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Question Number 59275 by Mr X pcx last updated on 07/May/19
let f(x)=∫_0 ^1 ln(1+xe^(−t) )dt 1)find a explicit form of f(x) 2)calculate ∫_0 ^1 ln(1+2e^(−t) )dt 3) developp f(x)at integr serie if ∣x∣<1
$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{xe}^{−{t}} \right){dt} \\ $$$$\left.\mathrm{1}\right){find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+\mathrm{2}{e}^{−{t}} \right){dt} \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\left({x}\right){at}\:{integr}\:{serie}\:{if}\:\mid{x}\mid<\mathrm{1} \\ $$
Commented by maxmathsup by imad last updated on 17/May/19
1) we have f^′ (x) =∫_0 ^1 (e^(−t) /(1+xe^(−t) )) dt = ∫_0 ^1 (1/(e^t +x)) dt⇒ f^′ (x) =_(e^t =u) ∫_1 ^e (1/(u+x)) (du/u) =(1/x)∫_1 ^e ((1/u) −(1/(u+x)))du (for x≠0) =(1/x)[ln∣(u/(u+x))∣]_1 ^e =(1/x) {ln((e/(e+x)))−ln((1/(1+x)))} =(1/x){1−ln(e+x) +ln(1+x)} =(1/x) −((ln(x+e))/x) +((ln(1+x))/x) ⇒ f(x) =ln∣x∣−∫ ((ln(x+e))/x) dx +∫ ((ln(1+x))/x) dx +C ....be continued...
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{e}^{−{t}} }{\mathrm{1}+{xe}^{−{t}} }\:{dt}\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{{e}^{{t}} \:+{x}}\:{dt}\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=_{{e}^{{t}} ={u}} \:\:\:\:\int_{\mathrm{1}} ^{{e}} \:\:\:\frac{\mathrm{1}}{{u}+{x}}\:\frac{{du}}{{u}}\:=\frac{\mathrm{1}}{{x}}\int_{\mathrm{1}} ^{{e}} \:\left(\frac{\mathrm{1}}{{u}}\:−\frac{\mathrm{1}}{{u}+{x}}\right){du}\:\:\:\:\:\:\left({for}\:{x}\neq\mathrm{0}\right) \\ $$$$=\frac{\mathrm{1}}{{x}}\left[{ln}\mid\frac{{u}}{{u}+{x}}\mid\right]_{\mathrm{1}} ^{{e}} \:=\frac{\mathrm{1}}{{x}}\:\left\{{ln}\left(\frac{{e}}{{e}+{x}}\right)−{ln}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{{x}}\left\{\mathrm{1}−{ln}\left({e}+{x}\right)\:+{ln}\left(\mathrm{1}+{x}\right)\right\}\:=\frac{\mathrm{1}}{{x}}\:−\frac{{ln}\left({x}+{e}\right)}{{x}}\:+\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:={ln}\mid{x}\mid−\int\:\:\frac{{ln}\left({x}+{e}\right)}{{x}}\:{dx}\:+\int\:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}\:{dx}\:+{C}\:….{be}\:{continued}… \\ $$
Commented by maxmathsup by imad last updated on 17/May/19
let take another essay by parts u^′ =1 and v =ln(1+xe^(−t) ) f(x) =[t ln(1+xe^(−t) )]_0 ^1 −∫_0 ^1 t .((−xe^(−t) )/(1+xe^(−t) )) dt =ln(1+x e^(−1) ) +x ∫_0 ^1 ((t e^(−t) )/(1+x e^(−t) )) dt but ∫_0 ^1 ((t e^(−t) )/(1+x e^(−t) ))dt =∫_0 ^1 (t/(e^t +x)) dt =_(e^t =u) ∫_1 ^e ((ln(u))/(u +x)) (du/u) =(1/x)∫_1 ^e lnu{(1/u) −(1/(u+x))}dx =(1/x) ∫_1 ^e ((ln(u))/u) du −(1/x) ∫_1 ^e ((ln(u))/(u+x)) dx ∫_1 ^e ((ln(u))/u) du =_(by parts) [ln^2 (u)]_1 ^e −∫_1 ^e ((lnu)/u) du ⇒ 2 ∫_1 ^e ((ln(u))/u) du =1 ⇒ ∫_1 ^e ((ln(u))/u) du =(1/2) also by parts ∫_1 ^e ((lnu)/(u+x)) dx =[ln∣u+x∣lnu]_1 ^e −∫_1 ^e ((ln∣u+x∣)/u) du =ln(e+x)−∫_1 ^e ((ln(u+x))/u) du if we suppose x>0 ....be continued ...
$${let}\:{take}\:{another}\:{essay}\:\:\:{by}\:{parts}\:\:\:{u}^{'} =\mathrm{1}\:\:\:{and}\:{v}\:={ln}\left(\mathrm{1}+{xe}^{−{t}} \right) \\ $$$${f}\left({x}\right)\:=\left[{t}\:{ln}\left(\mathrm{1}+{xe}^{−{t}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}\:.\frac{−{xe}^{−{t}} }{\mathrm{1}+{xe}^{−{t}} }\:{dt} \\ $$$$={ln}\left(\mathrm{1}+{x}\:{e}^{−\mathrm{1}} \right)\:+{x}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{t}\:{e}^{−{t}} }{\mathrm{1}+{x}\:{e}^{−{t}} }\:{dt}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}\:{e}^{−{t}} }{\mathrm{1}+{x}\:{e}^{−{t}} }{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}}{{e}^{{t}} \:+{x}}\:{dt}\:\:=_{{e}^{{t}} ={u}} \:\:\:\:\int_{\mathrm{1}} ^{{e}} \:\:\:\frac{{ln}\left({u}\right)}{{u}\:+{x}}\:\frac{{du}}{{u}} \\ $$$$=\frac{\mathrm{1}}{{x}}\int_{\mathrm{1}} ^{{e}} \:\:{lnu}\left\{\frac{\mathrm{1}}{{u}}\:−\frac{\mathrm{1}}{{u}+{x}}\right\}{dx}\:=\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{1}} ^{{e}} \:\frac{{ln}\left({u}\right)}{{u}}\:{du}\:−\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{1}} ^{{e}} \:\frac{{ln}\left({u}\right)}{{u}+{x}}\:{dx} \\ $$$$\int_{\mathrm{1}} ^{{e}} \:\:\frac{{ln}\left({u}\right)}{{u}}\:{du}\:=_{{by}\:{parts}} \:\:\:\:\:\left[{ln}^{\mathrm{2}} \left({u}\right)\right]_{\mathrm{1}} ^{{e}} \:−\int_{\mathrm{1}} ^{{e}} \:\frac{{lnu}}{{u}}\:{du}\:\Rightarrow\:\mathrm{2}\:\int_{\mathrm{1}} ^{{e}} \:\frac{{ln}\left({u}\right)}{{u}}\:{du}\:=\mathrm{1}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{{e}} \:\frac{{ln}\left({u}\right)}{{u}}\:{du}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\:{also}\:{by}\:{parts} \\ $$$$\int_{\mathrm{1}} ^{{e}} \:\:\frac{{lnu}}{{u}+{x}}\:{dx}\:=\left[{ln}\mid{u}+{x}\mid{lnu}\right]_{\mathrm{1}} ^{{e}} \:−\int_{\mathrm{1}} ^{{e}} \:\:\frac{{ln}\mid{u}+{x}\mid}{{u}}\:{du}\:={ln}\left({e}+{x}\right)−\int_{\mathrm{1}} ^{{e}} \:\frac{{ln}\left({u}+{x}\right)}{{u}}\:{du} \\ $$$${if}\:{we}\:{suppose}\:{x}>\mathrm{0}\:\:….{be}\:{continued}\:… \\ $$
Commented by maxmathsup by imad last updated on 17/May/19
2) let I =∫_0 ^1 ln(1+2 e^(−t) )dt chang. 2e^(−t) =x give e^(−t) =(x/2) ⇒t =−ln((x/2)) 3) we have ln^′ (1+u) =(1/(1+u)) =Σ_(n=0) ^∞ (−1)^n u^n ⇒ln(1+u) =Σ_(n=0) ^∞ (((−1)^n )/(n+1))u^(n+1) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) u^n ⇒ln(1+xe^(−t) ) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)(xe^(−t) )^n =Σ_(n=1) ^∞ (−1)^(n−1) (x^n /n) e^(−nt) ⇒∫_0 ^1 ln(1+xe^(−t) )dt =Σ_(n=1) ^∞ (−1)^(n−1) (x^n /n) ∫_0 ^1 e^(−nt) dt =Σ_(n=1) ^∞ (−1)^n (x^n /n^2 ){e^(−n) −1} ⇒f(x) =Σ_(n=1) ^∞ (((−1)^n (e^(−n) −1))/n^2 ) x^n .
$$\left.\mathrm{2}\right)\:{let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+\mathrm{2}\:{e}^{−{t}} \right){dt}\:\:\:{chang}.\:\mathrm{2}{e}^{−{t}} \:={x}\:{give}\:{e}^{−{t}} \:=\frac{{x}}{\mathrm{2}}\:\Rightarrow{t}\:=−{ln}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$\left.\mathrm{3}\right)\:{we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {u}^{{n}} \:\Rightarrow{ln}\left(\mathrm{1}+{u}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{u}^{{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{u}^{{n}} \:\Rightarrow{ln}\left(\mathrm{1}+{xe}^{−{t}} \right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\left({xe}^{−{t}} \right)^{{n}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{x}^{{n}} }{{n}}\:{e}^{−{nt}} \:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{xe}^{−{t}} \right){dt}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{x}^{{n}} }{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{nt}} \:{dt} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}} \:\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }\left\{{e}^{−{n}} −\mathrm{1}\right\}\:\Rightarrow{f}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \left({e}^{−{n}} −\mathrm{1}\right)}{{n}^{\mathrm{2}} }\:{x}^{{n}} \:. \\ $$

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