prove-thst-R-e-i-x-1-x-2-dx-pi-e-

Question Number 29000 by abdo imad last updated on 03/Feb/18

p r o v e t h s t \int_{R} \frac{e^{i ξ x}}{1 + x^{2}} d x = π e^{- ∣ ξ ∣} .

Commented by abdo imad last updated on 04/Feb/18

l e t p u t I = \int_{- \infty}^{+ \infty} \frac{e^{i ξ x}}{1 + x^{2}} d x

c a s e 1 f o r ξ ⩾ 0 l e t p u t f (z) = \frac{e^{i ξ z}}{1 + z^{2}} t h e p o l e s o f f a r e

i a n d - i (s i m p l e s) s o w e h a v e

\int_{- \infty}^{+ \infty} f (z) d z = 2 i π R e (f, i) = 2 i π \frac{e^{i (ξ i)}}{2 i} = π . e^{- ξ}

c a s e 2 i f ξ < o I = \int_{R} \frac{e^{- i ξ (- x)}}{1 + x^{2}} d x = \int_{R} \frac{e^{- i ξ t}}{1 + t^{2}} d t (c h . - x = t) t h e n

w e t a k e g (x) = \frac{e^{- i ξ t}}{1 + t^{2}} \Rightarrow I = \int_{- \infty}^{+ \infty} g (z) d z = 2 i π R e (g, i)

= 2 i π \frac{e^{ξ}}{2 i} = π . e^{ξ} i n b o t h c a s e \int_{R} \frac{e^{i ξ x}}{1 + x^{2}} d x = π e^{- ∣ ξ ∣} .