Question Number 202866 by mnjuly1970 last updated on 04/Jan/24
$$ \\ $$$$\:\:\:\:\:\:\:{calculate}\:… \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{tanh}^{\:−\mathrm{1}} \left({x}\right)}{\left(\mathrm{1}\:+\:{x}\:\right)^{\:\mathrm{2}} }\:{dx}\:=\:?\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by Mathspace last updated on 06/Jan/24
![x=th(t)=((sht)/(cht))=((e^t −e^(−t) )/(e^t +e^(−t) )) (dx/dt)=((ch^2 t−sh^2 t)/(ch^2 t))=(1/(ch^2 t)) Φ=∫_0 ^(argth1) (t/((1+((e^t −e^(−t) )/(e^t +e^(−t) )))^2 ))(dt/((((e^t +e^(−t) )/2))^2 )) =4∫_0 ^(argth(1)) (t/(4e^(2t) ))dt =∫_0 ^(argth(1)) t e^(−2t) dt =[−(1/2)t e^(−2t) ]_0 ^(argth(1)) +(1/2)∫_0 ^(argth1) e^(−2t) dt we have x=((e^t −e^(−t) )/(e^t +e^(−t) )) =((e^(2t) −1)/(e^(2t) +1)) ⇒xe^(2t) +x=e^(2t) −1 ⇒ (x−1)e^(2t) =−x−1 ⇒ e^(2t) =((1+x)/(1−x)) ⇒t=(1/2)ln(((1+x)/(1−x))) x→1^− ⇒t→+∞ ⇒ Φ=[−(1/2)te^(−2t) ]_0 ^∞ −(1/4)[e^(−2t) ]_0 ^∞ =(1/4)](https://www.tinkutara.com/question/Q202931.png)
$${x}={th}\left({t}\right)=\frac{{sht}}{{cht}}=\frac{{e}^{{t}} −{e}^{−{t}} }{{e}^{{t}} +{e}^{−{t}} } \\ $$$$\frac{{dx}}{{dt}}=\frac{{ch}^{\mathrm{2}} {t}−{sh}^{\mathrm{2}} {t}}{{ch}^{\mathrm{2}} {t}}=\frac{\mathrm{1}}{{ch}^{\mathrm{2}} {t}} \\ $$$$\Phi=\int_{\mathrm{0}} ^{{argth}\mathrm{1}} \:\frac{{t}}{\left(\mathrm{1}+\frac{{e}^{{t}} −{e}^{−{t}} }{{e}^{{t}} +{e}^{−{t}} }\right)^{\mathrm{2}} }\frac{{dt}}{\left(\frac{{e}^{{t}} +{e}^{−{t}} }{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{{argth}\left(\mathrm{1}\right)} \:\:\:\frac{{t}}{\mathrm{4}{e}^{\mathrm{2}{t}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{{argth}\left(\mathrm{1}\right)} {t}\:{e}^{−\mathrm{2}{t}} \:{dt} \\ $$$$=\left[−\frac{\mathrm{1}}{\mathrm{2}}{t}\:{e}^{−\mathrm{2}{t}} \right]_{\mathrm{0}} ^{{argth}\left(\mathrm{1}\right)} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{argth}\mathrm{1}} \:{e}^{−\mathrm{2}{t}} {dt} \\ $$$${we}\:{have}\:{x}=\frac{{e}^{{t}} −{e}^{−{t}} }{{e}^{{t}} +{e}^{−{t}} } \\ $$$$=\frac{{e}^{\mathrm{2}{t}} −\mathrm{1}}{{e}^{\mathrm{2}{t}} +\mathrm{1}}\:\Rightarrow{xe}^{\mathrm{2}{t}} +{x}={e}^{\mathrm{2}{t}} −\mathrm{1}\:\Rightarrow \\ $$$$\left({x}−\mathrm{1}\right){e}^{\mathrm{2}{t}} =−{x}−\mathrm{1}\:\Rightarrow \\ $$$${e}^{\mathrm{2}{t}} =\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\:\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$$${x}\rightarrow\mathrm{1}^{−} \:\Rightarrow{t}\rightarrow+\infty\:\Rightarrow \\ $$$$\Phi=\left[−\frac{\mathrm{1}}{\mathrm{2}}{te}^{−\mathrm{2}{t}} \right]_{\mathrm{0}} ^{\infty} −\frac{\mathrm{1}}{\mathrm{4}}\left[{e}^{−\mathrm{2}{t}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}} \\ $$