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Question-142870

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Question Number 142870 by mnjuly1970 last updated on 06/Jun/21
Answered by qaz last updated on 06/Jun/21
Θ=2∫_0 ^∞ ((ln(1+u)tan^(−1) (1/u))/u)du =2{∫_0 ^1 ((ln(1+u)tan^(−1) (1/u))/u)du+∫_0 ^1 (([ln(1+u)−lnu]tan^(−1) u)/u)du} =2∫_0 ^1 (((π/2)ln(1+u)−lnutan^(−1) u)/u)du =(π^3 /(12))−2∫_0 ^1 ((lnutan^(−1) u)/u)du =(π^3 /(12))−2{(1/2)ln^2 utan^(−1) u∣_0 ^1 −(1/2)∫_0 ^1 ((ln^2 u)/(1+u^2 ))du} =(π^3 /(12))+Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 u^(2n) ln^2 udu =(π^3 /(12))+2Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 )) =(π^3 /(12))+(π^3 /(16)) =(7/(48))π^3 −−−−−−−−−−−− ∫_0 ^1 x^a dx=(1/(a+1)) diff a for b times ⇒ ∫_0 ^1 x^a ln^b xdx=(((−1)^b b!)/((a+1)^(b+1) ))
$$\Theta=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{u}}}{\mathrm{u}}\mathrm{du} \\ $$$$=\mathrm{2}\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{u}}}{\mathrm{u}}\mathrm{du}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left[\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)−\mathrm{lnu}\right]\mathrm{tan}^{−\mathrm{1}} \mathrm{u}}{\mathrm{u}}\mathrm{du}\right\} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)−\mathrm{lnutan}^{−\mathrm{1}} \mathrm{u}}{\mathrm{u}}\mathrm{du} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{12}}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{lnutan}^{−\mathrm{1}} \mathrm{u}}{\mathrm{u}}\mathrm{du} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{12}}−\mathrm{2}\left\{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \mathrm{utan}^{−\mathrm{1}} \mathrm{u}\mid_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \mathrm{u}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\mathrm{du}\right\} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{12}}+\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{u}^{\mathrm{2n}} \mathrm{ln}^{\mathrm{2}} \mathrm{udu} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{12}}+\mathrm{2}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{12}}+\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\ $$$$=\frac{\mathrm{7}}{\mathrm{48}}\pi^{\mathrm{3}} \\ $$$$−−−−−−−−−−−− \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{a}} \mathrm{dx}=\frac{\mathrm{1}}{\mathrm{a}+\mathrm{1}} \\ $$$$\mathrm{diff}\:\mathrm{a}\:\mathrm{for}\:\mathrm{b}\:\mathrm{times}\:\Rightarrow\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{a}} \mathrm{ln}^{\mathrm{b}} \mathrm{xdx}=\frac{\left(−\mathrm{1}\right)^{\mathrm{b}} \mathrm{b}!}{\left(\mathrm{a}+\mathrm{1}\right)^{\mathrm{b}+\mathrm{1}} } \\ $$
Commented by Dwaipayan Shikari last updated on 06/Jun/21
(π/2)tan(π/2)x=(1/(1−x))−(1/(1+x))+(1/(3−x))−(1/(3+x))+... (π^2 /4)sec^2 ((π/2)x)=(1/((1−x)^2 ))+(1/((1+x)^2 ))+... (π^3 /8)sec^2 ((π/2)x)tan((π/2)x)=(1/((1−x)^3 ))−(1/((1+x)^3 ))+... One can get different values by D.B.W.R.x at x=(1/2)
$$\frac{\pi}{\mathrm{2}}{tan}\frac{\pi}{\mathrm{2}}{x}=\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{3}−{x}}−\frac{\mathrm{1}}{\mathrm{3}+{x}}+… \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \:}+… \\ $$$$\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}{x}\right){tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }+… \\ $$$${One}\:{can}\:{get}\:{different}\:{values}\:{by}\:{D}.{B}.{W}.{R}.{x}\:\:{at}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by qaz last updated on 06/Jun/21
β(1)=Σ_(n=0) ^∞ (((−1)^n )/(2n+1))=(π/4) β(2)=Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 ))=G β(3)=Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 ))=(π^3 /(32)) β(5)=Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^5 ))=((5π^5 )/(1536)) β(4)=? β(6)=? β(7)=? β(8)=? β(9)=? β(10)=?
$$\beta\left(\mathrm{1}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}+\mathrm{1}}=\frac{\pi}{\mathrm{4}} \\ $$$$\beta\left(\mathrm{2}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{G} \\ $$$$\beta\left(\mathrm{3}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\pi^{\mathrm{3}} }{\mathrm{32}} \\ $$$$\beta\left(\mathrm{5}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{5}} }=\frac{\mathrm{5}\pi^{\mathrm{5}} }{\mathrm{1536}} \\ $$$$\beta\left(\mathrm{4}\right)=?\:\:\:\beta\left(\mathrm{6}\right)=?\:\:\beta\left(\mathrm{7}\right)=?\:\:\beta\left(\mathrm{8}\right)=?\:\:\beta\left(\mathrm{9}\right)=?\:\:\beta\left(\mathrm{10}\right)=? \\ $$
Commented by mnjuly1970 last updated on 06/Jun/21
thank you so much...mr qaz..
$$\:\:\:{thank}\:{you}\:{so}\:{much}…{mr}\:{qaz}.. \\ $$

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