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0-1-lnxarctanx-x-dx-

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Question Number 222705 by MrGaster last updated on 05/Jul/25
∫_0 ^1 ((lnxarctanx)/x)dx
$$\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}\mathrm{arctan}{x}}{{x}}{dx} \\ $$
Answered by gabthemathguy25 last updated on 05/Jul/25
this is non elementary however i can still evaluate this tan^(−1) =Σ_(n = 0) ^∞ (−1)^n (x^(2n+1) /(2n+1)) ((ln(x)tan^(−1) (x))/x)=ln(x)Σ_(n = 0) ^∞ (−1)^n (x^(2n) /(2n+1)) ⇒Σ_(n=0) ^∞ (((−1)^n )/(2n+1))∫_0 ^1 x^k ln(x)dx ⇒∫_0 ^1 x^k ln(x)dx=−(1/((k+1)^2 )) therefore ∫_0 ^1 x^(2n) ln(x)dx=−(1/((2n+1)^2 )) Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 ))=(π^3 /(32)) thus ∫_0 ^1 ((ln(x)tan^(−1) )/x)dx=−(π^3 /(32))
$${this}\:{is}\:{non}\:{elementary}\:{however}\:{i}\:{can}\:{still}\:{evaluate}\:{this} \\ $$$$\mathrm{tan}^{−\mathrm{1}} =\underset{{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{ln}\left({x}\right)\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}}=\mathrm{ln}\left({x}\right)\underset{{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}} \mathrm{ln}\left({x}\right){dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}} \mathrm{ln}\left({x}\right){dx}=−\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${therefore}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}} \mathrm{ln}\left({x}\right){dx}=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\pi^{\mathrm{3}} }{\mathrm{32}} \\ $$$${thus}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}\right)\mathrm{tan}^{−\mathrm{1}} }{{x}}{dx}=−\frac{\pi^{\mathrm{3}} }{\mathrm{32}} \\ $$
Commented by Ghisom last updated on 06/Jul/25
nice! I tried this: ∫_0 ^1 ((arctan x ln x)/x)dx= [by parts] =[arctan x (ln x)^2 ]_0 ^1 −∫_0 ^1 ((((ln x)^2 )/(x^2 +1))+((arctan x ln x)/x))dx= =0−∫_0 ^1 ((((ln x)^2 )/(x^2 +1))+((arctan x ln x)/x))dx ⇒ ∫_0 ^1 ((arctan x ln x)/x)dx=(1/2)∫_0 ^1 (((ln x)^2 )/(x^2 +1))dx ∫(((ln x)^2 )/(x^2 +1))dx=(i/2)∫((((ln x)^2 )/(x+i))−(((ln x)^2 )/(x−i)))dx which can be solved but your method is faster
$$\mathrm{nice}! \\ $$$$\mathrm{I}\:\mathrm{tried}\:\mathrm{this}: \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{arctan}\:{x}\:\mathrm{ln}\:{x}}{{x}}{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=\left[\mathrm{arctan}\:{x}\:\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{1}} −\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{arctan}\:{x}\:\mathrm{ln}\:{x}}{{x}}\right){dx}= \\ $$$$=\mathrm{0}−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{arctan}\:{x}\:\mathrm{ln}\:{x}}{{x}}\right){dx} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{arctan}\:{x}\:\mathrm{ln}\:{x}}{{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\int\frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\frac{\mathrm{i}}{\mathrm{2}}\int\left(\frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{{x}+\mathrm{i}}−\frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{{x}−\mathrm{i}}\right){dx} \\ $$$$\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved} \\ $$$$\mathrm{but}\:\mathrm{your}\:\mathrm{method}\:\mathrm{is}\:\mathrm{faster} \\ $$
Commented by gabthemathguy25 last updated on 06/Jul/25
no problem
$${no}\:{problem} \\ $$
Answered by MrGaster last updated on 08/Jul/25
(1):u=arctanx⇒du=(1/(1−x^2 ))dx dv=((ln x)/x)dx⇒v=(((ln x)^2 )/2) I=(((ln x)^2 arctan x)/2)∣_0 ^1 −(1/2)∫_0 ^1 (((ln x)^2 )/(x^2 +1))dx ∵lim_(x→0) (ln x)^2 arctan x=0∧arctan=(π/4) ∴I=−(1/2)∫_0 ^1 (((ln x)^2 )/(x^2 +1))dx x=e^(−t) ,dx=−e^(−t) dt I=−(1/2)∫_0 ^∞ ((t^2 e^(−t) )/(1+e^(−2t) ))dt (1/(1+e^(−2t) ))=Σ_(n=0) ^∞ (−1)^n e^(−2nt) ∴I=−(1/2)(−1)^n ∫_0 ^∞ t^2 e^(−(2n+1)t) dt ∫_0 ^∞ t^2 e^(−at) dt=(2/a^3 ) I=−(1/2)Σ_(n=0) ^∞ (−1)^n (2/((2n+1)^3 ))=−Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 ))=(π^3 /(32)) I=−(π^3 /(32)) (2):arctan x=Σ_(k=0) ^∞ (((−1)^k x^(2k+1) )/(2k+1)) ∀∣x∣<1 ((ln x arctan)/x)=ln x∙Σ_(k=0) ^∞ (((−1)^k x^(2k) )/(2k+1)) ∫_0 ^1 ((ln x arctan)/x)dx=∫_0 ^1 ln x∙Σ_(k=0) ^∞ (((−1)^k x^(2k) )/(2k+1)) =Σ_(k=0) ^∞ (((−1)^k )/(2k+1))∫_0 ^1 x^(2k) ln x dx ∫_0 ^1 x^p ln x dx=−(1/((p+1)^2 )) ∀p>1 p=2k⇒∫_0 ^1 x^(2k) ln x dx=−(1/((2k−1)^2 )) =Σ_(k=0) ^∞ (((−1)^k )/(2k+1))(−(1/((2k+1)^2 )))=Σ_(k=0) ^∞ (−1)^(k+1) (1/((2k+1)^3 )) −Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 ))=(π^3 /(32)) =−Σ_(k=0) ^∞ (((−1)^k )/((2k+1)^3 ))=−(π^3 /(32))
$$\left(\mathrm{1}\right):{u}=\mathrm{arctan}{x}\Rightarrow{du}=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$${dv}=\frac{\mathrm{ln}\:{x}}{{x}}{dx}\Rightarrow{v}=\frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${I}=\frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} \mathrm{arctan}\:{x}}{\mathrm{2}}\mid_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\because\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} \mathrm{arctan}\:{x}=\mathrm{0}\wedge\mathrm{arctan}=\frac{\pi}{\mathrm{4}} \\ $$$$\therefore{I}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$${x}={e}^{−{t}} ,{dx}=−{e}^{−{t}} {dt} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} {e}^{−{t}} }{\mathrm{1}+{e}^{−\mathrm{2}{t}} }{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{e}^{−\mathrm{2}{t}} }=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {e}^{−\mathrm{2}{nt}} \\ $$$$\therefore{I}=−\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{2}} {e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){t}} {dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{2}} {e}^{−{at}} {dt}=\frac{\mathrm{2}}{{a}^{\mathrm{3}} } \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\pi^{\mathrm{3}} }{\mathrm{32}} \\ $$$${I}=−\frac{\pi^{\mathrm{3}} }{\mathrm{32}} \\ $$$$\left(\mathrm{2}\right):\mathrm{arctan}\:{x}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{2}{k}+\mathrm{1}} }{\mathrm{2}{k}+\mathrm{1}}\:\forall\mid{x}\mid<\mathrm{1} \\ $$$$\frac{\mathrm{ln}\:{x}\:\mathrm{arctan}}{{x}}=\mathrm{ln}\:{x}\centerdot\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{2}{k}} }{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\:{x}\:\mathrm{arctan}}{{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\:{x}\centerdot\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{2}{k}} }{\mathrm{2}{k}+\mathrm{1}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{k}} \mathrm{ln}\:{x}\:{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{p}} \mathrm{ln}\:{x}\:{dx}=−\frac{\mathrm{1}}{\left({p}+\mathrm{1}\right)^{\mathrm{2}} }\:\forall{p}>\mathrm{1} \\ $$$${p}=\mathrm{2}{k}\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{k}} \mathrm{ln}\:{x}\:{dx}=−\frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}\left(−\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\overset{\mathrm{3}} {\right)}} \\ $$$$−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\pi^{\mathrm{3}} }{\mathrm{32}} \\ $$$$=−\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{3}} }=−\frac{\pi^{\mathrm{3}} }{\mathrm{32}} \\ $$

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