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Question Number 55268 by maxmathsup by imad last updated on 20/Feb/19

$$calculate\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{4n+3}$$

Commented by maxmathsup by imad last updated on 21/Feb/19

$$letS\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{4n+3}{x}^{4n+3}with\mid x\mid <1wehave$$$$\frac{dS}{dx}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^{4n+2}=x^2\sum _{n=0}^{\mathrm{\infty }}{(-x^4)}^n=\frac{x^2}{1+x^4}\Rightarrow$$$$S\left(x\right)={\int }_0^x\frac{t^2}{1+t^4}dt+cbutc=S\left(0\right)=0\Rightarrow S\left(x\right)={\int }_0^x\frac{t^2}{t^4+1}dtand$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{4n+3}={lim}_{x\to 1}S\left(x\right)={\int }_0^1\frac{t^2}{1+t^4}dt=I$$$$wehave{\int }_0^{\mathrm{\infty }}\frac{t^2}{t^4+1}dt={\int }_0^1\frac{t^2}{t^4+1}dt+{\int }_1^{+\mathrm{\infty }}\frac{t^2}{t^4+1}dtand$$$${\int }_1^{+\mathrm{\infty }}\frac{t^2}{t^4+1}dt{=}_{t=\frac{1}{u}}-{\int }_0^1\frac{1}{u^2(\frac{1}{u^4}+1)}\frac{-du}{u^2}={\int }_0^1\frac{du}{1+u^4}\Rightarrow$$$$2{\int }_0^1\frac{t^2}{t^4+1}dt={\int }_0^{\mathrm{\infty }}\frac{t^2}{t^4+1}dt\Rightarrow {\int }_0^1\frac{t^2}{t^4+1}dt=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{t^2}{t^4+1}dtchangement$$$$t={\alpha }^{\frac{1}{4}}give{\int }_0^{\mathrm{\infty }}\frac{t^2}{t^4+1}dt={\int }_0^{\mathrm{\infty }}\frac{{\alpha }^{\frac{1}{2}}}{1+\alpha }\frac{1}{4}{\alpha }^{\frac{1}{4}-1}d\alpha =\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{{\alpha }^{\frac{3}{4}-1}}{1+\alpha }d\alpha$$$$=\frac{1}{4}\frac{\pi }{sin\left(\frac{3\pi }{4}\right)}=\frac{\pi }{4sin\left(\frac{\pi }{4}\right)}=\frac{\pi }{4\frac{\sqrt{2}}{2}}=\frac{\pi }{2\sqrt{2}}\Rightarrow {\int }_0^1\frac{t^2}{t^4+1}dt=\frac{\pi }{4\sqrt{2}}\Rightarrow$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{4n+3}=\frac{\pi }{4\sqrt{2}}.$$

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