Question Number 70909 by rajesh4661kumar@gmail.com last updated on 09/Oct/19
Commented by rajesh4661kumar@gmail.com last updated on 10/Oct/19
$$\mathrm{28} \\ $$
Commented by Abdo msup. last updated on 10/Oct/19
![f(t)=∫_0 ^1 ((x^t −1)/(lnx))dx changement lnx=−u give f(t)=−∫_0 ^∞ ((e^(−tu) −1)/(−u))(−e^(−u) )du =−∫_0 ^∞ ((e^(−(t+1)u) −e^(−u) )/u) du=∫_0 ^∞ ((e^(−u) −e^(−(t+1)u) )/u)du f^′ (t) = ∫_0 ^∞ e^(−(t+1)u) du =[−(1/(t+1))e^(−(t+1)u) ]_(u=0) ^∞ =(1/(t+1)) ⇒f(t)=ln(t+1) +c f(o)=0=c ⇒f(t)=ln(t+1) (t≥0)](https://www.tinkutara.com/question/Q70938.png)
$${f}\left({t}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{{t}} −\mathrm{1}}{{lnx}}{dx}\:\:{changement}\:{lnx}=−{u}\:{give} \\ $$$${f}\left({t}\right)=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{tu}} −\mathrm{1}}{−{u}}\left(−{e}^{−{u}} \right){du} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\left({t}+\mathrm{1}\right){u}} −{e}^{−{u}} }{{u}}\:{du}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{u}} −{e}^{−\left({t}+\mathrm{1}\right){u}} }{{u}}{du} \\ $$$${f}^{'} \left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\left({t}+\mathrm{1}\right){u}} {du}\:=\left[−\frac{\mathrm{1}}{{t}+\mathrm{1}}{e}^{−\left({t}+\mathrm{1}\right){u}} \right]_{{u}=\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{1}}{{t}+\mathrm{1}}\:\Rightarrow{f}\left({t}\right)={ln}\left({t}+\mathrm{1}\right)\:+{c} \\ $$$${f}\left({o}\right)=\mathrm{0}={c}\:\Rightarrow{f}\left({t}\right)={ln}\left({t}+\mathrm{1}\right)\:\:\:\left({t}\geqslant\mathrm{0}\right) \\ $$
Answered by Henri Boucatchou last updated on 10/Oct/19
$${What}\:\:{question}\left({s}\right)\:? \\ $$