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Question Number 223079 by Nicholas666 last updated on 14/Jul/25
Evaluate : ∫_0 ^1 ((arctan(x))/x) Li_2 (x) dx
$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\boldsymbol{\mathrm{Evaluate}}\::\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{arctan}\left({x}\right)}{{x}}\:\mathrm{Li}_{\mathrm{2}} \left({x}\right)\:\mathrm{d}{x} \\ $$$$ \\ $$
Answered by MrGaster last updated on 14/Jul/25
Li_2 (x)=Σ_(k=1) ^∞ (x^k /k^2 ) arctan x=Σ_(m=0) ^∞ (((−1)^m x^(2m+1) )/(2m+1)) ((arctan x)/x)=Σ_(m=0) ^∞ (((−1)^m x^(2m) )/(2m+1)) ((arctan x)/x)Li_2 (x)Σ_(k=1) ^∞ Σ_(m=1) ^∞ (((−1)^m x^(2m+k) )/((2m+1)k^2 )) ∫_0 ^1 ((arctan x)/x)Li_2 (x)dx=Σ_(k=1) ^∞ Σ_(m=1) ^∞ (((−1^m ))/((2m+1)k^2 (2m+k+1))) Σ_(m=1) ^∞ (((−1^m ))/((2m+1)k^2 (2m+k+1)))=(1/k)((π/2)−Σ_(m=0) ^∞ (((−1)^m )/(2m+k+1))) Σ_(m=1) ^∞ (((−1)^m )/(2m+k+1))=(1/2)(ψ(((k+2)/4))−ψ((k/4))) ∫_0 ^1 ((arctan(x))/x) Li_2 (x)dx=(π/2)Σ_(k=1) ^∞ (1/k^3 )_(S=ζ(3)) −(1/2)Σ_(k=1) ^∞ (1/k^2 )(ψ(((k+2)/4))−ψ((k/4))) Σ_(k=1) ^∞ (1/k^2 )(ψ(((k+2)/4))−ψ((k/4)))=((2π(13+10π))/(39+39π+34π^2 ))
$$\mathrm{Li}_{\mathrm{2}} \left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{k}} }{{k}^{\mathrm{2}} } \\ $$$$\mathrm{arctan}\:{x}=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} {x}^{\mathrm{2}{m}+\mathrm{1}} }{\mathrm{2}{m}+\mathrm{1}} \\ $$$$\frac{\mathrm{arctan}\:{x}}{{x}}=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} {x}^{\mathrm{2}{m}} }{\mathrm{2}{m}+\mathrm{1}} \\ $$$$\frac{\mathrm{arctan}\:{x}}{{x}}\mathrm{Li}_{\mathrm{2}} \left({x}\right)\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} {x}^{\mathrm{2}{m}+{k}} }{\left(\mathrm{2}{m}+\mathrm{1}\right){k}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arctan}\:{x}}{{x}}\mathrm{Li}_{\mathrm{2}} \left({x}\right){dx}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}^{{m}} \right)}{\left(\mathrm{2}{m}+\mathrm{1}\right){k}^{\mathrm{2}} \left(\mathrm{2}{m}+{k}+\mathrm{1}\right)} \\ $$$$\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}^{{m}} \right)}{\left(\mathrm{2}{m}+\mathrm{1}\right){k}^{\mathrm{2}} \left(\mathrm{2}{m}+{k}+\mathrm{1}\right)}=\frac{\mathrm{1}}{{k}}\left(\frac{\pi}{\mathrm{2}}−\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{\mathrm{2}{m}+{k}+\mathrm{1}}\right) \\ $$$$\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{\mathrm{2}{m}+{k}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{{k}+\mathrm{2}}{\mathrm{4}}\right)−\psi\left(\frac{{k}}{\mathrm{4}}\right)\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{arctan}\left({x}\right)}{{x}}\:\mathrm{Li}_{\mathrm{2}} \left({x}\right){dx}=\frac{\pi}{\mathrm{2}}\underset{{S}=\zeta\left(\mathrm{3}\right)} {\underbrace{\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{3}} }}}\:−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\left(\psi\left(\frac{{k}+\mathrm{2}}{\mathrm{4}}\right)−\psi\left(\frac{{k}}{\mathrm{4}}\right)\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\left(\psi\left(\frac{{k}+\mathrm{2}}{\mathrm{4}}\right)−\psi\left(\frac{{k}}{\mathrm{4}}\right)\right)=\frac{\mathrm{2}\pi\left(\mathrm{13}+\mathrm{10}\pi\right)}{\mathrm{39}+\mathrm{39}\pi+\mathrm{34}\pi^{\mathrm{2}} } \\ $$

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