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1-ln-x-x-2-1-dx-Convergent-or-Divergent-

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Question Number 82517 by naka3546 last updated on 22/Feb/20
∫_( 1) ^( ∞) ((ln x)/(x^2 + 1)) dx , Convergent or Divergent ?
$$\underset{\:\mathrm{1}} {\int}\overset{\:\infty} {\:}\:\frac{\mathrm{ln}\:{x}}{{x}^{\mathrm{2}} \:+\:\mathrm{1}}\:{dx}\:\:,\:\:\:{Convergent}\:\:{or}\:\:{Divergent}\:\:? \\ $$
Commented by Tony Lin last updated on 22/Feb/20
∫_1 ^∞ ((lnx)/(x^2 +1))dx<∫_1 ^∞ ((lnx)/x^2 )dx ∫ ((lnx)/x^2 )dx =−(( lnx)/x)−∫−(1/x^2 ) dx =((−lnx−1)/x)+C ∫_1 ^∞ ((lnx)/x^2 )dx =−lim_(t→∞) ((lnt+1)/t)+1 =−lim_(t→∞) (1/t)+1 =1 By Comparison Test ∫_1 ^∞ ((lnx)/(x^2 +1))dx is convergent
$$\int_{\mathrm{1}} ^{\infty} \frac{{lnx}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}<\int_{\mathrm{1}} ^{\infty} \:\frac{{lnx}}{{x}^{\mathrm{2}} }{dx} \\ $$$$\int\:\frac{{lnx}}{{x}^{\mathrm{2}} }{dx} \\ $$$$=−\frac{\:{lnx}}{{x}}−\int−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{dx} \\ $$$$=\frac{−{lnx}−\mathrm{1}}{{x}}+{C} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{{lnx}}{{x}^{\mathrm{2}} }{dx} \\ $$$$=−\underset{{t}\rightarrow\infty} {\mathrm{lim}}\frac{{lnt}+\mathrm{1}}{{t}}+\mathrm{1} \\ $$$$=−\underset{{t}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{t}}+\mathrm{1} \\ $$$$=\mathrm{1} \\ $$$${By}\:{Comparison}\:{Test} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{{lnx}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\:{is}\:{convergent} \\ $$
Commented by mathmax by abdo last updated on 23/Feb/20
letA =∫_1 ^∞ ((lnx)/(1+x^2 ))dx changement x=(1/t) give A =−∫_0 ^1 ((−ln(t))/(1+(1/t^2 )))×(−(dt/t^2 )) =−∫_0 ^1 ((lnt)/(1+t^2 ))dt =−∫_0 ^1 ln(t)(Σ_(n=0) ^∞ (−1)^n t^(2n) )dt =−Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 t^(2n) ln(t)dt by parts u_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) dt =−(1/(2n+1))[(1/(2n+1))t^(2n+1) ]_0 ^1 =−(1/((2n+1)^2 )) ⇒ A =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) and this serie is convergente.
$${letA}\:=\int_{\mathrm{1}} ^{\infty} \:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{changement}\:{x}=\frac{\mathrm{1}}{{t}}\:{give}\: \\ $$$${A}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{−{ln}\left({t}\right)}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}×\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right)\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{lnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({t}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{t}^{\mathrm{2}{n}} \right){dt} \\ $$$$=−\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} \:{ln}\left({t}\right){dt}\:{by}\:{parts} \\ $$$${u}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} {ln}\left({t}\right){dt}\:=\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{t}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{t}^{\mathrm{2}{n}} \:{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{t}^{\mathrm{2}{n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} =−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${A}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:{and}\:{this}\:{serie}\:{is}\:{convergente}. \\ $$
Commented by mathmax by abdo last updated on 24/Feb/20
let I =∫_1 ^∞ ((ln(x))/(x^2 +1))dx changement x=tanθ give I =∫_(π/4) ^(π/2) ((ln(tanθ))/(1+tan^2 θ))(1+tan^2 θ)dθ =∫_(π/4) ^(π/2) {ln(sinθ)−ln(cosθ)}dθ =∫_(π/4) ^(π/2) ln(sinθ)dθ− ∫_(π/4) ^(π/2) ln(cosθ)dθ =H−K H+K =∫_(π/4) ^(π/2) ln((1/2)sin(2θ)dθ =−ln(2)×(π/4) +∫_(π/4) ^(π/2) ln(sin(2θ))dθ =−(π/4)ln(2) +(1/2)∫_(π/2) ^π ln(sinu)du (2θ =u) ∫_(π/2) ^π ln(sinu)du =_(u=(π/2)+α) ∫_0 ^(π/2) ln(cosα)dα =−(π/2)ln(2) ⇒ H+K =−(π/2)ln(2) K =∫_(π/4) ^(π/2) ln(cosθ)dθ =_(θ =(π/2)−α) ∫_(π/4) ^0 ln(sinα)(−dα) =∫_0 ^(π/4) ln(sinα)dα ....be continued...
$${let}\:{I}\:=\int_{\mathrm{1}} ^{\infty} \:\frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:\:\:{changement}\:{x}={tan}\theta\:{give} \\ $$$${I}\:=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{ln}\left({tan}\theta\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta\:=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \left\{{ln}\left({sin}\theta\right)−{ln}\left({cos}\theta\right)\right\}{d}\theta \\ $$$$=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\theta\right){d}\theta−\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\theta\right){d}\theta\:={H}−{K} \\ $$$${H}+{K}\:=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}\theta\right){d}\theta\:=−{ln}\left(\mathrm{2}\right)×\frac{\pi}{\mathrm{4}}\:+\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({sin}\left(\mathrm{2}\theta\right)\right){d}\theta\right. \\ $$$$=−\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} {ln}\left({sinu}\right){du}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\theta\:={u}\right) \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:{ln}\left({sinu}\right){du}\:=_{{u}=\frac{\pi}{\mathrm{2}}+\alpha} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\alpha\right){d}\alpha\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${H}+{K}\:\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\ $$$${K}\:=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\theta\right){d}\theta\:=_{\theta\:=\frac{\pi}{\mathrm{2}}−\alpha} \int_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} \:{ln}\left({sin}\alpha\right)\left(−{d}\alpha\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({sin}\alpha\right){d}\alpha\:\:\:….{be}\:{continued}… \\ $$

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