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Question Number 196013 by pticantor last updated on 15/Aug/23

q u e l e s t l a t r a n s f o r m e r d e F o u r i e r d e l a f o n c t i o n

s u i v a n t e :

f (x) = \boldsymbol e^{- \frac{\boldsymbol x^{2}}{2}}

\boldsymbol F i n d t h e F o u r i e r t r a n s f o r m o f t h e

f o l l o w i n g f o n c t i o n .

Answered by witcher3 last updated on 16/Aug/23

\int_{- \infty}^{\infty} e^{- ixt} e^{- \frac{x^{2}}{2}} dx

= \int_{- \infty}^{\infty} e^{- \frac{1}{2} (x^{2} + 2 ixt)} dx

= \int_{- \infty}^{\infty} e^{- \frac{1}{2} {(x + it)}^{2} - \frac{t^{2}}{2}} dx

= e^{- \frac{t^{2}}{2}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} {(x + it)}^{2}} dx

e^{- \frac{t^{2}}{2}} \int_{- \infty + it}^{\infty + it} e^{- \frac{1}{2} y^{2}} dy

D =] - R, R [\cup [R, R + it] \cup [R + it, - R + it] \cup [- R + it, - R]

\int_{D} e^{- \frac{1}{2} y^{2}} dy = 0

\lim_{R \to \infty} ∣ \int_{\underline{+} R}^{\underline{+} R + it} e^{- \frac{y^{2}}{2}} dy ∣⩽ \lim_{R \to \infty} ∣ {te}^{- \frac{1}{2} (R^{2} + t^{2})} ∣= 0

\Rightarrow \int_{- \infty + it}^{\infty + it} e^{- \frac{y^{2}}{2}} dy = \int_{- \infty}^{\infty} e^{- \frac{y^{2}}{2}} =

2 \int_{0}^{\infty} e^{- \frac{y^{2}}{2}} dy = \sqrt{2} \int_{0}^{\infty} e^{- x} . x^{- \frac{1}{2}} dx

= \sqrt{2} . Γ (\frac{1}{2}) = \sqrt{2 π}

F (e^{- \frac{x^{2}}{2}}) = \sqrt{2 π} e^{- \frac{t^{2}}{2}} .

Commented by witcher3 last updated on 16/Aug/23

no - \frac{x^{2}}{2} - ixt = - \frac{1}{2} (x^{2} + 2 ixt)

Commented by pticantor last updated on 16/Aug/23

i n t h e 2 n d l i n e y o u m a k e s o m e m i s t a k e l o o k w e l l

Commented by pticantor last updated on 16/Aug/23

t h e f o u r i e r t r a n f o r m a t i o n f o r m u l e i s :

F (f) (t) = \int_{- \infty}^{+ \infty} e^{- 2 π i x t} f (x) d x!!!

Commented by witcher3 last updated on 17/Aug/23

Ah yes forget the formula

i thougth it e^{- ixt} sorry