Question Number 80612 by M±th+et£s last updated on 04/Feb/20
$$\:\Psi\left({x}\right)=\int_{\mathrm{1}} ^{{x}} \frac{\mathrm{1}}{\sqrt{\mathrm{1}−{e}^{{t}} }}\:{dt}\:\:\:\:\:\forall{x}\in\mathbb{R} \\ $$$${prove}\:{that} \\ $$$$\Psi\left({x}\right)=\mathrm{2}{ln}\left(\frac{\mathrm{1}−\sqrt{\mathrm{1}−{e}^{{x}} }}{\mathrm{1}−\sqrt{\mathrm{1}−{e}}}\right)−{x}+\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 04/Feb/20
$$\mathrm{1}−{e}<\mathrm{0}\:{so}\:{there}\:{is}\:{aerror}\:{in}\:{the}\:{question}! \\ $$
Commented by mathmax by abdo last updated on 04/Feb/20
$${perhaps}\:\Phi\left({x}\right)=\int_{\mathrm{1}} ^{{x}} \:\frac{{dt}}{\sqrt{\mathrm{1}−{e}^{−{t}} }}\:\:{or}\:\Phi\left({x}\right)=\int_{\mathrm{1}} ^{{x}} \:\frac{{dt}}{\sqrt{\mathrm{1}+{e}^{{t}} }}\:{or}\:\Phi\left({x}\right)=\int_{\mathrm{1}} ^{{x}} \:\frac{{dt}}{\sqrt{\mathrm{1}+{e}^{−{t}} }} \\ $$
Commented by mathmax by abdo last updated on 04/Feb/20
![let take Φ(x)=∫_1 ^(x ) (dt/(√(1−e^(−t) ))) changement (√(1−e^(−t) ))=u give 1−e^(−t) =u^2 ⇒e^(−t) =1−u^2 ⇒−t=ln(1−u^2 ) ⇒t=−ln(1−u^2 ) ⇒ Φ(x)= ∫_(√(1−e^(−1) )) ^(√(1−e^(−x) )) ((2u)/((1−u^2 )u))du =2 ∫_(√(1−e^(−1) )) ^(√(1−e^(−x) )) (du/((1−u)(1+u))) =∫_(√(1−e^(−1) )) ^(√(1−e^(−x) )) {(1/(1+u))+(1/(1−u))}du=[ln∣((1+u)/(1−u))∣]_(√(1−e^(−1) )) ^(√(1−e^(−x) )) =ln∣((1+(√(1−e^(−x) )))/(1−(√(1−e^(−x) ))))∣−ln∣((1+(√(1−e^(−1) )))/(1−(√(1−e^(−1) ))))∣★](Q80635.png)
$${let}\:{take}\:\:\Phi\left({x}\right)=\int_{\mathrm{1}} ^{{x}\:} \:\frac{{dt}}{\sqrt{\mathrm{1}−{e}^{−{t}} }}\:\:{changement}\:\sqrt{\mathrm{1}−{e}^{−{t}} }={u}\:{give} \\ $$$$\mathrm{1}−{e}^{−{t}} ={u}^{\mathrm{2}} \:\Rightarrow{e}^{−{t}} =\mathrm{1}−{u}^{\mathrm{2}} \:\Rightarrow−{t}={ln}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\:\Rightarrow{t}=−{ln}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\Phi\left({x}\right)=\:\int_{\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }} ^{\sqrt{\mathrm{1}−{e}^{−{x}} }} \:\:\:\:\:\:\frac{\mathrm{2}{u}}{\left(\mathrm{1}−{u}^{\mathrm{2}} \right){u}}{du}\:=\mathrm{2}\:\int_{\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }} ^{\sqrt{\mathrm{1}−{e}^{−{x}} }} \:\:\:\:\frac{{du}}{\left(\mathrm{1}−{u}\right)\left(\mathrm{1}+{u}\right)} \\ $$$$=\int_{\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }} ^{\sqrt{\mathrm{1}−{e}^{−{x}} }} \:\:\:\left\{\frac{\mathrm{1}}{\mathrm{1}+{u}}+\frac{\mathrm{1}}{\mathrm{1}−{u}}\right\}{du}=\left[{ln}\mid\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\mid\right]_{\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }} ^{\sqrt{\mathrm{1}−{e}^{−{x}} }} \\ $$$$={ln}\mid\frac{\mathrm{1}+\sqrt{\mathrm{1}−{e}^{−{x}} }}{\mathrm{1}−\sqrt{\mathrm{1}−{e}^{−{x}} }}\mid−{ln}\mid\frac{\mathrm{1}+\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }}{\mathrm{1}−\sqrt{\mathrm{1}−{e}^{−\mathrm{1}} }}\mid\bigstar \\ $$