Question Number 35628 by abdo mathsup 649 cc last updated on 21/May/18
$${find}\:{the}\:{value}\:{of}\:\:{I}\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt} \\ $$
Commented by abdo mathsup 649 cc last updated on 24/May/18
![we have ln^′ (1−t) = ((−1)/(1−t)) =−Σ_(n=0) ^∞ t^n for ∣t∣<1 ⇒ ln(1−t) =−Σ_(n=0) ^∞ (t^(n+1) /(n+1)) =−Σ_(n=1) ^∞ (t^n /n) ⇒ ((ln(1−t))/t) = −Σ_(n=1) ^∞ (t^(n−1) /n) ⇒ I = − ∫_0 ^1 {Σ_(n=1) ^∞ (t^(n−1) /n)}ln(t)dt =−Σ_(n=1) ^∞ (1/n)∫_0 ^1 t^(n−1) ln(t)dt by parts A_n = ∫_0 ^1 t^(n−1) ln(t)dt =[(1/n)t^n ln(t)]_0 ^1 −∫_0 ^1 (1/n)t^n (dt/t) = −(1/n) ∫_0 ^1 t^(n−1) dt = −(1/n^2 ) ⇒ I = Σ_(n=1) ^∞ (1/n^3 ) I =ξ(3) the approximate value of ξ(3) is known.](https://www.tinkutara.com/question/Q35854.png)
$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}−{t}\right)\:=\:\frac{−\mathrm{1}}{\mathrm{1}−{t}}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{{n}} \:\:{for}\:\mid{t}\mid<\mathrm{1} \\ $$$$\Rightarrow\:{ln}\left(\mathrm{1}−{t}\right)\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{t}^{{n}} }{{n}}\:\Rightarrow \\ $$$$\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:=\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{t}^{{n}−\mathrm{1}} }{{n}}\:\Rightarrow \\ $$$${I}\:\:=\:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\left\{\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{t}^{{n}−\mathrm{1}} }{{n}}\right\}{ln}\left({t}\right){dt} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{{n}−\mathrm{1}} {ln}\left({t}\right){dt}\:\:{by}\:{parts} \\ $$$${A}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{{n}−\mathrm{1}} \:{ln}\left({t}\right){dt}\:=\left[\frac{\mathrm{1}}{{n}}{t}^{{n}} {ln}\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{n}}{t}^{{n}} \:\frac{{dt}}{{t}} \\ $$$$=\:−\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{{n}−\mathrm{1}} {dt}\:=\:−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\Rightarrow\:{I}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$$${I}\:=\xi\left(\mathrm{3}\right)\:\:{the}\:{approximate}\:{value}\:{of}\:\xi\left(\mathrm{3}\right)\:{is}\:{known}. \\ $$