Question Number 79128 by ~blr237~ last updated on 22/Jan/20
$${Find}\:{out}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right){dt} \\ $$$${Then}\:{deduce}\:{the}\:{value}\:{of}\:\:\:{A}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{{n}}\end{pmatrix}} \\ $$
Commented by mathmax by abdo last updated on 24/Jan/20
![let I =∫_0 ^1 ln(1−t +t^2 )dt by parts I =[t ln(1−t +t^2 )]_0 ^1 −∫_0 ^1 t ×((2t−1)/(t^2 −t+1))dt =−∫_0 ^1 ((2t^2 −t)/(t^2 −t +1))dt =−∫_0 ^1 ((2(t^2 −t +1)+2t−2−t)/(t^2 −t +1))dt =−2 −∫_0 ^1 ((t−2)/(t^2 −t +1))dt =−2−(1/2) ∫_0 ^1 ((2t−1−3)/(t^2 −t +1))dt =−2 −(1/2)[ln(t^2 −t+1)]_0 ^1 +(3/2) ∫_0 ^1 (dt/((t−(1/2))^2 +(3/4))) =−2+(3/2) ∫_0 ^1 (dt/((t−(1/2))^2 +(3/4))) changement t−(1/2) =((√3)/2)u give ∫_0 ^1 (dt/((t−(1/2))^2 +(3/4))) =(4/3)∫_(−(1/( (√3)))) ^(1/( (√3))) (1/(u^2 +1))×((√3)/2)du =(2/( (√3))) ∫_(−(1/( (√3)))) ^(1/( (√3))) (du/(1+u^2 )) =(4/( (√3)))[arctan(u)]_0 ^(1/( (√3))) =(4/( (√3)))×(π/6) =((2π)/(3(√3))) ⇒ I =−2 +(3/2)×((2π)/(3(√3))) =−2+(π/3)](https://www.tinkutara.com/question/Q79268.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{t}\:+{t}^{\mathrm{2}} \right){dt}\:\:{by}\:{parts}\: \\ $$$${I}\:=\left[{t}\:{ln}\left(\mathrm{1}−{t}\:+{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}\:×\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{t}^{\mathrm{2}} −{t}}{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}}{dt}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}\left({t}^{\mathrm{2}} −{t}\:+\mathrm{1}\right)+\mathrm{2}{t}−\mathrm{2}−{t}}{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}}{dt} \\ $$$$=−\mathrm{2}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}−\mathrm{2}}{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}}{dt}\:=−\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{t}−\mathrm{1}−\mathrm{3}}{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}}{dt} \\ $$$$=−\mathrm{2}\:−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:+\frac{\mathrm{3}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=−\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:\:{changement}\:{t}−\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=\frac{\mathrm{4}}{\mathrm{3}}\int_{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\mathrm{1}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du}\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\int_{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\left[{arctan}\left({u}\right)\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}×\frac{\pi}{\mathrm{6}}\:=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:\Rightarrow\:{I}\:=−\mathrm{2}\:+\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:=−\mathrm{2}+\frac{\pi}{\mathrm{3}} \\ $$
Commented by mathmax by abdo last updated on 24/Jan/20
$${I}\:=−\mathrm{2}+\frac{\pi}{\:\sqrt{\mathrm{3}}} \\ $$
Commented by mathmax by abdo last updated on 24/Jan/20
$${we}\:{have}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−\left({t}−{t}^{\mathrm{2}} \right)\right){dt}\:\:\:\:{we}\:{have}\:\mid{t}−{t}^{\mathrm{2}} \mid=\mid{t}\mid.\mid\mathrm{1}−{t}^{\mathrm{2}} \mid<\mathrm{1}\:\Rightarrow \\ $$$${ln}^{'} \left(\mathrm{1}−{u}\right)\:=−\frac{\mathrm{1}}{\mathrm{1}−{u}}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{u}^{{n}} \:\Rightarrow{ln}\left(\mathrm{1}−{u}\right)=−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{u}^{{n}} }{{n}}\:\Rightarrow{ln}\left(\mathrm{1}−\left({t}−{t}^{\mathrm{2}} \right)\right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left({t}−{t}^{\mathrm{2}} \right)^{{n}} }{{n}}\:\Rightarrow \\ $$$${I}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} \left(\mathrm{1}−{t}\right)^{{n}} \:{dt}\:\:{let}\:{remember}\:{that} \\ $$$${B}\left({p},{q}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{p}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{q}−\mathrm{1}} \:{dx}\:=\frac{\Gamma\left({p}\right).\Gamma\left({q}\right)}{\Gamma\left({p}+{q}\right)}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} \left(\mathrm{1}−{t}\right)^{{n}} \:{dt}\:={B}\left({n}+\mathrm{1},{n}+\mathrm{1}\right)\:=\frac{\Gamma\left({n}+\mathrm{1}\right)\Gamma\left({n}+\mathrm{1}\right)}{\Gamma\left(\mathrm{2}{n}+\mathrm{2}\right)} \\ $$$$=\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\:\:\left(\:\:\:\:\Gamma\left({m}\right)=\left({m}−\mathrm{1}\right)!\right)\:\:\:{we}\:{have}\:{so}\:{I}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}×\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$${A}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right){C}_{\mathrm{2}{n}+\mathrm{1}} ^{{n}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{{n}!\left({n}+\mathrm{1}\right)!}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left({n}!\right)^{\mathrm{2}} }{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\Rightarrow\:{A}\:=−{I}\: \\ $$
Commented by mathmax by abdo last updated on 24/Jan/20
$$\Rightarrow\:{A}\:=\mathrm{2}−\frac{\pi}{\:\sqrt{\mathrm{3}}}\:. \\ $$
Answered by mind is power last updated on 22/Jan/20
![=∫_0 ^1 ln((1−t+t^2 )dt by part=[tln(1−t+t^2 )]_0 ^1 −∫_0 ^1 (((2t−1)t)/(1−t+t^2 ))dt =−∫_0 ^1 ((2t^2 −t)/(1−t+t^2 ))dt−∫_0 ^1 (((2t^2 −2t+2)−1+t)/(t^2 −t+1))dt =−2+∫_0 ^1 ((t−1)/(t^2 −t+1))=−2+∫_0 ^1 ((t−(1/2))/(t^2 −t+1))dt−(1/2)∫_0 ^1 (dt/((t−(1/2))^2 +(3/4))) =−2−[_0 ^1 (1/( (√3)))arctan(((2t−1)/( (√3))))]=−2−(2/( (√3)))arctan((1/( (√3))))=−2−(π/(3(√3))). ln(1−t+t^2 )=ln(1−(t−t^2 ))=−Σ_(n≥1) (((t−t^2 )^n )/n) ∫_0 ^1 ln(1−t+t^2 )dt=∫_0 ^1 −Σ_(n≥1) (((t−t^2 )^n )/n)dt =−∫_0 ^1 Σ_(n≥1) (((t−t^2 )^n )/n)dt=−Σ_(n≥1) ∫_0 ^1 ((t^n (1−t)^n )/n)dt =−Σ_(n≥1) (1/n)β(n+1,n+1)=−Σ_(n≥1) ((n!.n!)/(n(2n+1)!)) =−Σ_(n≥1) (1/(n.(((2n+1)!(n+1))/(n!.(2n+1−n)!.n!))))=−Σ_(n≥1) (1/(n(n+1).(((2n+1)!)/(n!(n+1)!)))) =−Σ_(n≥1) (1/(n(n+1) (((2n+1)),(n) )))=−2−(π/(3(√3))) by first ⇒A=2+(π/(3(√3)))](https://www.tinkutara.com/question/Q79133.png)
$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right){dt}\right. \\ $$$${by}\:{part}=\left[{tln}\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{2}{t}−\mathrm{1}\right){t}}{\mathrm{1}−{t}+{t}^{\mathrm{2}} }{dt} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{t}^{\mathrm{2}} −{t}}{\mathrm{1}−{t}+{t}^{\mathrm{2}} }{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}\right)−\mathrm{1}+{t}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$=−\mathrm{2}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}=−\mathrm{2}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}−\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=−\mathrm{2}−\left[_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right]=−\mathrm{2}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=−\mathrm{2}−\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}. \\ $$$${ln}\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right)={ln}\left(\mathrm{1}−\left({t}−{t}^{\mathrm{2}} \right)\right)=−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({t}−{t}^{\mathrm{2}} \right)^{{n}} }{{n}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right){dt}=\int_{\mathrm{0}} ^{\mathrm{1}} −\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({t}−{t}^{\mathrm{2}} \right)^{{n}} }{{n}}{dt} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({t}−{t}^{\mathrm{2}} \right)^{{n}} }{{n}}{dt}=−\underset{{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}} \left(\mathrm{1}−{t}\right)^{{n}} }{{n}}{dt} \\ $$$$=−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}\beta\left({n}+\mathrm{1},{n}+\mathrm{1}\right)=−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{n}!.{n}!}{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$=−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}.\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!\left({n}+\mathrm{1}\right)}{{n}!.\left(\mathrm{2}{n}+\mathrm{1}−{n}\right)!.{n}!}}=−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right).\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{{n}!\left({n}+\mathrm{1}\right)!}} \\ $$$$=−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{{n}}\end{pmatrix}}=−\mathrm{2}−\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:{by}\:{first} \\ $$$$\Rightarrow{A}=\mathrm{2}+\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by ~blr237~ last updated on 23/Jan/20
$${nice}\:{sir} \\ $$
Commented by mind is power last updated on 23/Jan/20
$${thanx} \\ $$