Question Number 65290 by mathmax by abdo last updated on 27/Jul/19
$${f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{x}+{e}^{{t}} }\:\:\:{with}\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{aexplicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{2}+{e}^{{t}} } \\ $$$$\left.\mathrm{3}\right)\:{find}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\left({x}+{e}^{{t}} \right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\left(\mathrm{1}+{e}^{{t}} \right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{5}\right)\:{give}\:{f}^{\left({n}\right)} \left({x}\right)\:{at}\:{form}\:{of}\:{integrals} \\ $$$$\left.\mathrm{6}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$
Commented by mathmax by abdo last updated on 28/Jul/19
![1) f(x) =∫_0 ^1 (dt/(x+e^t )) =∫_0 ^1 (e^(−t) /(1+xe^(−t) ))dt =−(1/x)∫_0 ^1 ((−xe^(−t) )/(1+xe^(−t) ))dt =−(1/x)[ln(1+xe^(−t) )]_(t=0) ^1 =−(1/x){ln(1+xe^(−1) )−ln(1+x)} ⇒ f(x) =((ln(1+x))/x) −((ln(1+xe^(−1) ))/x) if x≠0 2)∫_0 ^1 (dt/(2+e^t )) =f(2) =((ln(3))/2) −((ln(1+2e^(−1) ))/2) 3)we have f^′ (x) =−∫_0 ^1 (dt/((x+e^t )^2 )) =−g(x) ⇒g(x)=−f^′ (x) f^′ (x) =−(1/x^2 )ln(1+x)+(1/(x(1+x))) +(1/x^2 )ln(1+xe^(−1) )−(1/x)(e^(−1) /(1+xe^(−1) )) ⇒ g(x) =(1/x^2 )ln(1+x)−(1/(x(1+x)))−(1/x^2 )ln(1+xe^(−1) )+(e^(−1) /(x(1+xe^(−1) ))) 4)∫_0 ^1 (dt/((1+e^t )^2 )) =g(1) =ln(2)−(1/2) −ln(1+e^(−1) ) +(e^(−1) /(1+e^(−1) ))](https://www.tinkutara.com/question/Q65318.png)
$$\left.\mathrm{1}\right)\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{x}+{e}^{{t}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{e}^{−{t}} }{\mathrm{1}+{xe}^{−{t}} }{dt}\:\:=−\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{−{xe}^{−{t}} }{\mathrm{1}+{xe}^{−{t}} }{dt} \\ $$$$=−\frac{\mathrm{1}}{{x}}\left[{ln}\left(\mathrm{1}+{xe}^{−{t}} \right)\right]_{{t}=\mathrm{0}} ^{\mathrm{1}} \:=−\frac{\mathrm{1}}{{x}}\left\{{ln}\left(\mathrm{1}+{xe}^{−\mathrm{1}} \right)−{ln}\left(\mathrm{1}+{x}\right)\right\}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}\:−\frac{{ln}\left(\mathrm{1}+{xe}^{−\mathrm{1}} \right)}{{x}}\:\:\:{if}\:\:\:{x}\neq\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{2}+{e}^{{t}} }\:={f}\left(\mathrm{2}\right)\:=\frac{{ln}\left(\mathrm{3}\right)}{\mathrm{2}}\:−\frac{{ln}\left(\mathrm{1}+\mathrm{2}{e}^{−\mathrm{1}} \right)}{\mathrm{2}} \\ $$$$\left.\mathrm{3}\right){we}\:{have}\:{f}^{'} \left({x}\right)\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\left({x}+{e}^{{t}} \right)^{\mathrm{2}} }\:=−{g}\left({x}\right)\:\Rightarrow{g}\left({x}\right)=−{f}^{'} \left({x}\right) \\ $$$${f}^{'} \left({x}\right)\:=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\mathrm{1}+{x}\right)+\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}\right)}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\mathrm{1}+{xe}^{−\mathrm{1}} \right)−\frac{\mathrm{1}}{{x}}\frac{{e}^{−\mathrm{1}} }{\mathrm{1}+{xe}^{−\mathrm{1}} }\:\Rightarrow \\ $$$${g}\left({x}\right)\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\mathrm{1}+{x}\right)−\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}\right)}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\mathrm{1}+{xe}^{−\mathrm{1}} \right)+\frac{{e}^{−\mathrm{1}} }{{x}\left(\mathrm{1}+{xe}^{−\mathrm{1}} \right)} \\ $$$$\left.\mathrm{4}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{e}^{{t}} \right)^{\mathrm{2}} }\:={g}\left(\mathrm{1}\right)\:={ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\:−{ln}\left(\mathrm{1}+{e}^{−\mathrm{1}} \right)\:+\frac{{e}^{−\mathrm{1}} }{\mathrm{1}+{e}^{−\mathrm{1}} } \\ $$
Commented by mathmax by abdo last updated on 28/Jul/19
![5) we have f(x) =∫_0 ^1 (dt/(x+e^t )) ⇒f^((n)) (x) =∫_0 ^1 (((−1)^n n!)/((x+e^t )^(n+1) ))dt =(−1)^n n! ∫_0 ^1 (dt/((x+e^t )^(n+1) )) 6) f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n but f^((n)) (0) =(−1)^n n! ∫_0 ^1 (dt/e^((n+1)t) ) =(−1)^n n! ∫_0 ^1 e^(−(n+1)t) dt =(−1)^n n! [−(1/(n+1))e^(−(n+1)t) ]_0 ^1 =−(((−1)^n n!)/(n+1)){ e^(−(n+1)) −1} ⇒f(x) =Σ_(n=0) ^∞ (((−1)^(n+1) )/(n+1)){e^(−(n+1)) −1} ⇒f(x) =Σ_(n=0) ^∞ (((−1)^n )/(n+1)){1−e^(−(n+1)) } .](https://www.tinkutara.com/question/Q65322.png)
$$\left.\mathrm{5}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{{x}+{e}^{{t}} }\:\Rightarrow{f}^{\left({n}\right)} \left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}+{e}^{{t}} \right)^{{n}+\mathrm{1}} }{dt} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} {n}!\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left({x}+{e}^{{t}} \right)^{{n}+\mathrm{1}} } \\ $$$$\left.\mathrm{6}\right)\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\:\:\:{but}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\left(−\mathrm{1}\right)^{{n}} {n}!\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{e}^{\left({n}+\mathrm{1}\right){t}} } \\ $$$$=\left(−\mathrm{1}\right)^{{n}} {n}!\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\left({n}+\mathrm{1}\right){t}} \:{dt}\:=\left(−\mathrm{1}\right)^{{n}} {n}!\:\left[−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\left({n}+\mathrm{1}\right){t}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{n}+\mathrm{1}}\left\{\:\:{e}^{−\left({n}+\mathrm{1}\right)} −\mathrm{1}\right\}\:\Rightarrow{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\left\{{e}^{−\left({n}+\mathrm{1}\right)} −\mathrm{1}\right\} \\ $$$$\Rightarrow{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\left\{\mathrm{1}−{e}^{−\left({n}+\mathrm{1}\right)} \right\}\:. \\ $$
Commented by mathmax by abdo last updated on 28/Jul/19
$${f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\left\{\mathrm{1}−{e}^{−\left({n}+\mathrm{1}\right)} \right\}\:{x}^{{n}} \\ $$
Answered by Eminem last updated on 28/Jul/19
$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{−{t}} }{{xe}^{−{t}} +\mathrm{1}}{dt}=−\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{xe}^{−{t}} }{\mathrm{1}+{xe}^{−{t}} }{dt}\:\:\:\:\:\forall{x}\in{IR}^{\ast} \\ $$$${f}\left({x}\right)=−\frac{\mathrm{1}}{{x}}\left\{{ln}\left(\mathrm{1}+{xe}^{−\mathrm{1}} \right)−{ln}\left(\mathrm{1}+{x}\right)\:\right\}=\frac{\mathrm{1}}{{x}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}+{xe}^{−\mathrm{1}} }\right) \\ $$$$\mathrm{2}\rangle{f}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{3}}{\mathrm{1}+\mathrm{2}{e}^{−\mathrm{1}} }\right) \\ $$$$\left.\mathrm{3}\right){find}\:\:{g}\left({x}\right\}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\left({x}+{e}^{{t}} \right)^{\mathrm{2}} } \\ $$$$\forall{x}\in{IR}^{\ast} \:{f}\left({x}\right)=\frac{\mathrm{1}}{{x}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}+{xe}^{−\mathrm{1}} }\right) \\ $$$${f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{{x}+{e}^{{t}} }\right){dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−\mathrm{1}}{\left({x}+{e}^{{t}} \right)^{\mathrm{2}} }{dt}=−{g}\left({x}\right)==> \\ $$$${g}\left({x}\right)=−{f}^{'} \left({x}\right)=−\frac{{d}}{{dx}}\left\{\frac{\mathrm{1}}{{x}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}+{xe}^{−\mathrm{1}} }\right)\right\} \\ $$$$=−\left\{\frac{−\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}+{xe}^{−\mathrm{1}} }\right)+\frac{\mathrm{1}}{{x}}×\left(\frac{\mathrm{1}+{xe}^{−\mathrm{1}} −{e}^{−\mathrm{1}} \left(\mathrm{1}+{x}\right)}{\frac{\left(\mathrm{1}+{xe}^{−\mathrm{1}} \right)^{\mathrm{2}} }{\frac{\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}+{xe}^{−\mathrm{1}} \right)}}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}+{xe}^{−\mathrm{1}} }\right)−\frac{\mathrm{1}}{{x}}×\frac{\left(\mathrm{1}−{e}^{−\mathrm{1}} \right)}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{xe}^{−\mathrm{1}} \right)}={g}\left({x}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\left(\mathrm{1}+{e}^{{t}} \right)^{\mathrm{2}} }={g}\left(\mathrm{1}\right)={ln}\left(\frac{\mathrm{2}}{\mathrm{1}+{e}^{−\mathrm{1}} }\right)−\frac{\left(\mathrm{1}−{e}^{−\mathrm{1}} \right)}{\left(\mathrm{1}+{e}^{−\mathrm{1}} \right)\left(\mathrm{2}\right)} \\ $$$${f}^{{n}} \left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}+{e}^{{t}} \right)^{{n}+\mathrm{1}} } \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}}\left\{{ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}+{xe}^{−\mathrm{1}} \right)\right\} \\ $$$$=\frac{\mathrm{1}}{{x}}\left\{\underset{{k}=\mathrm{0}} {\sum}\left(\left(−\mathrm{1}\right)^{{k}} \frac{{x}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}−\left(−\mathrm{1}\right)^{{k}} \frac{\left({e}^{−\mathrm{1}} {x}\right)^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\right)\right. \\ $$$$=\underset{{k}=\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} \left(\mathrm{1}−{e}^{−\left({k}+\mathrm{1}\right)} \right)\frac{{x}^{{k}} }{{k}+\mathrm{1}} \\ $$$$ \\ $$