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Question Number 64677 by mathmax by abdo last updated on 20/Jul/19
let f(x) =∫_0 ^1 lnt ln(1−xt)dt with ∣x∣<1 1)determine a explicit form for f(x) 2) find also g(x) =∫_0 ^1 ((tlnt)/(1−xt))dt 3) give f^((n)) (x) at form of integral 4) calculate ∫_0 ^1 ln(t)ln(1−t)dt and ∫_0 ^1 ln(t)ln(2−t)dt 5) calculate ∫_0 ^1 ((tln(t))/(2−t)) dt .
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {lnt}\:{ln}\left(\mathrm{1}−{xt}\right){dt}\:\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$$\left.\mathrm{1}\right){determine}\:{a}\:{explicit}\:{form}\:{for}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{also}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{tlnt}}{\mathrm{1}−{xt}}{dt} \\ $$$$\left.\mathrm{3}\right)\:{give}\:{f}^{\left({n}\right)} \left({x}\right)\:{at}\:{form}\:{of}\:{integral} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right){dt}\:\:{and}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left({t}\right){ln}\left(\mathrm{2}−{t}\right){dt} \\ $$$$\left.\mathrm{5}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{tln}\left({t}\right)}{\mathrm{2}−{t}}\:{dt}\:. \\ $$
Commented by mathmax by abdo last updated on 21/Jul/19
1) f(x) =∫_0 ^1 lntln(1−xt)dt case 1 0<x<1 changement xt =u give f(x) =∫_0 ^x ln((u/x))ln(1−u)(du/x) =(1/x) ∫_0 ^x (lnu−lnx)ln(1−u)du =(1/x) ∫_0 ^x ln(u)ln(1−u)du−((lnx)/x) ∫_0 ^x ln(1−u)du ∫_0 ^x ln(1−u)du =_(1−u =z) ∫_1 ^(1−x) ln(z)(−dz)=∫_(1−x) ^1 ln(z)dz [zlnz−z]_(1−x) ^1 =−1−((1−x)ln(1−x)−(1−x)) =−1−(1−x)ln(1−x) +1−x =−x−(1−x)ln(1−x) also we have ln^′ (1−u) =((−1)/(1−u)) =−Σ_(n=0) ^∞ u^n ⇒ ln(1−u) =−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) +c (c=0) =−Σ_(n=1) ^∞ (u^n /n) ⇒ ∫_0 ^x ln(u)ln(1−u)du =−∫_0 ^x ln(u)Σ_(n=1) ^∞ (u^n /n) du =−Σ_(n=1) ^∞ (1/n) ∫_0 ^x u^n lnu du by parts f^′ =u^n and g =lnu ⇒ w_n =∫_0 ^x u^n ln(u)du =[(1/(n+1))u^(n+1) lnu]_0 ^x −∫_0 ^x (u^(n+1) /(n+1)) (du/u) =((lnx)/(n+1))x^(n+1) −(1/(n+1)) ∫_0 ^x u^n du =((lnx)/(n+1))x^(n+1) −(1/(n+1))[(u^(n+1) /(n+1))]_0 ^x =((lnx)/(n+1))x^(n+1) −(x^(n+1) /((n+1)^2 )) ⇒ ∫_0 ^x ln(u)ln(1−u)du =−Σ_(n=1) ^∞ (1/n){((lnx)/(n+1))x^(n+1) −(x^(n+1) /((n+1)^2 ))} =−lnx Σ_(n=1) ^∞ (x^(n+1) /(n(n+1))) +Σ_(n=1) ^∞ (x^(n+1) /(n(n+1)^2 )) ⇒ f(x) =−lnx Σ_(n=1) ^∞ (x^n /(n(n+1))) +Σ_(n=1) ^∞ (x^n /(n(n+1)^2 )) −((lnx)/x)(−x−(1−x)ln(1−x)) f(x) =−lnxΣ_(n=1) ^∞ (x^n /(n(n+1))) +Σ_(n=1) ^∞ (x^n /(n(n+1)^2 )) +lnx( 1+(1−x)ln(1−x)).
$$\left.\mathrm{1}\right)\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {lntln}\left(\mathrm{1}−{xt}\right){dt}\:\:{case}\:\mathrm{1}\:\:\mathrm{0}<{x}<\mathrm{1}\:\:{changement}\:{xt}\:={u} \\ $$$${give}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} {ln}\left(\frac{{u}}{{x}}\right){ln}\left(\mathrm{1}−{u}\right)\frac{{du}}{{x}}\:=\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{{x}} \left({lnu}−{lnx}\right){ln}\left(\mathrm{1}−{u}\right){du} \\ $$$$=\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{{x}} {ln}\left({u}\right){ln}\left(\mathrm{1}−{u}\right){du}−\frac{{lnx}}{{x}}\:\int_{\mathrm{0}} ^{{x}} \:{ln}\left(\mathrm{1}−{u}\right){du} \\ $$$$\int_{\mathrm{0}} ^{{x}} {ln}\left(\mathrm{1}−{u}\right){du}\:=_{\mathrm{1}−{u}\:={z}} \:\:\:\int_{\mathrm{1}} ^{\mathrm{1}−{x}} {ln}\left({z}\right)\left(−{dz}\right)=\int_{\mathrm{1}−{x}} ^{\mathrm{1}} {ln}\left({z}\right){dz} \\ $$$$\left[{zlnz}−{z}\right]_{\mathrm{1}−{x}} ^{\mathrm{1}} \:=−\mathrm{1}−\left(\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}\right)−\left(\mathrm{1}−{x}\right)\right) \\ $$$$=−\mathrm{1}−\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}\right)\:+\mathrm{1}−{x}\:=−{x}−\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}\right) \\ $$$${also}\:{we}\:{have}\:{ln}^{'} \left(\mathrm{1}−{u}\right)\:=\frac{−\mathrm{1}}{\mathrm{1}−{u}}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{u}^{{n}} \:\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{u}\right)\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:+{c}\:\:\left({c}=\mathrm{0}\right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{u}^{{n}} }{{n}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{{x}} {ln}\left({u}\right){ln}\left(\mathrm{1}−{u}\right){du}\:=−\int_{\mathrm{0}} ^{{x}} {ln}\left({u}\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{u}^{{n}} }{{n}}\:{du} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{{x}} \:{u}^{{n}} {lnu}\:{du}\:\:\:{by}\:{parts}\:\:{f}^{'} \:={u}^{{n}} \:{and}\:{g}\:={lnu}\:\Rightarrow \\ $$$${w}_{{n}} =\int_{\mathrm{0}} ^{{x}} {u}^{{n}} {ln}\left({u}\right){du}\:=\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}{u}^{{n}+\mathrm{1}} {lnu}\right]_{\mathrm{0}} ^{{x}} \:−\int_{\mathrm{0}} ^{{x}} \:\frac{{u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:\frac{{du}}{{u}} \\ $$$$=\frac{{lnx}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{{x}} \:{u}^{{n}} \:{du}\:=\frac{{lnx}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\left[\frac{{u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\right]_{\mathrm{0}} ^{{x}} \\ $$$$=\frac{{lnx}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:−\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{{x}} {ln}\left({u}\right){ln}\left(\mathrm{1}−{u}\right){du}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\left\{\frac{{lnx}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} −\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$$$=−{lnx}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}+\mathrm{1}} }{{n}\left({n}+\mathrm{1}\right)}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}+\mathrm{1}} }{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}\left({x}\right)\:=−{lnx}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}\left({n}+\mathrm{1}\right)}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{{lnx}}{{x}}\left(−{x}−\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}\right)\right) \\ $$$${f}\left({x}\right)\:=−{lnx}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}\left({n}+\mathrm{1}\right)}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:+{lnx}\left(\:\mathrm{1}+\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}\right)\right). \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 21/Jul/19
we have Σ_(n=1) ^∞ (x^n /(n(n+1))) =Σ_(n=1) ^∞ ((1/n)−(1/(n+1)))x^n =Σ_(n=1) ^∞ (x^n /n) −Σ_(n=1) ^∞ (x^n /(n+1)) =−ln(1−x)−Σ_(n=2) ^∞ (x^(n−1) /n) =−ln(1−x)−(1/x)( Σ_(n=1) ^∞ (x^n /n) −x) =−ln(1−x)+(1/x)ln(1−x) +1 =((1/x)−1)ln(1−x) +1 let decompose F(t) =(1/(t(t+1)^2 )) ⇒F(t) =(a/t) +(b/(t+1)) +(c/((t+1)^2 )) a =lim_(t→0) tF(t) =1 c =lim_(t→−1) (t+1)^2 F(t) =−1 ⇒F(t) =(1/t) +(b/(t+1)) −(1/((t+1)^2 )) lim_(t→+∞) tF(t) =0 =a+b ⇒b=−1 ⇒ F(t) =(1/t)−(1/(t+1)) −(1/((t+1)^2 )) ⇒ Σ_(n=1) ^∞ (x^n /(n(n+1)^2 )) =Σ_(n=1) ^∞ (x^n /n) −Σ_(n=1) ^∞ (x^n /(n+1)) −Σ_(n=1) ^∞ (x^n /((n+1)^2 )) =−ln(1−x)−Σ_(n=2) ^∞ (x^(n−1) /n) −Σ_(n=1) ^∞ (x^n /((n+1)^2 )) =−ln(1−x)−(1/x) {Σ_(n=1) ^∞ (x^n /n)−1} −Σ_(n=1) ^∞ (x^n /((n+1)^2 )) =−ln(1−x)+((ln(1−x))/x) +(1/x) −Σ_(n=1) ^∞ (x^n /((n+1)^2 ))
$${we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}\left({n}+\mathrm{1}\right)}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){x}^{{n}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}+\mathrm{1}}\:=−{ln}\left(\mathrm{1}−{x}\right)−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{x}^{{n}−\mathrm{1}} }{{n}} \\ $$$$=−{ln}\left(\mathrm{1}−{x}\right)−\frac{\mathrm{1}}{{x}}\left(\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:\:−{x}\right) \\ $$$$=−{ln}\left(\mathrm{1}−{x}\right)+\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}−{x}\right)\:+\mathrm{1}\:=\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right){ln}\left(\mathrm{1}−{x}\right)\:+\mathrm{1}\:\:{let}\: \\ $$$${decompose}\:\:{F}\left({t}\right)\:=\frac{\mathrm{1}}{{t}\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\Rightarrow{F}\left({t}\right)\:=\frac{{a}}{{t}}\:+\frac{{b}}{{t}+\mathrm{1}}\:+\frac{{c}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${a}\:={lim}_{{t}\rightarrow\mathrm{0}} \:{tF}\left({t}\right)\:=\mathrm{1} \\ $$$${c}\:={lim}_{{t}\rightarrow−\mathrm{1}} \left({t}+\mathrm{1}\right)^{\mathrm{2}} \:{F}\left({t}\right)\:=−\mathrm{1}\:\Rightarrow{F}\left({t}\right)\:=\frac{\mathrm{1}}{{t}}\:+\frac{{b}}{{t}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{t}\rightarrow+\infty} \:{tF}\left({t}\right)\:=\mathrm{0}\:={a}+{b}\:\Rightarrow{b}=−\mathrm{1}\:\Rightarrow\:{F}\left({t}\right)\:=\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}+\mathrm{1}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−{ln}\left(\mathrm{1}−{x}\right)−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{x}^{{n}−\mathrm{1}} }{{n}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−{ln}\left(\mathrm{1}−{x}\right)−\frac{\mathrm{1}}{{x}}\:\left\{\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}−\mathrm{1}\right\}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−{ln}\left(\mathrm{1}−{x}\right)+\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}\:+\frac{\mathrm{1}}{{x}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 21/Jul/19
Σ_(n=1) ^∞ (x^n /(n(n+1)^2 )) =−ln(1−x)+((ln(1−x))/x) +1 −Σ_(n=1) ^∞ (x^n /((n+1)^2 )) ⇒ f(x) =
$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=−{ln}\left(\mathrm{1}−{x}\right)+\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}\:+\mathrm{1}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}\left({x}\right)\:= \\ $$
Commented by mathmax by abdo last updated on 22/Jul/19
2) we have f^′ (x) =∫_0 ^1 ((−t)/(1−xt))lnt dt =−∫_0 ^1 ((tlnt)/(1−xt))dt =−g(x) ⇒ g(x)=−f^′ (x) the function f is known rest to calculate f^′ (x) 3) we have f^′ (x) =∫_0 ^1 ((tlnt)/(xt−1)) dt =∫_0 ^1 ((lnt)/(x−(1/t))) dt ⇒ f^((n)) (x) =∫_0 ^1 ((1/(x−(1/t))))^((n−1)) lntdt =∫_0 ^1 (((−1)^(n−1) (n−1)!)/((x−(1/t))^n ))lntdt f^((n)) (x)=(−1)^(n−1) (n−1)! ∫_0 ^1 ((t^n lnt)/((xt−1)^n ))dt
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{t}}{\mathrm{1}−{xt}}{lnt}\:{dt}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{tlnt}}{\mathrm{1}−{xt}}{dt}\:=−{g}\left({x}\right)\:\Rightarrow \\ $$$${g}\left({x}\right)=−{f}^{'} \left({x}\right)\:\:{the}\:{function}\:{f}\:{is}\:{known}\:\:{rest}\:{to}\:{calculate}\:{f}^{'} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{tlnt}}{{xt}−\mathrm{1}}\:{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{lnt}}{{x}−\frac{\mathrm{1}}{{t}}}\:{dt}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{{x}−\frac{\mathrm{1}}{{t}}}\right)^{\left({n}−\mathrm{1}\right)} {lntdt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}−\frac{\mathrm{1}}{{t}}\right)^{{n}} }{lntdt} \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}^{{n}} {lnt}}{\left({xt}−\mathrm{1}\right)^{{n}} }{dt} \\ $$

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