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Question Number 143860 by mathmax by abdo last updated on 19/Jun/21

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

Answered by mathmax by abdo last updated on 20/Jun/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}{\mathrm{xe}}^{-{\mathrm{x}}^2}\log (1+{\mathrm{e}}^{\mathrm{x}})\mathrm{dx}\Rightarrow \mathrm{\Phi }={[-\frac{1}{2}{\mathrm{e}}^{-{\mathrm{x}}^2}\log (1+{\mathrm{e}}^{\mathrm{x}})]}_0^{\mathrm{\infty }}$$$$-{\int }_0^{\mathrm{\infty }}-\frac{1}{2}{\mathrm{e}}^{-{\mathrm{x}}^2}\times \frac{{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}=\frac{\log 2}{2}+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{x}}^2}}{1+{\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}\mathrm{but}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{x}}^2}}{1+{\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{x}}^2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{nx}}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{x}}^2-\mathrm{nx}}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-({\mathrm{x}}^2+2\frac{\mathrm{n}}{2}\mathrm{x}+\frac{{\mathrm{n}}^2}{4}-\frac{{\mathrm{n}}^2}{4})}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{\frac{{\mathrm{n}}^2}{4}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{(\mathrm{x}+\frac{\mathrm{n}}{2})}^2}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{\frac{{\mathrm{n}}^2}{4}}{\int }_{\frac{\mathrm{n}}{2}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{z}}^2}\mathrm{dz}$$$$\mathrm{if}\mathrm{we}\mathrm{put}\mathrm{erf}\left(\lambda \right)={\int }_{\lambda }^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{z}}^2}\mathrm{dz}\mathrm{we}\mathrm{get}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{x}}^2}}{1+{\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{\frac{{\mathrm{n}}^2}{4}}\mathrm{erf}\left(\frac{\mathrm{n}}{2}\right)\Rightarrow$$$$\mathrm{\Phi }=\frac{\log 2}{2}+\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\mathrm{erf}\left(\frac{\mathrm{n}}{2}\right){\mathrm{e}}^{\frac{{\mathrm{n}}^2}{4}}$$

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