Question-86924

Question Number 86924 by TawaTawa1 last updated on 01/Apr/20

Answered by mind is power last updated on 01/Apr/20

=xΣ_(n≥1) .(((−2x^2 )^n )/n^2 )=U_n ∫((ln(1+x))/x)dx=∫Σ_(n≥0) (((−1)^n x^n )/(n+1))dx=Σ_(n≥1) (((−1)^n x^n )/n^2 )=S_n ∫_0 ^t ((ln(1+x))/x)dx=∫_0 ^(−t) ((ln(1−y))/y)dy=−Li_2 (−t) S_n =−Li_2 (−x) U_n (x)=−Li_2 (−2x^2 )

$$={x}\underset{{n}\geqslant\mathrm{1}} {\sum}.\frac{\left(−\mathrm{2}{x}^{\mathrm{2}} \right)^{{n}} }{{n}^{\mathrm{2}} }={U}_{{n}} \\ $$$$\int\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}=\int\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} }{{n}+\mathrm{1}}{dx}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} }{{n}^{\mathrm{2}} }={S}_{{n}} \\ $$$$\int_{\mathrm{0}} ^{{t}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}=\int_{\mathrm{0}} ^{−{t}} \frac{{ln}\left(\mathrm{1}−{y}\right)}{{y}}{dy}=−{Li}_{\mathrm{2}} \left(−{t}\right) \\ $$$${S}_{{n}} =−{Li}_{\mathrm{2}} \left(−{x}\right) \\ $$$${U}_{{n}} \left({x}\right)=−{Li}_{\mathrm{2}} \left(−\mathrm{2}{x}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by TawaTawa1 last updated on 01/Apr/20