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Question Number 49232 by Abdo msup. last updated on 04/Dec/18

l e t f (a) = \int_{- \infty}^{+ \infty} c o s (x^{2} + a x + 1) d x

1) c a l c u l a t e f (a) a n d f^{^{'}} (a)

2) f i n d f^{(n)} (a)

Commented by Abdo msup. last updated on 06/Dec/18

l e t g (a) = \int_{- \infty}^{+ \infty} s i n (x^{2} + a x + 1) d x w e h s v e

f (a) - i g (a) = \int_{- \infty}^{+ \infty} e^{- i (x^{2} + a x + 1)} d x

= \int_{- \infty}^{+ \infty} e^{- i (x^{2} + 2 \frac{a}{2} x + \frac{a^{2}}{4} + 1 - \frac{a^{2}}{4})} d x

= e^{- i (1 - \frac{a^{2}}{4})} \int_{- \infty}^{+ \infty} e^{- {(\sqrt{i} (x + \frac{a}{2}))}^{2}} d x

=_{\sqrt{i} (x + \frac{a}{2}) = u} e^{- i (1 - \frac{a^{2}}{4})} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{i}}

= \sqrt{π} e^{- i (1 - \frac{a^{2}}{4})} e^{- \frac{i π}{4}} = \sqrt{π} e^{- i (1 + \frac{π}{4} - \frac{a^{2}}{2})}

\sqrt{π} {c o s (1 + \frac{π}{4} - \frac{a^{2}}{4}) - i s i n (1 + \frac{π}{4} - \frac{a^{2}}{4})} \Rightarrow

f (a) = \sqrt{π} c o s (1 + \frac{π}{4} - \frac{a^{2}}{4}) a l s o w e h a v e

g (a) = \sqrt{π} s i n (1 + \frac{π}{4} - \frac{a^{2}}{4}) .

Commented by Abdo msup. last updated on 06/Dec/18

f^{^{'}} (a) = - \sqrt{π} (- \frac{a}{2}) s i n (1 + \frac{π}{4} - \frac{a^{2}}{4})

= \frac{a \sqrt{π}}{2} s i n (1 + \frac{π}{4} - \frac{a^{2}}{4}) .