Question Number 40884 by prof Abdo imad last updated on 28/Jul/18
$$\left.\mathrm{1}\right)\:{fond}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} −\mathrm{1}}{dt} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{4}} −\mathrm{1}}{dt} \\ $$
Answered by maxmathsup by imad last updated on 30/Jul/18
![1) let I = ∫_0 ^1 ((ln(t))/(t^2 −1))dt ⇒ I =−∫_0 ^1 lnt(Σ_(n=0) ^∞ t^(2n) )dt =−Σ_(n=0) ^∞ ∫_0 ^1 t^(2n) ln(t)dt =−Σ_(n=0) ^∞ A_n and by parts A_n =∫_0 ^1 t^(2n) ln(t)dt =[(1/(2n+1))t^(2n+1) ln(t)]_0 ^1 −∫_0 ^1 (1/(2n+1))t^(2n+1) (dt/t) =−(1/(2n+1)) ∫_0 ^1 t^(2n) dt =−(1/((2n+1)^2 )) ⇒ I =Σ_(n=0) ^∞ (1/((2n+1)^2 )) but Σ_(n=1) ^∞ (1/n^2 ) =(π^2 /6) =Σ_(n=1) ^∞ (1/((2n)^2 )) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(1/4) (π^2 /6) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(π^2 /6) −(π^2 /(24)) =((3π^2 )/(24)) =(π^2 /8) ⇒ I =(π^2 /8) .](https://www.tinkutara.com/question/Q40994.png)
$$\left.\mathrm{1}\right)\:{let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\:\Rightarrow\:{I}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:{lnt}\left(\sum_{{n}=\mathrm{0}} ^{\infty} {t}^{\mathrm{2}{n}} \right){dt} \\ $$$$=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} {ln}\left({t}\right){dt}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \:\:\:{and}\:{by}\:{parts} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} {ln}\left({t}\right){dt}\:=\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{t}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{t}^{\mathrm{2}{n}+\mathrm{1}} \frac{{dt}}{{t}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} {dt}\:=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\:{I}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:{but} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:=\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{24}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow\:{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:. \\ $$
Answered by maxmathsup by imad last updated on 30/Jul/18
![2) let J =∫_0 ^1 ((ln(t))/(t^4 −1)) J =−∫_0 ^1 ((ln(t))/((t^2 −1)(t^2 +1)))dt =−(1/2)∫_0 ^1 {(1/(t^2 −1)) −(1/(t^2 +1))}ln(t)dt =(1/2) ∫_0 ^1 ((lnt)/(1−t^2 ))dt +(1/2) ∫_0 ^1 ((ln(t))/(t^2 +1))dt =−(π^2 /(16)) +(1/2) ∫_0 ^1 ((ln(t))/(1+t^2 ))dt but ∫_0 ^1 ((ln(t))/(1+t^2 ))dt = ∫_0 ^1 ln(t)(Σ_(n=0) ^∞ (−1)^n t^(2n) )dt =Σ_(n=0) ^1 (−1)^n ∫_0 ^1 t^(2n) ln(t)dt =Σ_(n=0) ^∞ (−1)^n A_n by parts A_n =[(1/(2n+1))t^(2n+1) ]_0 ^1 −∫_0 ^1 (t^(2n) /(2n+1))dt =−(1/((2n+1)^2 )) ⇒∫_0 ^1 ((ln(t))/(1+t^2 ))dt =−Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^n )) =−λ_0 (λ_0 is known by fourier serie) ⇒ ∫_0 ^1 ((ln(t))/(t^4 −1))dt =−(π^2 /(16)) −(λ_0 /2)](https://www.tinkutara.com/question/Q40995.png)
$$\left.\mathrm{2}\right)\:{let}\:{J}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{4}} −\mathrm{1}} \\ $$$${J}\:\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left({t}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}\:=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\frac{\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}\:−\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\right\}{ln}\left({t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{lnt}}{\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{but} \\ $$$$\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({t}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {t}^{\mathrm{2}{n}} \right){dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\mathrm{1}} \:\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} {ln}\left({t}\right){dt}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{A}_{{n}} \:{by}\:{parts} \\ $$$${A}_{{n}} =\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{t}^{\mathrm{2}{n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}{dt}\:=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{n}} } \\ $$$$=−\lambda_{\mathrm{0}} \:\:\left(\lambda_{\mathrm{0}} {is}\:{known}\:{by}\:{fourier}\:{serie}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{4}} −\mathrm{1}}{dt}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:−\frac{\lambda_{\mathrm{0}} }{\mathrm{2}} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 30/Jul/18
$$\lambda_{\mathrm{0}} =\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$