Question Number 91220 by Ar Brandon last updated on 28/Apr/20
$$\int_{\mathrm{1}} ^{\mathrm{x}} \frac{\mathrm{lnt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$
Commented by abdomathmax last updated on 28/Apr/20
![let take a try if x>1 we do the changement t=(1/u) ⇒ I =−∫_(1/x) ^1 ((−lnu)/(1+(1/u^2 )))(−(du/u^2 )) =−∫_(1/x) ^1 ((lnu)/(1+u^2 ))du =−∫_(1/x) ^1 lnu(Σ_(n=0) ^∞ (−1)^n u^(2n) )du =Σ_(n=0) ^∞ (−1)^(n+1) ∫_(1/x) ^1 u^(2n) ln u du =Σ_(n=0) ^∞ (−1)^(n+1) U_n U_n =∫_(1/x) ^1 u^(2n) ln(u)du =_(by parts) [(u^(2n+1) /(2n+1))lnu]_(1/x) ^1 −∫_(1/x) ^1 (u^(2n) /(2n+1))du =(1/(2n+1))((lnx)/x^(2n+1) )−(1/(2n+1))[(1/(2n+1))u^(2n+1) ]_(1/x) ^1 =((ln(x))/((2n+1)x^(2n+1) ))−(1/((2n+1)^2 ))(1−(1/x^(2n+1) )) ⇒ I =−lnx Σ_(n=0) ^∞ (((−1)^n )/((2n+1)x^(2n+1) )) +Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) −Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 x^(2n+1) )) rest to calculate those sums if x<1 I =−∫_x ^1 ((lnt)/(1+t^2 ))dt =−∫_x ^1 lnt(Σ_(n=0) ^∞ (−1)^n t^(2n) )dt =−Σ_(n=0) ^∞ (−1)^n ∫_x ^1 t^(2n) lnt dt and we follow the same way ....be continued...](https://www.tinkutara.com/question/Q91232.png)
$${let}\:{take}\:{a}\:{try}\:\: \\ $$$${if}\:{x}>\mathrm{1}\:\:\:\:{we}\:{do}\:{the}\:{changement}\:{t}=\frac{\mathrm{1}}{{u}}\:\Rightarrow \\ $$$${I}\:=−\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:\frac{−{lnu}}{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}\left(−\frac{{du}}{{u}^{\mathrm{2}} }\right)\:=−\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:\frac{{lnu}}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$=−\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} {lnu}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{u}^{\mathrm{2}{n}} \right){du} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:{u}^{\mathrm{2}{n}} \:{ln}\:{u}\:{du}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{U}_{{n}} \\ $$$${U}_{{n}} =\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:{u}^{\mathrm{2}{n}} \:{ln}\left({u}\right){du}\:=_{{by}\:{parts}} \:\:\left[\frac{{u}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}{lnu}\right]_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \\ $$$$−\int_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \:\frac{{u}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}{du}\:=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\frac{{lnx}}{{x}^{\mathrm{2}{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{u}^{\mathrm{2}{n}+\mathrm{1}} \right]_{\frac{\mathrm{1}}{{x}}} ^{\mathrm{1}} \\ $$$$=\frac{{ln}\left({x}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right){x}^{\mathrm{2}{n}+\mathrm{1}} }−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}{n}+\mathrm{1}} }\right)\:\Rightarrow \\ $$$${I}\:=−{lnx}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){x}^{\mathrm{2}{n}+\mathrm{1}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} {x}^{\mathrm{2}{n}+\mathrm{1}} }\:\:{rest}\:{to}\:{calculate}\:{those}\:{sums} \\ $$$${if}\:{x}<\mathrm{1}\:\:\:{I}\:=−\int_{{x}} ^{\mathrm{1}} \:\:\frac{{lnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=−\int_{{x}} ^{\mathrm{1}} \:\:{lnt}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{t}^{\mathrm{2}{n}} \right){dt} \\ $$$$=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\int_{{x}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} \:{lnt}\:{dt}\:\:\:{and}\:{we}\:{follow}\:{the}\: \\ $$$${same}\:{way}\:….{be}\:{continued}… \\ $$
Commented by Ar Brandon last updated on 28/Apr/20
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Commented by mathmax by abdo last updated on 29/Apr/20
$${thankx} \\ $$