let f(x)=∫_(−∞) ^(+∞) cos(t^2 +xt +3)dt with x>0 1) find f(x) 2) calculate ∫_1 ^4 f(x)dx and ∫


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Question Number 56310 by maxmathsup by imad last updated on 13/Mar/19

$l e t f (x) = \int_{- \infty}^{+ \infty} c o s (t^{2} + x t + 3) d t w i t h x > 0$ $1) f i n d f (x)$ $2) c a l c u l a t e \int_{1}^{4} f (x) d x a n d \int_{1}^{+ \infty} f (x) d x$
Commented by maxmathsup by imad last updated on 14/Mar/19
$1) f(x) =Re( ∫_(−∞) ^(+∞) e^(−i(t^2 +xt +3)) dt) but ∫_(−∞) ^(+∞) e^(−i(t^2 +xt +3)) dt =e^(−3i) ∫_(−∞) ^(+∞) e^(−i(t^2 +2(x/2)t +(x^2 /4) −(x^2 /4))) dt =e^(−3i) ∫_(−∞) ^(+∞) e^(−i(t+(x/2))^2 +i(x^2 /4)) dt =e^(i((x^2 /4)−3)) ∫_(−∞) ^(+∞) e^(−{(√i)(t +(x/2))}^2 ) dt changement (√i)(t+(x/2)) =u give f(x) =e^(i((x^2 /4)−3)) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/(√i)) =(π/(√i)) e^(i((x^2 /4)−3)) =π e^(−((iπ)/4)) e^(i((x^2 /4)−3)) =π e^(i((x^2 /4)−3−(π/4)) ) ⇒f(x) =π cos((x^2 /4)−3−(π/4)) .$
$1) f (x) = R e (\int_{- \infty}^{+ \infty} e^{- i (t^{2} + x t + 3)} d t) b u t$ $\int_{- \infty}^{+ \infty} e^{- i (t^{2} + x t + 3)} d t = e^{- 3 i} \int_{- \infty}^{+ \infty} e^{- i (t^{2} + 2 \frac{x}{2} t + \frac{x^{2}}{4} - \frac{x^{2}}{4})} d t$ $= e^{- 3 i} \int_{- \infty}^{+ \infty} e^{- i {(t + \frac{x}{2})}^{2} + i \frac{x^{2}}{4}} d t = e^{i (\frac{x^{2}}{4} - 3)} \int_{- \infty}^{+ \infty} e^{- {\sqrt{i} (t + \frac{x}{2})}^{2}} d t$ $c h a n g e m e n t \sqrt{i} (t + \frac{x}{2}) = u g i v e$ $f (x) = e^{i (\frac{x^{2}}{4} - 3)} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{i}} = \frac{π}{\sqrt{i}} e^{i (\frac{x^{2}}{4} - 3)}$ $= π e^{- \frac{i π}{4}} e^{i (\frac{x^{2}}{4} - 3)} = π e^{i (\frac{x^{2}}{4} - 3 - \frac{π}{4}}) \Rightarrow f (x) = π c o s (\frac{x^{2}}{4} - 3 - \frac{π}{4}) .$
Commented by maxmathsup by imad last updated on 14/Mar/19

$e r r o r f r o m l i n e 5 w e h a v e \int_{- \infty}^{+ \infty} e^{- u^{2}} d u = \sqrt{π} \Rightarrow f (x) = R e (\frac{\sqrt{π}}{\sqrt{i}} e^{i (\frac{x^{2}}{4} - 3)}) \Rightarrow$ $f (x) = \sqrt{π} c o s (\frac{x^{2}}{4} - 3 - \frac{π}{4}) .$
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