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Question Number 99242 by abdomathmax last updated on 19/Jun/20

$$\mathrm{calculate}\mathrm{L}\left({\mathrm{e}}^{-\mathrm{ax}}\mathrm{ch}\left(3\mathrm{x}\right)\right)\mathrm{with}\mathrm{a}>0$$

Commented byPRITHWISH SEN 2 last updated on 20/Jun/20

$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{s}+\mathrm{a})\mathrm{x}}\mathrm{ch}\left(3\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{s}+\mathrm{a}}{{(\mathrm{s}+\mathrm{a})}^2-9}$$

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

$$\mathrm{L}\left({\mathrm{e}}^{-\mathrm{ax}}\mathrm{ch}\left(3\mathrm{x}\right)\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{at}}\mathrm{ch}\left(3\mathrm{t}\right){\mathrm{e}}^{-\mathrm{xt}}\mathrm{dt}$$ $$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{at}}({\mathrm{e}}^{3\mathrm{t}}+{\mathrm{e}}^{-3\mathrm{t}}){\mathrm{e}}^{-\mathrm{xt}}\mathrm{dt}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}({\mathrm{e}}^{(-\mathrm{a}+3-\mathrm{x})\mathrm{t}}+{\mathrm{e}}^{(-\mathrm{a}-3-\mathrm{x})\mathrm{t}})\mathrm{dt}$$ $$=\frac{1}{2}{[\frac{1}{3-\mathrm{a}-\mathrm{x}}{\mathrm{e}}^{(3-\mathrm{a}-\mathrm{x})\mathrm{t}}+\frac{1}{-3-\mathrm{a}-\mathrm{x}}{\mathrm{e}}^{(-3-\mathrm{a}-\mathrm{x})\mathrm{t}}]}_0^{+\mathrm{\infty }}$$ $$=\frac{1}{2}(-\frac{1}{3-\mathrm{a}-\mathrm{x}}+\frac{1}{3+\mathrm{a}+\mathrm{x}})=\frac{1}{2}(\frac{1}{\mathrm{a}+\mathrm{x}+3}+\frac{1}{\mathrm{a}+\mathrm{x}-3})$$ $$=\frac{1}{2}\left\{\frac{\mathrm{a}+\mathrm{x}-3+\mathrm{a}+\mathrm{x}+3}{{(\mathrm{a}+\mathrm{x})}^2-9}\right\}=\frac{\mathrm{a}+\mathrm{x}}{{(\mathrm{a}+\mathrm{x})}^2-9}\Rightarrow$$ $$\mathrm{L}\left({\mathrm{e}}^{-\mathrm{ax}}\mathrm{ch}\left(3\mathrm{x}\right)\right)=\frac{\mathrm{x}+\mathrm{a}}{{(\mathrm{x}+\mathrm{a})}^2-9}$$

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