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Question Number 102417 by mathmax by abdo last updated on 09/Jul/20

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\ln (1+{\mathrm{e}}^{\mathrm{x}})\mathrm{dx}$$

Answered by floor(10²Eta[1]) last updated on 09/Jul/20

$$\mathrm{u}=\ln ({\mathrm{e}}^{\mathrm{x}}+1)\Rightarrow \mathrm{du}=\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}+1}\mathrm{dx}$$$$\mathrm{dv}={\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}\Rightarrow \mathrm{v}=-{\mathrm{e}}^{-\mathrm{x}}$$$${[-{\mathrm{e}}^{-\mathrm{x}}\ln ({\mathrm{e}}^{\mathrm{x}}+1)]}_0^{\mathrm{\infty }}+\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{1}{{\mathrm{e}}^{\mathrm{x}}+1}\mathrm{dx}$$$$\ln \left(2\right)+{\int }_0^{\mathrm{\infty }}\frac{1}{{\mathrm{e}}^{\mathrm{x}}+1}\mathrm{dx}=\mathrm{I}$$$$\mathrm{u}={\mathrm{e}}^{\mathrm{x}}\Rightarrow \frac{\mathrm{du}}{\mathrm{u}}=\mathrm{dx}$$$$\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{1}{\mathrm{u}(\mathrm{u}+1)}\mathrm{du}=\underset{1}{\stackrel{\mathrm{\infty }}{\int }}(\frac{1}{\mathrm{u}}-\frac{1}{\mathrm{u}+1})\mathrm{du}$$$$\underset{\mathrm{a}\to \mathrm{\infty }}{\lim }{[\ln \left(\mathrm{u}\right)-\ln (\mathrm{u}+1)]}_1^{\mathrm{a}}=\ln \left(2\right))$$$$\Rightarrow \mathrm{I}=\ln \left(2\right)+\ln \left(2\right)=\ln \left(4\right).$$

Commented by mathmax by abdo last updated on 10/Jul/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by OlafThorendsen last updated on 09/Jul/20

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{e}^{-x}\ln (1+e^x)dx$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{e}^{-x}\ln \left[e^x(1+e^{-x})\right]dx$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}[{xe}^{-x}+e^{-x}\ln (1+e^{-x})]dx$$$$\mathrm{I}={[-(x+1)e^{-x}+(1+e^{-x})(1-\ln (1+e^{-x}))]}_0^{+\mathrm{\infty }}$$$$\mathrm{I}=(0+1)-(-1+2(1-\ln 2))$$$$\mathrm{I}=2\ln 2$$$$$$

Answered by mathmax by abdo last updated on 10/Jul/20

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\ln (1+{\mathrm{e}}^{\mathrm{x}})\mathrm{dx}\mathrm{by}\mathrm{parts}{\mathrm{u}}^{{}^{\prime }}={\mathrm{e}}^{-\mathrm{x}}\mathrm{and}\mathrm{v}=\ln (1+{\mathrm{e}}^{\mathrm{x}})$$$$\mathrm{I}={[-{\mathrm{e}}^{-\mathrm{x}}\ln (1+{\mathrm{e}}^{\mathrm{x}})]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\times \frac{{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}=\ln \left(2\right)+{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{e}}^{\mathrm{x}}}$$$$4\mathrm{hangement}{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}\mathrm{give}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{e}}^{\mathrm{x}}}={\int }_1^{+\mathrm{\infty }}\frac{\mathrm{dt}}{\mathrm{t}(1+\mathrm{t})}$$$$={\int }_1^{+\mathrm{\infty }}(\frac{1}{\mathrm{t}}-\frac{1}{\mathrm{t}+1})\mathrm{dt}={[\ln \mid \frac{\mathrm{t}}{\mathrm{t}+1}\mid ]}_1^{+\mathrm{\infty }}=-\ln \left(\frac{1}{2}\right)=\ln \left(2\right)\Rightarrow$$$$\mathrm{I}=2\ln \left(2\right)$$

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