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calculate-0-1-i-x-2-dx-




Question Number 36917 by maxmathsup by imad last updated on 07/Jun/18
calculate ∫_0 ^(+∞) (1+i)^(−x^2 ) dx
calculate0+(1+i)x2dx
Commented by math khazana by abdo last updated on 09/Jun/18
we have (1+i)=(√2) e^(i(π/4))  ⇒  I = ∫_0 ^(+∞)    ((√2) e^(i(π/4)) )^(−x^2 ) dx  = ∫_0 ^(+∞)  {e^(ln((√2))+i(π/4)) }^(−x^2 ) dx  =∫_0 ^(+∞)   e^(−{ln((√2)) +i(π/4)}x^2 ) dx  changement  (√(ln((√2))+i(π/4))) x =t give  I  = (1/( (√(ln((√2)) +i(π/4))))) ∫_0 ^∞     e^(−u^2 ) du  I = ((√π)/(2{(√( ln((√2))+i(π/4)))}))  .
wehave(1+i)=2eiπ4I=0+(2eiπ4)x2dx=0+{eln(2)+iπ4}x2dx=0+e{ln(2)+iπ4}x2dxchangementln(2)+iπ4x=tgiveI=1ln(2)+iπ40eu2duI=π2{ln(2)+iπ4}.

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