Question Number 40887 by prof Abdo imad last updated on 28/Jul/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{tln}\left({t}\right)}{{t}^{\mathrm{2}} −\mathrm{1}}{dt} \\ $$
Commented by prof Abdo imad last updated on 30/Jul/18
![let I = ∫_0 ^1 ((tln(t))/(t^2 −1))dt I =−∫_0 ^1 tln(t)(Σ_(n=0) ^∞ t^(2n) )dt =−Σ_(n=0) ^∞ ∫_0 ^1 t^(2n+1) ln(t)dt by parts A_n =∫_0 ^1 t^(2n+1) ln(t)dt =[(1/(2n+2))t^(2n+2) ln(t)]_0 ^1 −∫_0 ^1 (1/(2n+2)) t^(2n+1) dt =−(1/((2n+2)^2 )) ⇒ I =Σ_(n=0) ^∞ (1/((2n+2)^2 )) =(1/4) Σ_(n=0) ^∞ (1/((n+1)^2 )) =(1/4)Σ_(n=1) ^∞ (1/n^2 ) =(1/4) .(π^2 /6) ⇒ I =(π^2 /(24))](https://www.tinkutara.com/question/Q40978.png)
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tln}\left({t}\right)}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\: \\ $$$${I}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} {tln}\left({t}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{\mathrm{2}{n}} \right){dt} \\ $$$$=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({t}\right){dt}\:\:{by}\:{parts} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({t}\right){dt}\:=\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}{t}^{\mathrm{2}{n}+\mathrm{2}} {ln}\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}\:{t}^{\mathrm{2}{n}+\mathrm{1}} {dt}\:=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${I}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\Rightarrow\:{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$