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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) = 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}) .