∫_0 ^( ∞) e^(−x^2 ) cos(5x)dx


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Question Number 121972 by Lordose last updated on 13/Nov/20

$\int_{0}^{\infty} e^{- x^{2}} \cos (5 x) dx$
	Answered by Bird last updated on 13/Nov/20
	$2I=∫_(−∞) ^(+∞) e^(−x^2 ) cos(5x)dx =Re(∫_(−∞) ^(+∞) e^(−x^2 +5ix) dx) ∫_(−∞) ^(+∞) e^(−x^2 +5ix) dx =∫_(−∞) ^(+∞) e^(−(x^2 −5ix)) dx =∫_(−∞) ^(+∞) e^(−{x^2 −2((5i)/2)x +(((5i)/2))^2 −(((5i)/2))^2 }) dx =∫_(−∞) ^(+∞) e^(−{x−((5i)/2)}^2 −((25)/4)) dx =_(x−((5i)/2)=t) e^(−((25)/4)) ∫_(−∞) ^(+∞) e^(−t^2 ) dt =(√π) e^(−((25)/4)) ⇒I =((√π)/2) e^(−((25)/4))$
	$2 I = \int_{- \infty}^{+ \infty} e^{- x^{2}} c o s (5 x) d x$ $= R e (\int_{- \infty}^{+ \infty} e^{- x^{2} + 5 i x} d x)$ $\int_{- \infty}^{+ \infty} e^{- x^{2} + 5 i x} d x = \int_{- \infty}^{+ \infty} e^{- (x^{2} - 5 i x)} d x$ $= \int_{- \infty}^{+ \infty} e^{- {x^{2} - 2 \frac{5 i}{2} x + {(\frac{5 i}{2})}^{2} - {(\frac{5 i}{2})}^{2}}} d x$ $= \int_{- \infty}^{+ \infty} e^{- {x - \frac{5 i}{2}}^{2} - \frac{25}{4}} d x$ $=_{x - \frac{5 i}{2} = t} e^{- \frac{25}{4}} \int_{- \infty}^{+ \infty} e^{- t^{2}} d t$ $= \sqrt{π} e^{- \frac{25}{4}} \Rightarrow I = \frac{\sqrt{π}}{2} e^{- \frac{25}{4}}$
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