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Question Number 42260 by math khazana by abdo last updated on 21/Aug/18
letf(a)=∫−∞+∞cos(ax2)dxwitha>0 1)calculatef(a)intermsofa )calculate∫−∞+∞cos(2x2)dx 3)findthevalueof∫−∞+∞cos(x2+x+1)dx.
Commented bymaxmathsup by imad last updated on 21/Aug/18
3)∫−∞+∞cos(x2+x+1)dx=∫−∞+∞cos((x+12)2+34)dx =x+12=32t∫−∞+∞cos(34(t2+1))32dt =32∫−∞+∞cos(34t2+34)dt =32∫−∞+∞{cos(34t2)cos(34)−sin(34t2)sin(34)}dt =32cos(34)∫−∞+∞cos(34t2)dt−32sin(34)∫−∞+∞sin(34t2)dt =32cos(34)2π234−32sin(34)2π234 =2π2cos(34)−2π2sin(34)⇒ ∫−∞+∞cos(x2+x+1)dx=π2{cos(34)−sin(34)}.
wehave∫−∞+∞cos(ax2)dx−i∫−∞+∞sin(ax2)dx =∫−∞+∞e−iax2dx=∫−∞+∞e−(iax)2dx=iax=t∫−∞+∞e−t2dtia =1aeiπ4π=πae−iπ4=πa{cos(π4)−isin(π4)}⇒ f(a)=πa22=2π2a.alsowehave∫−∞+∞sin(ax2)dx=2π2a 2)∫−∞+∞cos(2x2)dx=f(2)=2π22=π2.
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