) we have f^′ (α) =∫_0 ^∞ (x/((1+α^2 x^2 )(1+x^2 )))dx =_(αx =t) ∫_0 ^∞ (t/(α(1+t^2 )(1+(t^2 /α^2 ))))(dt/α) = ∫_0 ^∞ ((tdt)/(α^2 (t^2 +1)(1+(t^2 /α^2 )))) =∫_0 ^∞ ((tdt)/((t^2 +1)(t^2 +α^2 ))) let decompose F(t) =(t/((t^2 +1)(t^2 +α^2 ))) F(t) =((at +b)/(t^2 +1)) +((ct +d)/(t^2 +α^2 )) F(−t)=−F(t) ⇒ ((−at +b)/(t^2 +1)) +((−ct +d)/(t^2 +α^2 )) =((−at−b)/(t^2 +1)) +((−ct−d)/(t^2 +α^2 )) ⇒ b=d =0 ⇒F(t) =((at)/(t^2 +1)) +((ct)/(t^2 +α^2 )) lim_(t→+∞) tF(t)=0 =a+c ⇒c =−a ⇒F(t) =((at)/(t^2 +1)) −((at)/(t^2 +α^2 )) F(1) =(1/(2(1+α^2 ))) =(a/2) −(a/(1+α^2 )) =(((1+α^2 −2)/(2(1+α^2 ))))a =((α^2 −1)/(2(1+α^2 )))a ⇒ a =(1/(α^2 −1)) (so we suppose α≠+^− 1) ⇒ F(t) =(1/(α^2 −1)){ (t/(t^2 +1)) −(t/(t^2 +α^2 ))} ⇒f^′ (α) =(1/(2(α^2 −1))){∫_0 ^∞ ((2t)/(t^2 +1))−((2t)/(t^2 +α^2 ))}dt =(1/(2(α^2 −1)))[ln(((t^2 +1)/(t^2 +α^2 )))]_0 ^(+∞) =(1/(2(α^2 −1))){1−ln((1/α^2 ))} if α> ⇒f^′ (α) =(1/(2(α^2 −1))){1+2ln(α)} =(1/(2(α^2 −1))) +((ln(α))/(α^2 −1)) ⇒ f(α) =∫ (dα/(2(α^2 −1))) +∫ ((ln(α))/(α^2 −1))dα +C ∫ (dα/(2(α^2 −1))) =(1/4)∫ ((1/(1+α)) +(1/(1−α)))dα =(1/4)ln∣((1+α)/(1−α))∣ ⇒ f(α) =(1/4)ln∣((1+α)/(1−α))∣ +∫ ((ln(α))/(α^2 −1))dα +C ....be continued...](https://www.tinkutara.com/question/Q64800.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}}{\left(\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=_{\alpha{x}\:={t}} \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}}{\alpha\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{{t}^{\mathrm{2}} }{\alpha^{\mathrm{2}} }\right)}\frac{{dt}}{\alpha}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{tdt}}{\alpha^{\mathrm{2}} \left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{1}+\frac{{t}^{\mathrm{2}} }{\alpha^{\mathrm{2}} }\right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{tdt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} \right)}\:\:{let}\:{decompose}\:{F}\left({t}\right)\:=\frac{{t}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} \right)} \\ $$$${F}\left({t}\right)\:=\frac{{at}\:+{b}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} } \\ $$$${F}\left(−{t}\right)=−{F}\left({t}\right)\:\Rightarrow\:\frac{−{at}\:+{b}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:=\frac{−{at}−{b}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{ct}−{d}}{{t}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:\Rightarrow \\ $$$${b}={d}\:=\mathrm{0}\:\Rightarrow{F}\left({t}\right)\:=\frac{{at}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ct}}{{t}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} } \\ $$$${lim}_{{t}\rightarrow+\infty} {tF}\left({t}\right)=\mathrm{0}\:={a}+{c}\:\Rightarrow{c}\:=−{a}\:\Rightarrow{F}\left({t}\right)\:=\frac{{at}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{at}}{{t}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\:=\frac{{a}}{\mathrm{2}}\:−\frac{{a}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:=\left(\frac{\mathrm{1}+\alpha^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\right){a}\:=\frac{\alpha^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}{a}\:\Rightarrow \\ $$$${a}\:=\frac{\mathrm{1}}{\alpha^{\mathrm{2}} −\mathrm{1}}\:\:\:\:\left({so}\:\:{we}\:{suppose}\:\alpha\neq\overset{−} {+}\mathrm{1}\right)\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{1}}{\alpha^{\mathrm{2}} −\mathrm{1}}\left\{\:\frac{{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{t}}{{t}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\right\}\:\Rightarrow{f}^{'} \left(\alpha\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\left\{\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}−\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\right\}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\left[{ln}\left(\frac{{t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\right)\right]_{\mathrm{0}} ^{+\infty} \:=\frac{\mathrm{1}}{\mathrm{2}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\left\{\mathrm{1}−{ln}\left(\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\right)\right\} \\ $$$${if}\:\alpha>\:\Rightarrow{f}^{'} \left(\alpha\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\left\{\mathrm{1}+\mathrm{2}{ln}\left(\alpha\right)\right\}\:=\frac{\mathrm{1}}{\mathrm{2}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:+\frac{{ln}\left(\alpha\right)}{\alpha^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow \\ $$$${f}\left(\alpha\right)\:=\int\:\:\frac{{d}\alpha}{\mathrm{2}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:+\int\:\:\frac{{ln}\left(\alpha\right)}{\alpha^{\mathrm{2}} −\mathrm{1}}{d}\alpha\:+{C} \\ $$$$\int\:\:\:\frac{{d}\alpha}{\mathrm{2}\left(\alpha^{\mathrm{2}} −\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}}\int\:\left(\frac{\mathrm{1}}{\mathrm{1}+\alpha}\:+\frac{\mathrm{1}}{\mathrm{1}−\alpha}\right){d}\alpha\:\:=\frac{\mathrm{1}}{\mathrm{4}}{ln}\mid\frac{\mathrm{1}+\alpha}{\mathrm{1}−\alpha}\mid\:\Rightarrow \\ $$$${f}\left(\alpha\right)\:=\frac{\mathrm{1}}{\mathrm{4}}{ln}\mid\frac{\mathrm{1}+\alpha}{\mathrm{1}−\alpha}\mid\:+\int\:\:\frac{{ln}\left(\alpha\right)}{\alpha^{\mathrm{2}} −\mathrm{1}}{d}\alpha\:+{C}\:\:\:….{be}\:{continued}… \\ $$