Question Number 38125 by maxmathsup by imad last updated on 22/Jun/18
$${let}\:\alpha>\mathrm{0}\:{find} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{e}^{−\alpha{x}} }{\sqrt{\mathrm{1}+{e}^{−\mathrm{2}\alpha{x}} }}{dx}\:. \\ $$
Commented bymath khazana by abdo last updated on 26/Jun/18
![let A(α) =∫_0 ^∞ (e^(−αx) /(√(1+e^(−2αx) )))dx changement e^(−αx) =t give −αx =ln(t) ⇒dx=−(1/(αt)) and A(α)=− ∫_0 ^1 (t/(√(1+t^2 ))) ((−1)/(αt)) dt =(1/α) ∫_0 ^1 (dt/(√(1+t^2 ))) =(1/α)[ ln(t +(√(1+t^2 )) ]_0 ^1 = (1/α)ln(1+(√2)).](Q38508.png)
$${let}\:{A}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−\alpha{x}} }{\sqrt{\mathrm{1}+{e}^{−\mathrm{2}\alpha{x}} }}{dx}\:{changement} \\ $$ $${e}^{−\alpha{x}} ={t}\:{give}\:−\alpha{x}\:={ln}\left({t}\right)\:\Rightarrow{dx}=−\frac{\mathrm{1}}{\alpha{t}}\:\:{and} \\ $$ $${A}\left(\alpha\right)=−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:\:\frac{−\mathrm{1}}{\alpha{t}}\:{dt} \\ $$ $$=\frac{\mathrm{1}}{\alpha}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:=\frac{\mathrm{1}}{\alpha}\left[\:{ln}\left({t}\:+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:\right]_{\mathrm{0}} ^{\mathrm{1}} \right. \\ $$ $$=\:\frac{\mathrm{1}}{\alpha}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right). \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18
$${e}^{−\alpha{x}} ={t}\:\:{dt}=−\alpha{e}^{−\alpha{x}} {dx} \\ $$ $$=\frac{−\mathrm{1}}{\alpha}\int_{\mathrm{1}} ^{\mathrm{0}} \frac{{dt}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} \:}} \\ $$ $$=\frac{\mathrm{1}}{\alpha}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}\:} \:}} \\ $$ $${use}\:{formula}\:\int\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }} \\ $$