Question Number 38127 by maxmathsup by imad last updated on 22/Jun/18
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} \sqrt{\mathrm{1}+{e}^{−\mathrm{2}{x}} }{dx} \\ $$
Commented by abdo mathsup 649 cc last updated on 23/Jun/18
![let I = ∫_0 ^∞ e^(−x) (√(1+e^(−2x) ))dx changement e^(−x) =t give −x=ln(t) ⇒x=−ln(t) and I = −∫_0 ^1 t(√(1+t^2 )) ((−dt)/t) = ∫_0 ^1 (√(1+t^2 )) dt also chsng. t=shu give I = ∫_0 ^(argsh(1)) ch(u)ch(u)du = ∫_0 ^(ln(1+(√2))) ((1+ch(2u))/2)du =(1/2)ln(1+(√2)) +(1/4)[sh(2u)]_0 ^(ln(1+(√2))) =(1/2)ln(1+(√2)) +(1/4)[ ((e^(2u) −e^(−2u) )/2)]_0 ^(ln(1+(√2))) =(1/2)ln(1+(√2)) +(1/8){ (1+(√2))^2 −(1+(√2))^(−2) }](https://www.tinkutara.com/question/Q38219.png)
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} \sqrt{\mathrm{1}+{e}^{−\mathrm{2}{x}} }{dx}\:{changement}\:{e}^{−{x}} ={t} \\ $$$${give}\:\:−{x}={ln}\left({t}\right)\:\Rightarrow{x}=−{ln}\left({t}\right)\:{and} \\ $$$${I}\:=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:\frac{−{dt}}{{t}}\: \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:\:{also}\:{chsng}.\:{t}={shu}\:{give} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{{argsh}\left(\mathrm{1}\right)} \:\:{ch}\left({u}\right){ch}\left({u}\right){du} \\ $$$$=\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{u}\right)}{\mathrm{2}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sh}\left(\mathrm{2}{u}\right)\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:\:+\frac{\mathrm{1}}{\mathrm{4}}\left[\:\frac{{e}^{\mathrm{2}{u}} \:−{e}^{−\mathrm{2}{u}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{−\mathrm{2}} \right\} \\ $$