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Question Number 124141 by nico last updated on 01/Dec/20

Answered by Lordose last updated on 01/Dec/20

  ∫_( 0) ^( ∞) e^(−ix^2 ) dx       Set u=x(√i)  ⇒ du = (√i)dx  (1/( (√i)))∫_( 0) ^( ∞) e^(−u^2 ) du  (1/( (√i)))∙((√π)/2) = −((i(√(iπ)))/2)

$$ \\ $$$$\int_{\:\mathrm{0}} ^{\:\infty} \mathrm{e}^{−\mathrm{ix}^{\mathrm{2}} } \mathrm{dx}\:\:\:\:\: \\ $$$$\mathrm{Set}\:\mathrm{u}=\mathrm{x}\sqrt{\mathrm{i}}\:\:\Rightarrow\:\mathrm{du}\:=\:\sqrt{\mathrm{i}}\mathrm{dx} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{i}}}\int_{\:\mathrm{0}} ^{\:\infty} \mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \mathrm{du} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{i}}}\centerdot\frac{\sqrt{\pi}}{\mathrm{2}}\:=\:−\frac{\mathrm{i}\sqrt{\mathrm{i}\pi}}{\mathrm{2}} \\ $$

Answered by Bird last updated on 01/Dec/20

∫_0 ^∞  e^(−ix^2 ) dx=∫_0 ^∞  e^(−((√i)x)^2 ) dx  =_((√i)x=t)   ∫_0 ^∞   e^(−t^2 ) (dt/( (√i))) =(1/e^((iπ)/4) )×((√π)/2)  =((√π)/2)e^(−((iπ)/4))  =((√π)/2)(cos((π/4))−isin((π/4)))  =((√π)/2)(((√2)/2)−i((√2)/2))=((√(2π))/4)−i((√(2π))/4)

$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{ix}^{\mathrm{2}} } {dx}=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\sqrt{{i}}{x}\right)^{\mathrm{2}} } {dx} \\ $$$$=_{\sqrt{{i}}{x}={t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \frac{{dt}}{\:\sqrt{{i}}}\:=\frac{\mathrm{1}}{{e}^{\frac{{i}\pi}{\mathrm{4}}} }×\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:=\frac{\sqrt{\pi}}{\mathrm{2}}\left({cos}\left(\frac{\pi}{\mathrm{4}}\right)−{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right) \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{4}}−{i}\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{4}} \\ $$

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