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Question Number 96658 by mathmax by abdo last updated on 03/Jun/20

$$\mathrm{determine}\mathrm{L}\left({\mathrm{e}}^{-{\mathrm{x}}^2-\mathrm{x}}\right)\mathrm{with}\mathrm{L}\mathrm{laplace}\mathrm{transform}$$

Answered by mathmax by abdo last updated on 04/Jun/20

$$\mathrm{L}\left({\mathrm{e}}^{-{\mathrm{x}}^2-\mathrm{x}}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2-\mathrm{t}}{\mathrm{e}}^{-\mathrm{xt}}\mathrm{dt}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2-(\mathrm{x}+1)\mathrm{t}}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-({\mathrm{t}}^2+2\mathrm{t}\times \frac{\mathrm{x}+1}{2}+\frac{{(\mathrm{x}+1)}^2}{4}-\frac{{(\mathrm{x}+1)}^2}{4})}={\mathrm{e}}^{\frac{{(\mathrm{x}+1)}^2}{4}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\{\mathrm{t}+\frac{\mathrm{x}+1}{2}\}}^2}\mathrm{dt}$$$${=}_{\mathrm{t}+\frac{\mathrm{x}+1}{2}=\mathrm{u}}{\mathrm{e}}^{\frac{{(\mathrm{x}+1)}^2}{4}}{\int }_{\frac{\mathrm{x}+1}{2}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}={\mathrm{e}}^{\frac{{(\mathrm{x}+1)}^2}{4}}\{{\int }_{\frac{\mathrm{x}+1}{2}}^0{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}+{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}\}$$$$={\mathrm{e}}^{\frac{{(\mathrm{x}+1)}^2}{4}}\{\frac{\pi }{2}-{\int }_0^{\frac{\mathrm{x}+1}{2}}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}\}\Rightarrow$$$$\mathrm{L}\left({\mathrm{e}}^{-{\mathrm{x}}^2-\mathrm{x}}\right)=\frac{\pi }{2}{\mathrm{e}}^{\frac{{(\mathrm{x}+1)}^2}{4}}-{\mathrm{e}}^{\frac{{(\mathrm{x}+1)}^2}{4}}{\int }_0^{\frac{\mathrm{x}+1}{2}}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}$$

Commented by mathmax by abdo last updated on 04/Jun/20

$$\mathrm{sorry}\mathrm{L}\left({\mathrm{e}}^{-{\mathrm{x}}^2-\mathrm{x}}\right)={\mathrm{e}}^{\frac{{(\mathrm{x}+1)}^2}{2}}\{\frac{\sqrt{\pi }}{2}-{\int }_0^{\frac{\mathrm{x}+1}{2}}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}\}\Rightarrow$$$$\mathrm{L}\left({\mathrm{e}}^{-{\mathrm{x}}^2-\mathrm{x}}\right)=\frac{\sqrt{\pi }}{2}{\mathrm{e}}^{\frac{{(\mathrm{x}+1)}^2}{4}}-{\mathrm{e}}^{\frac{{(\mathrm{x}+1)}^2}{4}}{\int }_0^{\frac{\mathrm{x}+1}{2}}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}$$

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