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Question Number 45235 by maxmathsup by imad last updated on 10/Oct/18
let ∣a∣<1 and f(a)=∫_0 ^1 ln(x)ln(1+ax)dx 1) find a explicit form of f(a) 2) calculate g(a) =∫_0 ^1 ((xln(x))/(1+ax))dx 3) calculate ∫_0 ^1 ln(x)ln(2+x)dx 4) calculate ∫_0 ^1 ((xln(x))/(2+x))dx 5) calculate u_n =∫_0 ^1 ((xln(x))/(n+x))dx with n integr and n>1 find nature of the serie Σ u_n
$${let}\:\mid{a}\mid<\mathrm{1}\:{and}\:\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}+{ax}\right){dx} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{xln}\left({x}\right)}{\mathrm{1}+{ax}}{dx} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{2}+{x}\right){dx} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{xln}\left({x}\right)}{\mathrm{2}+{x}}{dx}\: \\ $$$$\left.\mathrm{5}\right)\:{calculate}\:{u}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{xln}\left({x}\right)}{{n}+{x}}{dx}\:{with}\:{n}\:{integr}\:{and}\:{n}>\mathrm{1} \\ $$$${find}\:{nature}\:{of}\:{the}\:{serie}\:\Sigma\:{u}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 13/Oct/18
1) we have ln^′ (1+u)=Σ_(n=0) ^∞ (−1)^n u^n with ∣u∣<1 ⇒ ln(1+u) =Σ_(n=0) ^∞ (((−1)^n u^(n+1) )/(n+1)) =Σ_(n=1) ^∞ (−1)^(n−1) (u^n /n) ⇒ ln(1+ax) =Σ_(n=1) ^∞ (−1)^(n−1) ((a^n x^n )/n) ⇒f(a) =∫_0 ^1 ln(x){Σ_(n=1) ^∞ (−1)^(n−1) ((a^n x^n )/n)}dx =Σ_(n=1) ^∞ (−1)^(n−1) (a^n /n) ∫_0 ^1 x^n ln(x)dx by parts A_n =∫_0 ^1 x^n ln(x)dx =[(1/(n+1))x^(n+1) ln(x)]_0 ^1 −∫_0 ^1 (1/(n+1)) x^n dx =−(1/((n+1)^2 )) ⇒f(a) = Σ_(n=1) ^∞ (−1)^n a^n (1/(n(n+1)^2 )) let decompose F(x) =(1/(x(x+1)^2 )) ⇒F(x) = (a/x) +(b/(x+1)) +(c/((x+1)^2 )) c =lim_(x→−1) (x+1)^2 F(x) =−1 a =lim_(x→0) xF(x) =1 ⇒F(x) =(1/x) +(b/(x+1)) −(1/((x+1)^2 )) F(2) =(1/(18)) =(1/2) +(b/3) −(1/9) ⇒1=9+6b−2 ⇒1=7+6b ⇒1−7=6b ⇒b=−1 ⇒F(x)=(1/x)−(1/(x+1)) −(1/((x+1)^2 )) ⇒ f(a) =Σ_(n=1) ^∞ (−a)^n {(1/n) −(1/(n+1)) −(1/((n+1)^2 ))} =Σ_(n=1) ^∞ (((−a)^n )/n) −Σ_(n=1) ^∞ (((−a)^n )/(n+1)) −Σ_(n=1) ^∞ (((−a)^n )/((n+1)^2 )) but Σ_(n=1) ^∞ (((−a)^n )/n) = −ln(1+a) Σ_(n=1) ^∞ (((−a)^n )/(n+1)) =Σ_(n=2) ^∞ (((−a)^(n−1) )/n) =−(1/a){Σ_(n=1) ^∞ (((−a)^n )/n) −1} =−(1/a){−ln(1+a)−1} =(1/a) +((ln(1+a))/a) ⇒ f(a) =−ln(1+a)−((ln(1+a))/a) −(1/a) −Σ_(n=2) ^∞ (((−a)^(n−1) )/n^2 ) f(a) =−(1+(1/a))ln(1+a) −(1/a) −Σ_(n=2) ^∞ (((−a)^(n−1) )/n^2 )
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {u}^{{n}} \:\:{with}\:\mid{u}\mid<\mathrm{1}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{u}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} {u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{u}^{{n}} }{{n}}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{ax}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{a}^{{n}} {x}^{{n}} }{{n}}\:\Rightarrow{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right)\left\{\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{a}^{{n}} {x}^{{n}} }{{n}}\right\}{dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{a}^{{n}} }{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} {ln}\left({x}\right){dx}\:\:{by}\:{parts} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} {ln}\left({x}\right){dx}\:=\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} {ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{n}+\mathrm{1}}\:{x}^{{n}} {dx} \\ $$$$=−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{f}\left({a}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {a}^{{n}} \:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:{let}\:{decompose} \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{F}\left({x}\right)\:=\:\frac{{a}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{c}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${c}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)\:=−\mathrm{1} \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{0}} \:{xF}\left({x}\right)\:=\mathrm{1}\:\:\Rightarrow{F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{2}\right)\:=\frac{\mathrm{1}}{\mathrm{18}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{b}}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{9}}\:\Rightarrow\mathrm{1}=\mathrm{9}+\mathrm{6}{b}−\mathrm{2}\:\Rightarrow\mathrm{1}=\mathrm{7}+\mathrm{6}{b}\:\Rightarrow\mathrm{1}−\mathrm{7}=\mathrm{6}{b}\:\Rightarrow{b}=−\mathrm{1} \\ $$$$\Rightarrow{F}\left({x}\right)=\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\left(−{a}\right)^{{n}} \left\{\frac{\mathrm{1}}{{n}}\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−{a}\right)^{{n}} }{{n}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−{a}\right)^{{n}} }{{n}+\mathrm{1}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−{a}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{but} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−{a}\right)^{{n}} }{{n}}\:=\:−{ln}\left(\mathrm{1}+{a}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−{a}\right)^{{n}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−{a}\right)^{{n}−\mathrm{1}} }{{n}}\:=−\frac{\mathrm{1}}{{a}}\left\{\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−{a}\right)^{{n}} }{{n}}\:−\mathrm{1}\right\} \\ $$$$=−\frac{\mathrm{1}}{{a}}\left\{−{ln}\left(\mathrm{1}+{a}\right)−\mathrm{1}\right\}\:=\frac{\mathrm{1}}{{a}}\:+\frac{{ln}\left(\mathrm{1}+{a}\right)}{{a}}\:\Rightarrow \\ $$$${f}\left({a}\right)\:=−{ln}\left(\mathrm{1}+{a}\right)−\frac{{ln}\left(\mathrm{1}+{a}\right)}{{a}}\:−\frac{\mathrm{1}}{{a}}\:−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−{a}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} } \\ $$$${f}\left({a}\right)\:=−\left(\mathrm{1}+\frac{\mathrm{1}}{{a}}\right){ln}\left(\mathrm{1}+{a}\right)\:−\frac{\mathrm{1}}{{a}}\:−\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−{a}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 13/Oct/18
2) we have f(a) =∫_0 ^1 ln(x)ln(1+ax)dx ⇒ f^′ (a) = ∫_0 ^1 ((xln(x))/(1+ax)) dx =g(a) but f(a)=−(1+(1/a))ln(1+a)−(1/a) −Σ_(n=2) ^∞ (((−a)^(n−1) )/n^2 ) ⇒f^′ (a) =(1/a^2 )ln(1+a)−(((a+1)/a))(1/(1+a)) +(1/a^2 ) −w^′ (a) withw(a)=Σ_(n=2) ^∞ (((−a)^(n−1) )/n^2 ) =((ln(1+a))/a^2 ) −(1/a) +(1/a^2 ) −w^′ (a) but w(a) =Σ_(n=2) ^∞ (−1)^(n−1) (a^(n−1) /n^2 ) due to uniform convegence we get w^′ (a) =Σ_(n=2) ^∞ (n−1)(−1)^(n−1) (a^(n−2) /n^2 ) =Σ_(n=2) ^∞ (−1)^(n−1) (a^(n−2) /n) −Σ_(n=2) ^∞ (−1)^(n−1) (a^(n−2) /n^2 ) ....be continued...
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}+{ax}\right){dx}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)\:=\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{xln}\left({x}\right)}{\mathrm{1}+{ax}}\:{dx}\:={g}\left({a}\right)\:\:{but}\:{f}\left({a}\right)=−\left(\mathrm{1}+\frac{\mathrm{1}}{{a}}\right){ln}\left(\mathrm{1}+{a}\right)−\frac{\mathrm{1}}{{a}}\:−\sum_{{n}=\mathrm{2}} ^{\infty} \frac{\left(−{a}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} } \\ $$$$\Rightarrow{f}^{'} \left({a}\right)\:=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }{ln}\left(\mathrm{1}+{a}\right)−\left(\frac{{a}+\mathrm{1}}{{a}}\right)\frac{\mathrm{1}}{\mathrm{1}+{a}}\:+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:−{w}^{'} \left({a}\right)\:{withw}\left({a}\right)=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−{a}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} } \\ $$$$=\frac{{ln}\left(\mathrm{1}+{a}\right)}{{a}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{{a}}\:+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:−{w}^{'} \left({a}\right)\:\:{but} \\ $$$${w}\left({a}\right)\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\:\:\frac{{a}^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:\:{due}\:{to}\:{uniform}\:{convegence}\:{we}\:{get} \\ $$$${w}^{'} \left({a}\right)\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\left({n}−\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\:\frac{{a}^{{n}−\mathrm{2}} }{{n}^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{2}} ^{\infty} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\frac{{a}^{{n}−\mathrm{2}} }{{n}} \\ $$$$−\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\:\:\frac{{a}^{{n}−\mathrm{2}} }{{n}^{\mathrm{2}} }\:….{be}\:{continued}… \\ $$

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