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Question Number 38118 by maxmathsup by imad last updated on 21/Jun/18

$$provethat{\int }_0^1\frac{1}{1+\frac{t^a}{2}}dt=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2^n(na+1)}$$$$2)findthevalueof\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2^n(3n+1)}$$

Commented by abdo mathsup 649 cc last updated on 05/Jul/18

$$wehave\mid \frac{t^a}{2}\mid <1\Rightarrow {\int }_0^1\frac{1}{1+\frac{t^a}{2}}dt$$$$={\int }_0^1\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{t^{na}}{2^n}\right)dt$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2^n}{\int }_0^1{t}^{na}dt$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2^n}\frac{1}{na+1}$$$$2)wehave\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2^n(3n+1)}={\int }_0^1\frac{dt}{1+\frac{t^3}{2}}$$$$=2{\int }_0^1\frac{dt}{2+t^3}let\alpha /{\alpha }^3=2\Rightarrow \alpha {=}^3\sqrt{2}$$$${\int }_0^1\frac{dt}{t^3+2}={\int }_0^1\frac{dt}{t^3+{\alpha }^3}letddcompose$$$$F\left(t\right)=\frac{1}{t^3+{\alpha }^3}$$$$F\left(t\right)=\frac{1}{(t+\alpha )(t^2-\alpha t+{\alpha }^2)}=\frac{a}{t+\alpha }+\frac{bt+c}{t^2-\alpha t+{\alpha }^2}$$$$...becontinued...$$

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