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Question Number 26107 by abdo imad last updated on 19/Dec/17
answer to 26024    let  put c= ∫_0 ^∞  cos(ax^2 )dx   and   c =  ∫_0 ^∞  sin(ax^2 )dx  ew have  c−is  =   ∫_0 ^∞   e^(−iax^2 ) dx   =2^(−1)   ∫_R  e^(−iax^2 ) dx   and  i put  x^(1/2)  =r(x)(notation)  so  2(c−is)  =   ∫_R   e^(−(r(ia)x)^2 ) dx  and by the changement   t= r(ia) x   we find  2(c+is)  =   (r(ia))^(−1)  ∫_R  e^(−t^2 ) dt   =  r(π)/r(ia)  but  r(ia)  =r(i) r(a)  =  r(a) e^    −−>2(c+is)    =  r(π) r(a)^(−1)   e^(−iπ/4)  ^)   −−>  c =   r(2π)/_(4r(a))   and   s  =  r(2π)/_(4r(a))
answerto26024letputc=0cos(ax2)dxandc=0sin(ax2)dxewhavecis=0eiax2dx=21Reiax2dxandiputx1/2=r(x)(notation)so2(cis)=Re(r(ia)x)2dxandbythechangementt=r(ia)xwefind2(c+is)=(r(ia))1Ret2dt=r(π)/r(ia)butr(ia)=r(i)r(a)=r(a)e>2(c+is)=r(π)r(a)1eiπ/4)>c=r(2π)/4r(a)ands=r(2π)/4r(a)

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