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)/


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Question Number 26107 by abdo imad last updated on 19/Dec/17

$a n s w e r t o 26024 l e t p u t c = \int_{0}^{\infty} c o s ({a x}^{2}) d x a n d c = \int_{0}^{\infty} s i n ({a x}^{2}) d x$ $e w h a v e c - i s = \int_{0}^{\infty} e^{- {i a x}^{2}} d x = 2^{- 1} \int_{R} e^{- {i a x}^{2}} d x a n d i p u t$ $x^{1 / 2} = r (x) (n o t a t i o n) s o 2 (c - i s) = \int_{R} e^{- {(r (i a) x)}^{2}} d x$ $a n d b y t h e c h a n g e m e n t t = r (i a) x w e f i n d$ $2 (c + i s) = {(r (i a))}^{- 1} \int_{R} e^{- t^{2}} d t = r (π) / r (i a) b u t$ $r (i a) = r (i) r (a) = r (a) e^{} - - > 2 (c + i s) = r (π) r {(a)}^{- 1} e^{- i π / 4}^{)}$ $- - > c = r (2 π) /_{4 r (a)} a n d s = r (2 π) /_{4 r (a)}$
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