Question Number 212164 by mnjuly1970 last updated on 04/Oct/24
$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:−\:\mathrm{ln}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\right)\:=\:? \\ $$$$\: \\ $$$$ \\ $$
Answered by Berbere last updated on 04/Oct/24
![=Σ_(n≥0) ∫_0 ^1 (−x^3 )^n dx−((ln(2))/3)=∫_0 ^1 (1/(1+x^3 ))dx−((ln(2))/3) =∫_0 ^1 (1/((x+1)(x^2 −x+1)))dx−((ln(2))/3) =∫_0 ^1 (((x+1)(x−2)−(x^2 −x+1))/(−3(x+1)(x^2 −x+1)))dx−((ln(2))/3) =∫_0 ^1 ((x−2)/(−3(x^2 −x+1)))+(1/(3(x+1)))dx−((ln(2))/3) =(1/3)∫_0 ^1 ((2−x)/(x^2 −x+1)))dx=−(1/6)∫_0 ^1 ((2x−1)/(x^2 −x+1))dx+(1/2)∫_0 ^1 (dx/(x^2 −x+1)) =(1/2)∫_0 ^1 (dx/((x−(1/2))^2 +(3/4)))=(1/( (√3)))tan^(−1) (((2x)/( (√3)))−(1/( (√3))))]_0 ^1 =(2/( (√3))).(π/6)=(π/(3(√3)))](https://www.tinkutara.com/question/Q212166.png)
$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−{x}^{\mathrm{3}} \right)^{{n}} {dx}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{3}} }{dx}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)−\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{−\mathrm{3}\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−\mathrm{2}}{−\mathrm{3}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)}{dx}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}−{x}}{\left.{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx}=−\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\left.=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}.\frac{\pi}{\mathrm{6}}=\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$
Commented by mnjuly1970 last updated on 04/Oct/24
$$\:\cancel{ } \\ $$