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Question Number 34296 by math khazana by abdo last updated on 03/May/18
find ∫_(−∞) ^(+∞) e^(−jx^2 ) with j =e^(i((2π)/3))
$${find}\:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{jx}^{\mathrm{2}} } \:\:\:\:{with}\:\:{j}\:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$
Commented by candre last updated on 04/May/18
j=cos ((2π)/3)+isin ((2π)/3)=−(1/2)+i((√3)/2)
$${j}=\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{3}}+{i}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Commented by abdo mathsup 649 cc last updated on 04/May/18
we have j =cos(((2π)/3))+i sin(((2π)/3))=−(1/2) +i((√3)/2) ⇒ ∫_(−∞) ^(+∞) e^(−jx^2 ) dx = ∫_(−∞) ^(+∞) e^(−((√j)x)^2 ) dx ch.(√j) x=t give ∫_(−∞) ^(+∞) e^(−jx^2 ) dx = ∫_(−∞) ^(+∞) e^(−t^2 ) (dt/( (√j))) =((√π)/( (√j))) = ((√π)/e^(i(π/3)) ) = (√(π )) e^(−i(π/3)) =(√π)( cos((π/3))+isin((π/3))) =(√π)( (1/2) +i((√3)/2)) ⇒ ∫_(−∞) ^(+∞) e^(−jx^2 ) dx = ((√π)/2) +i((√(3π))/2) .
$${we}\:{have}\:{j}\:={cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)+{i}\:{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{e}^{−{jx}^{\mathrm{2}} } {dx}\:=\:\int_{−\infty} ^{+\infty} \:\:\:{e}^{−\left(\sqrt{{j}}{x}\right)^{\mathrm{2}} } {dx}\:\:\:{ch}.\sqrt{{j}}\:{x}={t}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{e}^{−{jx}^{\mathrm{2}} } {dx}\:=\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \:\frac{{dt}}{\:\sqrt{{j}}} \\ $$$$=\frac{\sqrt{\pi}}{\:\sqrt{{j}}}\:\:=\:\frac{\sqrt{\pi}}{{e}^{{i}\frac{\pi}{\mathrm{3}}} }\:=\:\sqrt{\pi\:}\:{e}^{−{i}\frac{\pi}{\mathrm{3}}} =\sqrt{\pi}\left(\:{cos}\left(\frac{\pi}{\mathrm{3}}\right)+{isin}\left(\frac{\pi}{\mathrm{3}}\right)\right) \\ $$$$=\sqrt{\pi}\left(\:\frac{\mathrm{1}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{e}^{−{jx}^{\mathrm{2}} } {dx}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{3}\pi}}{\mathrm{2}}\:\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 04/May/18
∫_(−∞) ^(+∞) e^(−jx^2 ) dx =(√π)( cos((π/3))−i sin((π/3))) =(√π)( (1/2) −i((√3)/2)) ⇒ ∫_(−∞ ) ^(+∞) e^(−jx^2 ) dx = ((√π)/2) −i((√(3π))/2)
$$\int_{−\infty} ^{+\infty} \:{e}^{−{jx}^{\mathrm{2}} } {dx}\:=\sqrt{\pi}\left(\:{cos}\left(\frac{\pi}{\mathrm{3}}\right)−{i}\:{sin}\left(\frac{\pi}{\mathrm{3}}\right)\right) \\ $$$$=\sqrt{\pi}\left(\:\frac{\mathrm{1}}{\mathrm{2}}\:−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\int_{−\infty\:} ^{+\infty} \:{e}^{−{jx}^{\mathrm{2}} } {dx}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}}\:\:−{i}\frac{\sqrt{\mathrm{3}\pi}}{\mathrm{2}} \\ $$

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