please-prove-it-0-e-ax-2-cos-bx-dx-1-2-pi-a-e-b-2-4a-

Question Number 97489 by ali_golmakani last updated on 08/Jun/20

p l e a s e p r o v e i t

\int_{0}^{\infty} e^{- {a x}^{2}} \cos b x d x = \frac{1}{2} \sqrt{\frac{π}{a}} . e^{- \frac{b^{2}}{4 a}}

Answered by smridha last updated on 08/Jun/20

= \frac{1}{2} R e \int_{- \infty}^{\infty} e^{- {a x}^{2} + i b x} d x

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

w e k n o w \int_{- \infty}^{+ \infty} e^{- α x^{2} + β x} d x = \sqrt{\frac{π}{α}} . e^{\frac{β^{2}}{4 α}}

t h i s i s o n e o f t h e r e s u l t s u s e d

i n q u a n t u m m e c h a n i c s f o r f i n d i n g

t h e p r o b a l i t y d e n s i t y o f p a r t i c l e .

i a l s o p r o v e t h e r e s u l t \dots w h i c h

i d i r e c t l y u s e d h e r e .

Answered by mathmax by abdo last updated on 08/Jun/20

\int_{0}^{\infty} e^{- {ax}^{2}} \cos (bx) dx = \frac{1}{2} Re (\int_{- \infty}^{\infty} e^{- {ax}^{2} - ibx} dx) but

\int_{- \infty}^{+ \infty} e^{- {ax}^{2} - ibx} dx = \int_{- \infty}^{+ \infty} e^{- a (x^{2} + 2 \frac{ib}{2 a} x + {(\frac{ib}{2 a})}^{2} - {(\frac{ib}{2 a})}^{2})} dx

= \int_{- \infty}^{+ \infty} e^{- a {{(x + \frac{ib}{2 a})}^{2} + \frac{b^{2}}{{4 a}^{2}}}} dx = e^{- \frac{b 2}{4 a}} \int_{- \infty}^{+ \infty} e^{- a {(x + \frac{ib}{2 a})}^{2}} dx

=_{x + \frac{ib}{2 a} = t} e^{- \frac{b^{2}}{4 a}} \int_{- \infty}^{+ \infty} e^{- {at}^{2}} dt =_{\sqrt{a} t = u} e^{- \frac{b^{2}}{4 a}} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{du}{\sqrt{a}}

= \frac{\sqrt{π}}{2 \sqrt{a}} e^{- \frac{b^{2}}{4 a}} = \frac{1}{2} \sqrt{\frac{π}{a}} e^{- \frac{b^{2}}{4 a}} (a > 0)

Commented by ali_golmakani last updated on 08/Jun/20

t h a n k y o u d e a r f r i e n d

Commented by mathmax by abdo last updated on 08/Jun/20

you are welcome friend