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Question Number 123676 by mathmax by abdo last updated on 27/Nov/20

$$\mathrm{find}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\ln (1+{\mathrm{e}}^{2\mathrm{x}})\mathrm{dx}$$

Answered by mnjuly1970 last updated on 27/Nov/20

$$solution$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{e}^{-x}ln\left[e^{2x}(1+e^{-2x})\right]dx$$$$=2{\int }_0^{\mathrm{\infty }}{xe}^{-x}dx+{\int }_0^{\mathrm{\infty }}{e}^{-x}ln(1+e^{-2x})dx$$$$=2\mathrm{\Gamma }\left(2\right)+{\int }_0^{\mathrm{\infty }}{e}^{-x}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}{e}^{-2nx}}{n}dx$$$$=2+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{n}{\int }_0^{\mathrm{\infty }}{e}^{-x(2n+1)}dx$$$$=2+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n}{\left[\frac{e^{-x(2n+1)}}{2n+1}\right]}_0^{\mathrm{\infty }}$$$$=2+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{n(2n+1)}=2-[\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n}-\frac{2{(-1)}^n}{2n+1}]$$$$=2+ln\left(2\right)+2[-1+\frac{\pi }{4}]$$$$ln\left(2\right)+\frac{\pi }{2}\approx 2.26$$

Commented by mnjuly1970 last updated on 27/Nov/20

Commented by harckinwunmy last updated on 27/Nov/20

$$\mathrm{can}\mathrm{i}\mathrm{know}\mathrm{the}\mathrm{name}\mathrm{og}$$$$\mathrm{this}\mathrm{app}.\mathrm{thanks}$$

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

$$Wolframalpha$$

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

https://wolframalpha.com/

Commented by mnjuly1970 last updated on 27/Nov/20

$$thankyoumrpayan$$

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

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

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

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

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\ln (1+{\mathrm{e}}^{2\mathrm{x}})\mathrm{dx}\mathrm{changement}{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}\mathrm{give}$$$$\mathrm{I}={\int }_1^{\mathrm{\infty }}{\mathrm{t}}^{-1}\ln (1+{\mathrm{t}}^2)\frac{\mathrm{dt}}{\mathrm{t}}={\int }_1^{\mathrm{\infty }}\frac{\ln (1+{\mathrm{t}}^2)}{{\mathrm{t}}^2}\mathrm{dt}$$$${=}_{\mathrm{bypsrts}}{[-\frac{1}{\mathrm{t}}\ln (1+{\mathrm{t}}^2)]}_1^{\mathrm{\infty }}+{\int }_1^{\mathrm{\infty }}\frac{1}{\mathrm{t}}\times \frac{2\mathrm{t}}{1+{\mathrm{t}}^2}\mathrm{dt}$$$$=\ln \left(2\right)+2{\int }_1^{\mathrm{\infty }}\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}=\ln \left(2\right)+2{\left[\mathrm{arctant}\right]}_1^{\mathrm{\infty }}$$$$=\ln \left(2\right)+2\{\frac{\pi }{2}-\frac{\pi }{4}\}=\ln \left(2\right)+2\left(\frac{\pi }{4}\right)=\frac{\pi }{2}+\ln \left(2\right)$$

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