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Question Number 129576 by greg_ed last updated on 16/Jan/21

$$\mathbf{please},\mathbf{how}\mathbf{to}\mathbf{show}\mathbf{that}$$$$\mathit{f}:[0,\mathit{a}]\times {\mathbb{R}}_{+}\to \mathbb{R}$$$$(\mathit{x},\mathit{y}){\mathit{e}}^{-\mathit{x}\mathit{y}}\mathit{s}\mathit{i}\mathit{n}\mathit{x}$$$$\mathbf{is}\mathbf{integrable}???$$

Answered by mathmax by abdo last updated on 18/Jan/21

$$\mathrm{D}=[0,\mathrm{a}]\times {\mathrm{R}}^{+}\mathrm{and}\int {\int }_{\mathrm{D}}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{dxdy}$$$$={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{a}}{\mathrm{e}}^{-\mathrm{xy}}\mathrm{sinxdx}\right)\mathrm{dy}={\int }_0^{\mathrm{\infty }}\mathrm{A}\left(\mathrm{y}\right)\mathrm{dy}$$$$\mathrm{A}\left(\mathrm{y}\right)={\int }_0^{\mathrm{a}}{\mathrm{e}}^{-\mathrm{yx}}\mathrm{sinxdx}=\mathrm{Im}\left({\int }_0^{\mathrm{a}}{\mathrm{e}}^{-\mathrm{yx}+\mathrm{ix}}\mathrm{dx}\right)$$$${\int }_0^{\mathrm{a}}{\mathrm{e}}^{(-\mathrm{y}+\mathrm{i})\mathrm{x}}\mathrm{dx}={\left[\frac{1}{-\mathrm{y}+\mathrm{i}}{\mathrm{e}}^{(-\mathrm{y}+\mathrm{i})\mathrm{x}}\right]}_0^{\mathrm{a}}=-\frac{1}{\mathrm{y}-\mathrm{i}}\{{\mathrm{e}}^{(-\mathrm{y}+\mathrm{i})\mathrm{a}}-1\}$$$$=-\frac{\mathrm{y}+\mathrm{i}}{{\mathrm{y}}^2+1}({\mathrm{e}}^{-\mathrm{y}}(\mathrm{cosa}+\mathrm{isina})-1)$$$$=-\frac{1}{{\mathrm{y}}^2+1}(\mathrm{y}+\mathrm{i})({\mathrm{e}}^{-\mathrm{y}}\mathrm{cosa}+{\mathrm{ie}}^{-\mathrm{y}}\mathrm{sina}-1)$$$$=-\frac{1}{{\mathrm{y}}^2+1}({\mathrm{ye}}^{-\mathrm{y}}\mathrm{cosa}+{\mathrm{iye}}^{-\mathrm{y}}\mathrm{sina}-\mathrm{y}+{\mathrm{ie}}^{-\mathrm{y}}\mathrm{cosa}-{\mathrm{e}}^{-\mathrm{y}}\mathrm{sina}-\mathrm{i})\Rightarrow$$$$\mathrm{Im}(...)=-\frac{1}{{\mathrm{y}}^2+1}({\mathrm{ye}}^{-\mathrm{y}}\mathrm{sina}+{\mathrm{e}}^{-\mathrm{y}}\mathrm{cosa}-1)=\mathrm{A}\left(\mathrm{y}\right)\Rightarrow$$$$\int {\int }_{\mathrm{D}}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{dxdy}=-\mathrm{sina}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{ye}}^{-\mathrm{y}}}{{\mathrm{y}}^2+1}\mathrm{dy}-\mathrm{cosa}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{y}}}{{\mathrm{y}}^2+1}\mathrm{dy}+{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dy}}{1+{\mathrm{y}}^2}$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{dy}}{1+{\mathrm{y}}^2}=\frac{\pi }{2}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{y}}}{1+{\mathrm{y}}^2}\mathrm{dy}⩽{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}<+\mathrm{\infty }$$$$\mathrm{\exists }\mathrm{m}>0/{\int }_0^{\mathrm{\infty }}\frac{\mathrm{y}}{{\mathrm{y}}^2+1}{\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}⩽\mathrm{m}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}<+\mathrm{\infty }\Rightarrow \mathrm{f}\mathrm{is}\mathrm{integrable}\mathrm{on}\mathrm{D}$$

Commented by greg_ed last updated on 20/Jan/21

$$\mathbf{thank}\mathbf{u}\mathbf{very}\mathbf{much},\mathbf{sir}\mathbf{mathmax}\mathbf{by}\mathbf{abdo}!$$

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