Question Number 73336 by mathmax by abdo last updated on 10/Nov/19
$${find}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}\left(\mathrm{1}+{e}^{{t}} \right){dt} \\ $$
Commented by mathmax by abdo last updated on 10/Nov/19
![let I =∫_0 ^∞ e^(−t) ln(1+e^t )dt by parts u^′ =e^(−t) and v=ln(1+e^t ) we get I =[−e^(−t) ln(1+e^t )]_0 ^(+∞) −∫_0 ^∞ (−e^(−t) )(e^t /(1+e^t ))dt =ln(2) +∫_0 ^∞ (dt/(1+e^t )) and ∫_0 ^∞ (dt/(1+e^t )) =_(e^t =u) ∫_1 ^(+∞) (du/(u(1+u))) =∫_1 ^(+∞) ((1/u)−(1/(1+u)))du =[ln∣(u/(1+u))∣]_1 ^(+∞) =−ln((1/2))=ln(2) ⇒ I =2ln(2).](https://www.tinkutara.com/question/Q73369.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}} {ln}\left(\mathrm{1}+{e}^{{t}} \right){dt}\:\:\:{by}\:{parts}\:{u}^{'} ={e}^{−{t}} \:{and}\:{v}={ln}\left(\mathrm{1}+{e}^{{t}} \right) \\ $$$${we}\:{get}\:\:{I}\:=\left[−{e}^{−{t}} {ln}\left(\mathrm{1}+{e}^{{t}} \right)\right]_{\mathrm{0}} ^{+\infty} \:−\int_{\mathrm{0}} ^{\infty} \:\left(−{e}^{−{t}} \right)\frac{{e}^{{t}} }{\mathrm{1}+{e}^{{t}} }{dt} \\ $$$$={ln}\left(\mathrm{2}\right)\:+\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\mathrm{1}+{e}^{{t}} }\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{e}^{{t}} }\:=_{{e}^{{t}} ={u}} \:\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{du}}{{u}\left(\mathrm{1}+{u}\right)} \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \left(\frac{\mathrm{1}}{{u}}−\frac{\mathrm{1}}{\mathrm{1}+{u}}\right){du}\:=\left[{ln}\mid\frac{{u}}{\mathrm{1}+{u}}\mid\right]_{\mathrm{1}} ^{+\infty} \:=−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)={ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${I}\:=\mathrm{2}{ln}\left(\mathrm{2}\right). \\ $$
Answered by mind is power last updated on 10/Nov/19
![by part =[−e^(−t) ln(1+e^t )]_0 ^(+∞) +∫(dt/(1+e^t )) =ln(2)+∫_0 ^(+∞) (e^(−t) /(1+e^(−t) ))dt=ln(2)+[−ln(1+e^(−t) )]_0 ^(+∞) =2ln(2)](https://www.tinkutara.com/question/Q73355.png)
$$\mathrm{by}\:\mathrm{part}\:=\left[−\mathrm{e}^{−\mathrm{t}} \mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{\mathrm{t}} \right)\right]_{\mathrm{0}} ^{+\infty} +\int\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{e}^{\mathrm{t}} } \\ $$$$=\mathrm{ln}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{e}^{−\mathrm{t}} }{\mathrm{1}+\mathrm{e}^{−\mathrm{t}} }\mathrm{dt}=\mathrm{ln}\left(\mathrm{2}\right)+\left[−\mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{t}} \right)\right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\mathrm{2ln}\left(\mathrm{2}\right) \\ $$