Question Number 109952 by mathdave last updated on 26/Aug/20
Answered by maths mind last updated on 30/Aug/20
![(1/(x+1))=^ Σ_(k≥0) (((−1)^k )/(1+k))x^k ∫_0 ^1 ((x^(10) ln(k))/(1+x))dx=∫_0 ^1 x^(10) ln(x).Σ_(k≥0) (((−1)^k x^k )/(k+1))dx =∫_0 ^1 Σ_(k≥0) (−1)^k x^(10+k) ln(x)dx=A since ∀a∈]−1,1[ f(x)=Σ_(k≥0) (−1)^k x^(10+k) ln(x) cv absolutly for x∈[−a,a]⇒ A=Σ_(k≥0) ∫_0 ^1 (−1)^k x^(k+10) ln(x)dx u_n =∫_0 ^1 x^n ln(x)dx=[(x^(n+1) /(n+1))ln(k)]−∫_0 ^1 ((x^n )/(n+1)) dx =−(1/((n+1)^2 )) A=Σ_(k≥0) (−1)^k u_(k+10) =−Σ_(k≥0) (−1)^k .(1/((k+11)^2 )) A cv sequencese ⇒A=−Σ_(k≥0) (−1)^(2k) .(1/((2k+11)^2 ))−Σ_(k≥0) (((−1)^(2k+1) )/((2k+12)^2 )) =−Σ_(k≥) (1/((2k+11)^2 ))+Σ_(k≥0) (1/((2k+12)^2 ))=Σ_(k≥0) (1/(4(k+6)^2 ))−Σ_(k≥0) (1/(4(k+((11)/2))^2 )) =(1/4)(Σ_(k≥0) (1/((k+6)^2 ))−Σ_(k≥0) (1/((k+((11)/2))^2 )))=(1/4)(ζ(2,6)−ζ(2,((11)/2)))](https://www.tinkutara.com/question/Q110763.png)
$$\frac{\mathrm{1}}{{x}+\mathrm{1}}\hat {=}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{1}+{k}}{x}^{{k}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{10}} {ln}\left({k}\right)}{\mathrm{1}+{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{10}} {ln}\left({x}\right).\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{{k}} }{{k}+\mathrm{1}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{10}+{k}} {ln}\left({x}\right){dx}={A} \\ $$$$\left.{since}\:\forall{a}\in\right]−\mathrm{1},\mathrm{1}\left[\:{f}\left({x}\right)=\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{10}+{k}} {ln}\left({x}\right)\:{cv}\:{absolutly}\:{for}\right. \\ $$$${x}\in\left[−{a},{a}\right]\Rightarrow \\ $$$${A}=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} {x}^{{k}+\mathrm{10}} {ln}\left({x}\right){dx} \\ $$$${u}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {ln}\left({x}\right){dx}=\left[\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{ln}\left({k}\right)\right]−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} \:}{{n}+\mathrm{1}}\:{dx} \\ $$$$=−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${A}=\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} {u}_{{k}+\mathrm{10}} =−\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} .\frac{\mathrm{1}}{\left({k}+\mathrm{11}\right)^{\mathrm{2}} } \\ $$$${A}\:{cv}\:{sequencese}\:\Rightarrow{A}=−\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{2}{k}} .\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{11}\right)^{\mathrm{2}} }−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{k}+\mathrm{1}} }{\left(\mathrm{2}{k}+\mathrm{12}\right)^{\mathrm{2}} } \\ $$$$=−\underset{{k}\geqslant} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{11}\right)^{\mathrm{2}} }+\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{12}\right)^{\mathrm{2}} }=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{4}\left({k}+\mathrm{6}\right)^{\mathrm{2}} }−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{4}\left({k}+\frac{\mathrm{11}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({k}+\mathrm{6}\right)^{\mathrm{2}} }−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{11}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\zeta\left(\mathrm{2},\mathrm{6}\right)−\zeta\left(\mathrm{2},\frac{\mathrm{11}}{\mathrm{2}}\right)\right) \\ $$$$ \\ $$