prove-that-0-e-t-t-dt-e-i-pi-4-0-e-ix-x-dx-

Question Number 28999 by abdo imad last updated on 03/Feb/18

p r o v e t h a t \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} d t = e^{i \frac{π}{4}} \int_{0}^{\infty} \frac{e^{- i x}}{\sqrt{x}} d x .

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

t h e c h . \sqrt{x} = t g i v e \int_{0}^{\infty} \frac{e^{- i x}}{\sqrt{x}} d x = \int_{0}^{\infty} \frac{e^{- {i t}^{2}}}{t} 2 t d t

= 2 \int_{0}^{\infty} e^{- {(\sqrt{i} t)}^{2}} d t t h e c h . \sqrt{i} t = u

= 2 \int_{0}^{\infty} e^{- u^{2}} \frac{d u}{\sqrt{i}} (c h . \sqrt{i} t = u)

\frac{2}{\sqrt{i}} \frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{\sqrt{i}} b u t i = e^{i \frac{π}{2}} \Rightarrow \sqrt{i} = e^{i \frac{π}{4}} \Rightarrow

e^{i \frac{π}{4}} \int_{0}^{\infty} \frac{e^{- i x}}{\sqrt{x}} d x = e^{i \frac{π}{4}} e^{- \frac{π}{4}} \sqrt{π} = \sqrt{π} f r o m a n o t h e r s i d e

t h e c h . \sqrt{t} = u g i v e \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} d t = \int_{0}^{\infty} \frac{e^{- u^{2}}}{u} (2 u) d u

= 2 \int_{0}^{\infty} e^{- u^{2}} d u = 2 \frac{\sqrt{π}}{2} = \sqrt{π} s o

\int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} d t = e^{i \frac{π}{4}} \int_{0}^{\infty} \frac{e^{- i x}}{\sqrt{x}} d x .