Question Number 36412 by abdo.msup.com last updated on 01/Jun/18
$${calculate}\:{I}_{\lambda} \:=\int_{\mathrm{0}} ^{\lambda} \:{e}^{−{x}} {ln}\left(\mathrm{1}+{e}^{{x}} \right){dx} \\ $$
Commented by abdo.msup.com last updated on 02/Jun/18
![changement e^x =t give I_λ = ∫_1 ^e^λ ((ln(1+t))/t) (dt/t) = ∫_1 ^e^λ (1/t^2 )ln(1+t)dt and by parts we get I_λ = [−(1/t)ln(1+t)]_1 ^e^λ +∫_1 ^e^λ (1/(t(1+t)))dt =ln(2) −e^(−λ) ln(1+e^λ ) + ∫_1 ^e^λ { (1/t) −(1/(t+1))}dt =[ln((t/(t+1)))]_1 ^e^λ = ln( (e^λ /(e^λ +1))) −ln((1/2)) =λ −ln(1+e^λ ) +ln(2) ⇒ I_λ =ln(2)−e^(−λ) ln(1+e^λ ) +λ −ln(1+e^λ ) +ln(2) I_λ = 2ln(2) +λ −(1+e^(−λ) )ln(1+e^λ ).](Q36509.png)
$${changement}\:\:{e}^{{x}} ={t}\:{give} \\ $$$${I}_{\lambda} \:=\:\int_{\mathrm{1}} ^{{e}^{\lambda} } \:\frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}\:\frac{{dt}}{{t}}\:\:=\:\int_{\mathrm{1}} ^{{e}^{\lambda} } \:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{ln}\left(\mathrm{1}+{t}\right){dt} \\ $$$${and}\:{by}\:{parts}\:{we}\:{get} \\ $$$${I}_{\lambda} \:=\:\left[−\frac{\mathrm{1}}{{t}}{ln}\left(\mathrm{1}+{t}\right)\right]_{\mathrm{1}} ^{{e}^{\lambda} } \:\:+\int_{\mathrm{1}} ^{{e}^{\lambda} } \:\:\frac{\mathrm{1}}{{t}\left(\mathrm{1}+{t}\right)}{dt} \\ $$$$={ln}\left(\mathrm{2}\right)\:−{e}^{−\lambda} {ln}\left(\mathrm{1}+{e}^{\lambda} \right) \\ $$$$+\:\int_{\mathrm{1}} ^{{e}^{\lambda} } \left\{\:\frac{\mathrm{1}}{{t}}\:−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right\}{dt} \\ $$$$=\left[{ln}\left(\frac{{t}}{{t}+\mathrm{1}}\right)\right]_{\mathrm{1}} ^{{e}^{\lambda} } \:=\:{ln}\left(\:\frac{{e}^{\lambda} }{{e}^{\lambda} \:+\mathrm{1}}\right)\:−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\lambda\:−{ln}\left(\mathrm{1}+{e}^{\lambda} \right)\:+{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${I}_{\lambda} \:={ln}\left(\mathrm{2}\right)−{e}^{−\lambda} {ln}\left(\mathrm{1}+{e}^{\lambda} \right)\:+\lambda\:−{ln}\left(\mathrm{1}+{e}^{\lambda} \right) \\ $$$$+{ln}\left(\mathrm{2}\right) \\ $$$${I}_{\lambda} \:=\:\mathrm{2}{ln}\left(\mathrm{2}\right)\:+\lambda\:−\left(\mathrm{1}+{e}^{−\lambda} \right){ln}\left(\mathrm{1}+{e}^{\lambda} \right). \\ $$
Answered by sma3l2996 last updated on 02/Jun/18
![u=e^(−x) ⇒du=−e^(−x) dx I_λ =−∫_1 ^e^(−λ) ln(1+(1/u))du =∫_e^(−λ) ^1 (ln(1+u)−ln(u))du ∫ln(u)du=ulnu−∫(u/u)du+k=ulnu−u+K ∫ln(1+u)du=(u+1)ln(1+u)−∫((u+1)/(1+u))+c =(u+1)ln(1+u)−u+C So I_λ =∫_e^(−λ) ^1 (ln(u+1)−ln(u))du =[(u+1)ln(u+1)−uln(u)]_e^(−λ) ^1 =(e^(−λ) +1)ln(e^(−λ) +1)+λe^(−λ) −2ln2 =(e^(−λ) +1)ln(e^λ +1)−λ−2ln2](Q36496.png)
$${u}={e}^{−{x}} \Rightarrow{du}=−{e}^{−{x}} {dx} \\ $$$${I}_{\lambda} =−\int_{\mathrm{1}} ^{{e}^{−\lambda} } {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{u}}\right){du} \\ $$$$=\int_{{e}^{−\lambda} } ^{\mathrm{1}} \left({ln}\left(\mathrm{1}+{u}\right)−{ln}\left({u}\right)\right){du} \\ $$$$\int{ln}\left({u}\right){du}={ulnu}−\int\frac{{u}}{{u}}{du}+{k}={ulnu}−{u}+{K} \\ $$$$\int{ln}\left(\mathrm{1}+{u}\right){du}=\left({u}+\mathrm{1}\right){ln}\left(\mathrm{1}+{u}\right)−\int\frac{{u}+\mathrm{1}}{\mathrm{1}+{u}}+{c} \\ $$$$=\left({u}+\mathrm{1}\right){ln}\left(\mathrm{1}+{u}\right)−{u}+{C} \\ $$$${So}\:\:\:{I}_{\lambda} =\int_{{e}^{−\lambda} } ^{\mathrm{1}} \left({ln}\left({u}+\mathrm{1}\right)−{ln}\left({u}\right)\right){du} \\ $$$$=\left[\left({u}+\mathrm{1}\right){ln}\left({u}+\mathrm{1}\right)−{uln}\left({u}\right)\right]_{{e}^{−\lambda} } ^{\mathrm{1}} \\ $$$$=\left({e}^{−\lambda} +\mathrm{1}\right){ln}\left({e}^{−\lambda} +\mathrm{1}\right)+\lambda{e}^{−\lambda} −\mathrm{2}{ln}\mathrm{2} \\ $$$$=\left({e}^{−\lambda} +\mathrm{1}\right){ln}\left({e}^{\lambda} +\mathrm{1}\right)−\lambda−\mathrm{2}{ln}\mathrm{2} \\ $$