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Question Number 97616 by mathmax by abdo last updated on 08/Jun/20

$$\mathrm{give}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{x}}}{{(1+\mathrm{x})}^2}\mathrm{dx}\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{serie}$$

Answered by mathmax by abdo last updated on 09/Jun/20

$$\mathrm{we}\mathrm{have}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{x}}}{{(1+\mathrm{x})}^2}\mathrm{dx}{=}_{\mathrm{by}\mathrm{parts}}{[-\frac{1}{1+\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}(-\frac{1}{\mathrm{x}+1}){\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}$$$$=1-{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{x}}}{1+\mathrm{x}}\mathrm{dx}=1-{\int }_0^{\mathrm{\infty }}\frac{1}{1+\mathrm{x}}\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-\mathrm{x})}^{\mathrm{n}}}{\mathrm{n}!}\right)\mathrm{dx}$$$$=1-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{1+\mathrm{x}}\mathrm{dx}\mathrm{but}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{1+\mathrm{x}}\mathrm{dx}={\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}}{1+\mathrm{x}}\mathrm{dx}+{\int }_1^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{1+\mathrm{x}}\mathrm{dx}(\to \mathrm{x}=\frac{1}{\mathrm{t}})$$$$={\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}}{1+\mathrm{x}}\mathrm{dx}+{\int }_0^1\frac{1}{{\mathrm{t}}^{\mathrm{n}}(1+\frac{1}{\mathrm{t}})}\left(\frac{\mathrm{dt}}{{\mathrm{t}}^2}\right)={\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}}{1+\mathrm{x}}\mathrm{dx}+{\int }_0^1\frac{\mathrm{dt}}{{\mathrm{t}}^{\mathrm{n}}({\mathrm{t}}^2+1)}$$$$={\int }_0^1{\mathrm{x}}^{\mathrm{n}}\left(\sum _{\mathrm{p}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{p}}{\mathrm{x}}^{\mathrm{p}}\right)\mathrm{dx}+{\int }_0^1{\mathrm{x}}^{-\mathrm{n}}\left(\sum _{\mathrm{p}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{p}}{\mathrm{x}}^{2\mathrm{p}}\right)$$$$=\sum _{\mathrm{p}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{p}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}+\mathrm{p}}\mathrm{dx}+\sum _{\mathrm{p}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{p}}{\int }_0^1{\mathrm{x}}^{2\mathrm{p}-\mathrm{n}}\mathrm{dx}$$$$=\sum _{\mathrm{p}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{p}}}{\mathrm{n}+\mathrm{p}+1}+\sum _{\mathrm{p}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{p}}}{2\mathrm{p}-\mathrm{n}+1}\Rightarrow$$$$\mathrm{I}=1-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!}(\sum _{\mathrm{p}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{p}}}{\mathrm{n}+\mathrm{p}+1}+\sum _{\mathrm{p}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{p}}}{2\mathrm{p}-\mathrm{n}+1})$$$$=1-\sum _{\mathrm{n},\mathrm{p}}\frac{{(-1)}^{\mathrm{n}+\mathrm{p}}}{\mathrm{n}!(\mathrm{n}+\mathrm{p}+1)}-\sum _{\mathrm{n},\mathrm{p}}\frac{{(-1)}^{\mathrm{n}+\mathrm{p}}}{\mathrm{n}!(2\mathrm{p}-\mathrm{n}+1)}$$

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