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Question Number 145634 by mathmax by abdo last updated on 06/Jul/21

$$\mathrm{let}\mathrm{s}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{({2\mathrm{x}}^2+2\mathrm{x}\sqrt{1+{\mathrm{x}}^2}+1)}^{\mathrm{n}}}$$$$1)\mathrm{explicite}\mathrm{s}\left(\mathrm{x}\right)$$$$2)\mathrm{calculate}{\int }_0^1\mathrm{s}\left(\mathrm{x}\right)\mathrm{dx}$$

Answered by Olaf_Thorendsen last updated on 06/Jul/21

$$\mathrm{S}\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(2x^2+2x\sqrt{1+x^2}+1)}^n}$$$$\mathrm{S}\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(x^2+2x\sqrt{1+x^2}+(1+x^2))}^n}$$$$\mathrm{S}\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(x+\sqrt{1+x^2})}^{2n}}$$$$\mathrm{S}\left(x\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(x+\sqrt{1+x^2})}^{2n}}-1$$$$\mathrm{S}\left(x\right)=\frac{1}{1+\frac{1}{{(x+\sqrt{1+x^2})}^2}}-1$$$$\mathrm{S}\left(x\right)=-\frac{1}{{(x+\sqrt{1+x^2})}^2+1}$$$${\int }_0^1\mathrm{S}\left(x\right)dx=-{\int }_0^1\frac{dx}{{(x+\sqrt{1+x^2})}^2+1}$$$$=-{\int }_0^{{\mathrm{sh}}^{-1}\left(1\right)}\frac{\mathrm{ch}u}{{(\mathrm{sh}u+\sqrt{1+{\mathrm{sh}}^{2}u})}^2+1}du$$$$=-{\int }_0^{{\mathrm{sh}}^{-1}\left(1\right)}\frac{\mathrm{ch}u}{{(\mathrm{sh}u+\mathrm{ch}u)}^2+1}du$$$$=-{\int }_0^{{\mathrm{sh}}^{-1}\left(1\right)}\frac{\mathrm{ch}u}{{2\mathrm{ch}}^{2}u+2\mathrm{sh}u\mathrm{ch}u}du$$$$=-\frac{1}{2}{\int }_0^{{\mathrm{sh}}^{-1}\left(1\right)}\frac{du}{\mathrm{ch}u+\mathrm{sh}u}$$$$=-\frac{1}{2}{\int }_0^{{\mathrm{sh}}^{-1}\left(1\right)}\frac{du}{e^u}$$$$=\frac{1}{2}{\left[e^{-u}\right]}_0^{{\mathrm{sh}}^{-1}\left(1\right)}$$$$=\frac{1}{2}(e^{-{\mathrm{sh}}^{-1}\left(1\right)}-1)$$$$=\frac{1}{2}(\frac{1}{1+\sqrt{1+1^2}}-1)$$$$=\frac{1}{2}(\frac{1}{1+\sqrt{2}}-1)$$$$=\frac{1}{2}(\sqrt{2}-2)=\frac{1}{\sqrt{2}}-1$$

Answered by mathmax by abdo last updated on 07/Jul/21

$$1)\mathrm{we}\mathrm{have}\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}{=}_{\mathrm{x}=\mathrm{sht}}\frac{\mathrm{sht}}{\mathrm{cht}}=\frac{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}{{\mathrm{e}}^{\mathrm{t}}+{\mathrm{e}}^{-\mathrm{t}}}=\frac{1-{\mathrm{e}}^{-2\mathrm{t}}}{1+{\mathrm{e}}^{-2\mathrm{t}}}$$$$=(1-{\mathrm{e}}^{-2\mathrm{t}})\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{e}}^{-2\mathrm{nt}}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{e}}^{-2\mathrm{nt}}-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{e}}^{-(2\mathrm{n}+2)\mathrm{t}}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{e}}^{-2\mathrm{nt}}+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{e}}^{-2\mathrm{nt}}$$$$=1+2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{e}}^{2\mathrm{nt}}}=1+2\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{e}}^{2\mathrm{nlog}(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})}}$$$$=1+2\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})}^{2\mathrm{n}}}=1+2\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{({\mathrm{x}}^2+2\mathrm{x}\sqrt{1+{\mathrm{x}}^2}+1+{\mathrm{x}}^2)}^{\mathrm{n}}}$$$$=1+2\mathrm{s}\left(\mathrm{x}\right)\Rightarrow 2\mathrm{s}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}-1\Rightarrow \mathrm{s}\left(\mathrm{x}\right)=\frac{1}{2}(\frac{\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}}-1)$$$$$$

Commented by mathmax by abdo last updated on 07/Jul/21

$$2){\int }_0^1\mathrm{s}\left(\mathrm{x}\right)\mathrm{dx}=\frac{1}{2}{\int }_0^1\frac{\mathrm{xdx}}{\sqrt{1+{\mathrm{x}}^2}}-\frac{1}{2}$$$$=\frac{1}{2}{\left[\sqrt{1+{\mathrm{x}}^2}\right]}_0^1-\frac{1}{2}=\frac{1}{2}(\sqrt{2}-1)-\frac{1}{2}=\frac{\sqrt{2}}{2}-1$$

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