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Question Number 123715 by Engr_Jidda last updated on 27/Nov/20

Answered by Dwaipayan Shikari last updated on 27/Nov/20

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{2x}}{e^{3x}+1}dx$$$$=2{\int }_0^{\mathrm{\infty }}\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{e}^{2x}{e}^{-3nx}dx$$$$=2\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{\int }_0^{\mathrm{\infty }}{e}^{-x(3n-2)}dxx(3n-2)=u$$$$=2\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}\frac{1}{(3n-2)}{\int }_0^{\mathrm{\infty }}{e}^{-u}du$$$$=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{3n-2}\mathrm{\Gamma }\left(1\right)$$$$=2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{3n-2}=2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{3n+1}=2\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{\int }_0^1{x}^{3n}$$$$=2{\int }_0^1\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{x}^{3n}$$$$=2{\int }_0^1\frac{1}{1+x^3}dx$$$$=2{\int }_0^1\frac{1}{3(1+x)}+\frac{-\frac{x}{3}+\frac{2}{3}}{1-x+x^2}dx$$$$=-\frac{1}{3}{\int }_0^1\frac{2x-1}{x^2-x+1}+\int \frac{1}{x^2-x+1}dx$$$$=-\frac{1}{3}{\left[log(x^2-x+1)\right]}_0^1+\int \frac{1}{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx$$$$={\left[\frac{2}{\sqrt{3}}{tan}^{-1}\frac{2x-1}{\sqrt{3}}\right]}_0^1$$$$=\frac{2\pi }{3\sqrt{3}}$$

Answered by mathmax by abdo last updated on 28/Nov/20

$$\mathrm{A}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{2\mathrm{x}}}{{\mathrm{e}}^{3\mathrm{x}}+1}\mathrm{dx}={\int }_{-\mathrm{\infty }}^0\frac{{\mathrm{e}}^{2\mathrm{x}}}{{\mathrm{e}}^{3\mathrm{x}}+1}\mathrm{dx}(\to \mathrm{x}=-\mathrm{t})+{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{2\mathrm{x}}}{{\mathrm{e}}^{3\mathrm{x}}+1}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-2\mathrm{t}}}{{\mathrm{e}}^{-3\mathrm{t}}+1}\mathrm{dt}+{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{2\mathrm{x}}}{{\mathrm{e}}^{3\mathrm{x}}+1}\mathrm{dx}\mathrm{we}\mathrm{have}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-2\mathrm{x}}}{{\mathrm{e}}^{-3\mathrm{x}}+1}\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{x}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{e}}^{-3\mathrm{nx}}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(3\mathrm{n}+2)\mathrm{x}}\mathrm{dx}{=}_{(3\mathrm{n}+2)\mathrm{x}=\mathrm{u}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{u}}\frac{\mathrm{du}}{3\mathrm{n}+2}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{3\mathrm{n}+2}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{2\mathrm{x}}}{{\mathrm{e}}^{3\mathrm{x}}+1}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{x}}}{1+{\mathrm{e}}^{-3\mathrm{x}}}\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{e}}^{-3\mathrm{nx}}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(3\mathrm{n}+1)\mathrm{x}}\mathrm{dx}{=}_{(3\mathrm{n}+1)\mathrm{x}=\mathrm{u}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{u}}\frac{\mathrm{du}}{3\mathrm{n}+1}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{3\mathrm{n}+1}\Rightarrow \mathrm{A}=\sum _{\mathrm{n}0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{3\mathrm{n}+2}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{3\mathrm{n}+1}$$$$\mathrm{rest}\mathrm{calculus}\mathrm{those}\mathrm{series}...\mathrm{be}\mathrm{continued}...$$

Commented by Engr_Jidda last updated on 30/Nov/20

$$thanks{i}^{\prime }mwaitingsir$$

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