Question Number 43918 by maxmathsup by imad last updated on 17/Sep/18
$$\left.\mathrm{1}\right)\:{find}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} {ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right){dt}\:\:\:{with}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right){dt}\:. \\ $$
Commented by maxmathsup by imad last updated on 19/Sep/18
![1) we have for 0<t≤x<1 ln^′ (1−t) =−(1/(1−t)) =−Σ_(n=0) ^∞ t^n ⇒ ln(1−t) =−Σ_(n=0) ^∞ (1/(n+1))t^(n+1) =−Σ_(n=1) ^∞ (t^n /n) ⇒ f(x) =−∫_0 ^x ln(t)(Σ_(n=1) ^∞ (t^n /n))dt =−Σ_(n=1) ^∞ (1/n) ∫_0 ^x t^n ln(t) dt by parts A_n (x)= ∫_0 ^x t^n ln(t)dt =[(1/(n+1))t^(n+1) ln(t)]_0 ^x −∫_0 ^x (1/(n+1))t^(n+1) (dt/t) =(1/(n+1)) x^(n+1) ln(x) −(1/(n+1))∫_0 ^x t^n dt = (1/(n+1))x^(n+1) ln(x)−(1/((n+1)^2 )) x^(n+1) ⇒ f(x) =−Σ_(n=1) ^∞ (1/n){ (1/(n+1)) x^(n+1) ln(x)−(1/((n+1)^2 )) x^(n+1) } = −Σ_(n=1) ^∞ (1/(n(n+1))) x^(n+1) ln(x) +Σ_(n=1) ^∞ (1/(n(n+1)^2 )) x^(n+1) =H(x)−K(x) K(x) =ln(x) Σ_(n=1) ^∞ {(1/n)−(1/(n+1))}x^(n+1) =ln(x)Σ_(n=1) ^∞ (x^(n+1) /n) −ln(x)Σ_(n=1) ^∞ (x^(n+1) /(n+1)) =xln(x) Σ_(n=1) ^∞ (x^n /n) −ln(x) Σ_(n=2) ^∞ (x^n /n) =−xln(x)ln(1−x) −ln(x)(−ln∣1−x∣−x) =−xln(x)ln∣1−x∣+ln(x)(ln∣1−x∣ +x) =(1−x)ln(x)ln∣1−x∣ +xln(x) .also we have (d/dx)K(x) =Σ_(n=1) ^∞ (1/(n(n+1))) x^n =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 =(((1−x)ln∣1−x∣)/x) +1 ⇒ K(x) = x + ∫ (((1−x)ln∣1−x∣)/x) dx +k ....be continued...](https://www.tinkutara.com/question/Q44009.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{for}\:\mathrm{0}<{t}\leqslant{x}<\mathrm{1}\:\:\:\:{ln}^{'} \left(\mathrm{1}−{t}\right)\:=−\frac{\mathrm{1}}{\mathrm{1}−{t}}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{{n}} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{t}\right)\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}+\mathrm{1}}{t}^{{n}+\mathrm{1}} \:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{t}^{{n}} }{{n}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=−\int_{\mathrm{0}} ^{{x}} \:{ln}\left({t}\right)\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{t}^{{n}} }{{n}}\right){dt}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{{x}} {t}^{{n}} {ln}\left({t}\right)\:{dt} \\ $$$${by}\:{parts}\:\:{A}_{{n}} \left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} \:{t}^{{n}} {ln}\left({t}\right){dt}\:=\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}{t}^{{n}+\mathrm{1}} {ln}\left({t}\right)\right]_{\mathrm{0}} ^{{x}} \:−\int_{\mathrm{0}} ^{{x}} \:\frac{\mathrm{1}}{{n}+\mathrm{1}}{t}^{{n}+\mathrm{1}} \:\frac{{dt}}{{t}} \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{x}^{{n}+\mathrm{1}} {ln}\left({x}\right)\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{{x}} \:\:{t}^{{n}} {dt}\:=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} {ln}\left({x}\right)−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:{x}^{{n}+\mathrm{1}} \:\Rightarrow \\ $$$${f}\left({x}\right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\left\{\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{x}^{{n}+\mathrm{1}} {ln}\left({x}\right)−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:{x}^{{n}+\mathrm{1}} \right\} \\ $$$$=\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\:{x}^{{n}+\mathrm{1}} {ln}\left({x}\right)\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:{x}^{{n}+\mathrm{1}} \:={H}\left({x}\right)−{K}\left({x}\right) \\ $$$${K}\left({x}\right)\:={ln}\left({x}\right)\:\sum_{{n}=\mathrm{1}} ^{\infty} \left\{\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\}{x}^{{n}+\mathrm{1}} \\ $$$$={ln}\left({x}\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}+\mathrm{1}} }{{n}}\:−{ln}\left({x}\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$={xln}\left({x}\right)\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:−{ln}\left({x}\right)\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}}\:\:\:\:\:\:\:\:\:\: \\ $$$$=−{xln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)\:−{ln}\left({x}\right)\left(−{ln}\mid\mathrm{1}−{x}\mid−{x}\right) \\ $$$$=−{xln}\left({x}\right){ln}\mid\mathrm{1}−{x}\mid+{ln}\left({x}\right)\left({ln}\mid\mathrm{1}−{x}\mid\:+{x}\right) \\ $$$$=\left(\mathrm{1}−{x}\right){ln}\left({x}\right){ln}\mid\mathrm{1}−{x}\mid\:+{xln}\left({x}\right)\:.{also}\:{we}\:{have} \\ $$$$\frac{{d}}{{dx}}{K}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\:{x}^{{n}} \:\:=\sum_{{n}=\mathrm{1}} ^{\infty} \left\{\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\}{x}^{{n}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}+\mathrm{1}}\:=−{ln}\mid\mathrm{1}−{x}\mid−\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{{x}^{{n}−\mathrm{1}} }{{n}} \\ $$$$\left.=−{ln}\mid\mathrm{1}−{x}\mid\:−\frac{\mathrm{1}}{{x}}\:\left\{\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}}\:−{x}\right\}\:=−{ln}\mid\mathrm{1}−{x}\mid+\frac{\mathrm{1}}{{x}}{ln}\mid\mathrm{1}−{x}\right)\:+\mathrm{1} \\ $$$$=\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right){ln}\mid\mathrm{1}−{x}\mid\:+\mathrm{1}\:\:=\frac{\left(\mathrm{1}−{x}\right){ln}\mid\mathrm{1}−{x}\mid}{{x}}\:+\mathrm{1}\:\Rightarrow \\ $$$${K}\left({x}\right)\:=\:{x}\:+\:\int\:\:\:\:\frac{\left(\mathrm{1}−{x}\right){ln}\mid\mathrm{1}−{x}\mid}{{x}}\:{dx}\:+{k}\:\:….{be}\:{continued}… \\ $$
Commented by maxmathsup by imad last updated on 19/Sep/18
$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right){dt}\:\:={lim}_{{x}\rightarrow\mathrm{0}} \:\:\int_{\mathrm{0}} ^{{x}} \:{ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right){dt} \\ $$$$={lim}_{{x}\rightarrow\mathrm{1}} \left\{\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:−\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{x}^{{n}+\mathrm{1}} {lnx}}{{n}\left({n}+\mathrm{1}\right)}\right\}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\mathrm{1}\:. \\ $$
Commented by maxmathsup by imad last updated on 19/Sep/18
$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right)\:{dt}\:={lim}_{{x}\rightarrow\mathrm{1}} \int_{\mathrm{0}} ^{{x}} {ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right){dt}\:=… \\ $$$$ \\ $$