Question Number 128841 by mnjuly1970 last updated on 10/Jan/21
![...nice calculus ... Evaluation of :: Φ= ∫_0 ^( 1) ln(x).arctan(x)dx solution:: note 1:: Σ_(n=1) ^∞ (((−1)^(n−1) )/(2n−1))=arctan(1)=(π/4) note 2 :: Σ_(n=1) ^∞ (((−1)^(n−1) )/n) =ln(1+1)=ln(2) note 3:: Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =_(eta function) ^(Drichlet) η(2)=(π^2 /(12)) ......start....... Φ=∫_0 ^( 1) {ln(x)Σ_(n=1 ) ^∞ (((−1)^(n−1) )/(2n−1)))x^(2n−1) }dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/((2n−1)))∫_0 ^( 1) ln(x)x^(2n−1) dx =Σ_(n=1 ) ^∞ (((−1)^(n−1) )/((2n−1))){[(x^(2n) /(2n))ln(x)]_0 ^1 −(1/(2n))∫_0 ^( 1) x^(2n−1) dx} =(1/4)Σ_(n=1) ^∞ (((−1)^n )/((2n−1)n^2 ))=(1/4)Σ_(n=1) ^∞ (−1)^n [((2n−(2n−1))/((2n−1)n^2 ))] =(1/2)Σ_(n=1) ^∞ (((−1)^n )/((2n−1)n)) −(1/4)Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(1/2)Σ_(n=1) ^∞ ((2(−1)^n )/(2n−1))−(1/2)Σ_(n=1) ^∞ (((−1)^n )/n)+(1/4)η(2) =((−π)/4)+(1/2)ln(2)+(π^2 /(48)) ... ...Φ=(π^2 /(48)) −(π/4) +(1/2)ln(2) ... ...m.n.july.1970...](Q128841.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:{calculus}\:... \\ $$$$\:\:\:\:\mathscr{E}{valuation}\:{of}\:::\:\Phi=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({x}\right).{arctan}\left({x}\right){dx} \\ $$$${solution}:: \\ $$$$\:\:\:\:{note}\:\mathrm{1}::\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{2}{n}−\mathrm{1}}={arctan}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}} \\ $$$${note}\:\mathrm{2}\:::\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:={ln}\left(\mathrm{1}+\mathrm{1}\right)={ln}\left(\mathrm{2}\right) \\ $$$${note}\:\mathrm{3}::\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:\underset{{eta}\:{function}} {\overset{{Drichlet}} {=}}\:\eta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\:......{start}....... \\ $$$$\:\:\:\Phi=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{{ln}\left({x}\right)\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left.\mathrm{2}{n}−\mathrm{1}\right)}{x}^{\mathrm{2}{n}−\mathrm{1}} \right\}{dx} \\ $$$$\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({x}\right){x}^{\mathrm{2}{n}−\mathrm{1}} {dx} \\ $$$$\:\:=\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)}\left\{\left[\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}}{ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{2}{n}−\mathrm{1}} {dx}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}−\mathrm{1}\right){n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left[\frac{\mathrm{2}{n}−\left(\mathrm{2}{n}−\mathrm{1}\right)}{\left(\mathrm{2}{n}−\mathrm{1}\right){n}^{\mathrm{2}} }\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}−\mathrm{1}\right){n}}\:−\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}+\frac{\mathrm{1}}{\mathrm{4}}\eta\left(\mathrm{2}\right) \\ $$$$=\frac{−\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{48}}\:\:... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\Phi=\frac{\pi^{\mathrm{2}} }{\mathrm{48}}\:−\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.{july}.\mathrm{1970}... \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 10/Jan/21
$$\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left({x}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} {log}\left({x}\right){dx}\:\:\:\:{log}\left({x}\right)={t} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} {e}^{−\left(\mathrm{2}{n}+\mathrm{2}\right){t}} {tdt} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)}−\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$={log}\left(\sqrt{\mathrm{2}}\right)−\frac{\pi}{\mathrm{4}}−\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$
Commented by mnjuly1970 last updated on 11/Jan/21
$${thank}\:{you}\:{mr}\:{dwaipayan}... \\ $$$${grateful}\:. \\ $$
Answered by mindispower last updated on 10/Jan/21
![1) by part=[ln(x).(xarctan(x)−((ln(1+x^2 ))/2))]_0 ^1 −∫_0 ^1 arctan(x)dx+∫_0 ^1 ((ln(1+x^2 ))/(2x))dx ∫_0 ^1 arctan(x)=(π/4)−((ln(2))/2) ∫_0 ^1 ((ln(1+x^2 ))/(2x))dx=∫_0 ^1 ((ln(1+x^2 ))/(4x^2 )).d(x^2 ) =(1/4)∫_0 ^1 ((ln(1+t))/t)dt=(1/4)∫_0 ^1 Σ_(n≥1) (((−1)^(n−1) t^(n−1) )/n) =(1/4)Σ_(n≥1) (((−1)^(n−1) )/n^2 )=−(1/(16))ζ(2)+(1/4)((3/4))ζ(2) =((ζ(2))/8)=(π^2 /(48))...we get (π^2 /(48))−(π/4)+((ln(2))/2) 2)ln(x)=∂_y x^y ∣_(y=0) f(y)=∫_0 ^1 x^y arctan(x)dx by part=(π/(4(y+1)))−(1/(y+1))∫_0 ^1 (x^(y+1) /(1+x^2 )) =(π/(4(y+1)))−(1/(y+1))∫_0 ^1 ((x^(y+1) −x^(y+3) )/(1−x^4 ))dx x^4 =t =(π/(4(y+1)))−(1/(y+1))∫_0 ^1 ((t^((y+1)/4) −t^((y+3)/4) )/(4(1−t))).t^(−(3/4)) dt =(1/(y+1))((π/4)−∫_0 ^1 ((t^((y−2)/4) −t^((y−1)/4) )/(1−t))dt) Ψ(s+1)=−γ+∫_0 ^1 ((1−x^s )/(1−x))dx we get(1/(y+1))((π/4)−Ψ(((y+3)/4))+Ψ(((y+2)/4)))=f(y) f′(y)=−(1/((y+1)^2 ))((π/4)+Ψ(((y+2)/4))−Ψ(((y+3)/4)))+((1/4)(Ψ′(((y+2)/4))−Ψ′(((y+3)/4))).(1/(y+1))](Q128872.png)
$$\left.\mathrm{1}\right)\:{by}\:{part}=\left[{ln}\left({x}\right).\left({xarctan}\left({x}\right)−\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{2}{x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({x}\right)=\frac{\pi}{\mathrm{4}}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{2}{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{4}{x}^{\mathrm{2}} }.{d}\left({x}^{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}{dt}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {t}^{{n}−\mathrm{1}} }{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{16}}\zeta\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\zeta\left(\mathrm{2}\right) \\ $$$$=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{8}}=\frac{\pi^{\mathrm{2}} }{\mathrm{48}}...{we}\:{get}\:\frac{\pi^{\mathrm{2}} }{\mathrm{48}}−\frac{\pi}{\mathrm{4}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){ln}\left({x}\right)=\partial_{{y}} {x}^{{y}} \mid_{{y}=\mathrm{0}} \\ $$$${f}\left({y}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{y}} {arctan}\left({x}\right){dx} \\ $$$${by}\:{part}=\frac{\pi}{\mathrm{4}\left({y}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{y}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{y}+\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\frac{\pi}{\mathrm{4}\left({y}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{y}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{y}+\mathrm{1}} −{x}^{{y}+\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{4}} }{dx} \\ $$$${x}^{\mathrm{4}} ={t} \\ $$$$=\frac{\pi}{\mathrm{4}\left({y}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{y}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{{y}+\mathrm{1}}{\mathrm{4}}} −{t}^{\frac{{y}+\mathrm{3}}{\mathrm{4}}} }{\mathrm{4}\left(\mathrm{1}−{t}\right)}.{t}^{−\frac{\mathrm{3}}{\mathrm{4}}} {dt} \\ $$$$=\frac{\mathrm{1}}{{y}+\mathrm{1}}\left(\frac{\pi}{\mathrm{4}}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{{y}−\mathrm{2}}{\mathrm{4}}} −{t}^{\frac{{y}−\mathrm{1}}{\mathrm{4}}} }{\mathrm{1}−{t}}{dt}\right) \\ $$$$\Psi\left({s}+\mathrm{1}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{s}} }{\mathrm{1}−{x}}{dx} \\ $$$${we}\:{get}\frac{\mathrm{1}}{{y}+\mathrm{1}}\left(\frac{\pi}{\mathrm{4}}−\Psi\left(\frac{{y}+\mathrm{3}}{\mathrm{4}}\right)+\Psi\left(\frac{{y}+\mathrm{2}}{\mathrm{4}}\right)\right)={f}\left({y}\right) \\ $$$${f}'\left({y}\right)=−\frac{\mathrm{1}}{\left({y}+\mathrm{1}\right)^{\mathrm{2}} }\left(\frac{\pi}{\mathrm{4}}+\Psi\left(\frac{{y}+\mathrm{2}}{\mathrm{4}}\right)−\Psi\left(\frac{{y}+\mathrm{3}}{\mathrm{4}}\right)\right)+\left(\frac{\mathrm{1}}{\mathrm{4}}\left(\Psi'\left(\frac{{y}+\mathrm{2}}{\mathrm{4}}\right)−\Psi'\left(\frac{{y}+\mathrm{3}}{\mathrm{4}}\right)\right).\frac{\mathrm{1}}{{y}+\mathrm{1}}\right. \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 11/Jan/21
$${thanks}\:{alot}\:{sir}\:{power} \\ $$$$\:{powerful}\:{and}\:{mighty}\:{as}\:{always} \\ $$$$\:.{peace}\:{be}\:{upon}\:{you}... \\ $$