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Question Number 173069 by mnjuly1970 last updated on 06/Jul/22

$$$$$$$$$$\mathrm{\Theta }={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}xy{e}^{-(x+y)}cos(x+y)dxdy=\frac{1}{\sigma }$$$$findthevalueof{}^{″}\sigma {}^{″}.$$$$$$

Answered by aleks041103 last updated on 07/Jul/22

$$\mathrm{\Theta }=Re\left({\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}xy{e}^{-(x+y)}{e}^{i(x+y)}dxdy\right)$$$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}xy{e}^{-(x+y)}{e}^{i(x+y)}dxdy=$$$$={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}xy{e}^{(i-1)(x+y)}dxdy=$$$$={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{xe}^{(i-1)x}{ye}^{(i-1)y}dxdy=$$$$={\left({\int }_0^{\mathrm{\infty }}x{e}^{(i-1)x}dx\right)}^2$$$${\int }_0^{\mathrm{\infty }}x{e}^{-ax}dx=\frac{1}{a^2}{\int }_0^{\mathrm{\infty }}\left(ax\right)e^{-\left(ax\right)}d\left(ax\right)=$$$$=\frac{1}{a^2}\left({\int }_0^{\mathrm{\infty }}{x}^1{e}^{-x}dx\right)=\frac{1!}{a^2}=\frac{1}{a^2}$$$$\Rightarrow \mathrm{\Theta }=Re\left(\frac{1}{{(1-i)}^4}\right)$$$$1-i=\sqrt{1^2+1^2}{e}^{-iarctg(-1/1)}=\sqrt{2}{e}^{-i\pi /4}$$$$\Rightarrow {(1-i)}^4={\left(\sqrt{2}\right)}^4{e}^{-i\pi }=-4$$$$\Rightarrow \mathrm{\Theta }=Re\left(\frac{1}{-4}\right)=\frac{1}{-4}=\frac{1}{\sigma }$$$$\Rightarrow \sigma =-4$$

Commented by aleks041103 last updated on 07/Jul/22

$$a^2{\int }_0^{\mathrm{\infty }}x{e}^{-ax}dx={\int }_0^{a\mathrm{\infty }}{x}^1{e}^{-x}dx=1!,ifRe\left(a\right)>0$$

Answered by Eulerian last updated on 07/Jul/22

$$\mathrm{\Theta }=\mathrm{\Re }{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}xy{e}^{-(x+y)}{\mathrm{e}}^{\mathrm{i}(\mathrm{x}+\mathrm{y})}dxdy$$$$=\mathrm{\Re }{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}xy{e}^{-(1-\mathrm{i})(x+y)}dxdy$$$$=\mathrm{\Re }{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}xy{e}^{-(1-\mathrm{i})x-(1-\mathrm{i})y}dxdy$$$$=\mathrm{\Re }\left({\int }_0^{\mathrm{\infty }}y{\mathrm{e}}^{-(1-\mathrm{i})y}dy\right)\left({\int }_0^{\mathrm{\infty }}x{e}^{-(1-\mathrm{i})x}dx\right)$$$$=\mathrm{\Re }{\left(\frac{1!}{{(1-\mathrm{i})}^{1+1}}\right)}^2$$$$=-\frac{1}{4}$$$$$$$$\mathrm{Hence},\sigma =-4$$

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