please prove it ∫_0 ^∞ e^(−ax^2 ) cos bx dx= (1/2)(√(π/a)).e^(−(b^2 /(4a)))


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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

	$= \frac{1}{2} R e \int_{- \infty}^{\infty} e^{- {a x}^{2} + i b x} d x$ $= \frac{1}{2} R e \sqrt{\frac{π}{a}} . e^{\frac{{(- i b)}^{2}}{4 a}} = \frac{1}{2} \sqrt{\frac{π}{a}} . e^{- \frac{b^{2}}{4 a}}$ $w e k n o w \int_{- \infty}^{+ \infty} e^{- α x^{2} + β x} d x = \sqrt{\frac{π}{α}} . e^{\frac{β^{2}}{4 α}}$ $t h i s i s o n e o f t h e r e s u l t s u s e d$ $i n q u a n t u m m e c h a n i c s f o r f i n d i n g$ $t h e p r o b a l i t y d e n s i t y o f p a r t i c l e .$ $i a l s o p r o v e t h e r e s u l t . . . w h i c h$ $i d i r e c t l y u s e d h e r e .$
	Answered by mathmax by abdo last updated on 08/Jun/20
	$∫_0 ^∞ e^(−ax^2 ) cos(bx)dx =(1/2)Re(∫_(−∞) ^∞ e^(−ax^2 −ibx) dx) but ∫_(−∞) ^(+∞) e^(−ax^2 −ibx) dx =∫_(−∞) ^(+∞) e^(−a( x^2 +2((ib)/(2a))x +(((ib)/(2a)))^2 −(((ib)/(2a)))^2 )) dx =∫_(−∞) ^(+∞) e^(−a{ (x+((ib)/(2a)))^2 +(b^2 /(4a^2 ))}) dx = e^(−((b2)/(4a))) ∫_(−∞) ^(+∞) e^(−a(x+((ib)/(2a)))^2 ) dx =_(x+((ib)/(2a))=t) e^(−(b^2 /(4a))) ∫_(−∞) ^(+∞) e^(−at^2 ) dt =_((√a)t =u) e^(−(b^2 /(4a))) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/(√a)) =((√π)/(2(√a))) e^(−(b^2 /(4a))) =(1/2)(√(π/a))e^(−(b^2 /(4a))) (a>0)$
	$\int_{0}^{\infty} e^{- {ax}^{2}} \cos (bx) dx = \frac{1}{2} Re (\int_{- \infty}^{\infty} e^{- {ax}^{2} - ibx} dx) but$ $\int_{- \infty}^{+ \infty} e^{- {ax}^{2} - ibx} dx = \int_{- \infty}^{+ \infty} e^{- a (x^{2} + 2 \frac{ib}{2 a} x + {(\frac{ib}{2 a})}^{2} - {(\frac{ib}{2 a})}^{2})} dx$ $= \int_{- \infty}^{+ \infty} e^{- a {{(x + \frac{ib}{2 a})}^{2} + \frac{b^{2}}{{4 a}^{2}}}} dx = e^{- \frac{b 2}{4 a}} \int_{- \infty}^{+ \infty} e^{- a {(x + \frac{ib}{2 a})}^{2}} dx$ $=_{x + \frac{ib}{2 a} = t} e^{- \frac{b^{2}}{4 a}} \int_{- \infty}^{+ \infty} e^{- {at}^{2}} dt =_{\sqrt{a} t = u} e^{- \frac{b^{2}}{4 a}} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{du}{\sqrt{a}}$ $= \frac{\sqrt{π}}{2 \sqrt{a}} e^{- \frac{b^{2}}{4 a}} = \frac{1}{2} \sqrt{\frac{π}{a}} e^{- \frac{b^{2}}{4 a}} (a > 0)$
	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

	$you are welcome friend$
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