∫_0 ^(+∞) e^(−x) cos(x^2 )dx. Discuss the convergence of this generalised intergral. Please help


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Question Number 97563 by 675480065 last updated on 08/Jun/20

$\int_{0}^{+ \infty} e^{- x} \cos (x^{2}) dx .$ $Discuss the convergence of this$ $generalised intergral .$ $Please help$
	Answered by mathmax by abdo last updated on 08/Jun/20
	$I =∫_0 ^∞ e^(−x) cos(x^2 )dx for all a>0 x→e^(−x) cos(x^2 )is continue on [0,a] so integrable on[0,a] at [a,+∞[ lim_(x→+∞) e^(−x) cos(x^2 )dx =0 ⇒ ∫_a ^(+∞) e^(−x) cos(x^2 )dx cv ⇒∫_0 ^∞ e^(−x) cos(x^2 )dx converge. the vslue I =Re( ∫_0 ^∞ e^(−x+ix^2 ) dx) and ∫_0 ^∞ e^(−x+ix^2 ) dx =∫_0 ^∞ e^(((√i)x)^2 −2(√i)x×(1/(2(√i))) +((1/(2(√i))))^2 −((1/(2(√i))))^2 ) dx =∫_0 ^∞ e^(((√i)x−(1/(2(√i))))^2 −(1/(4i))) dx =e^(i/4) ∫_0 ^∞ e^(((√i)x−(1/(2(√i))))^2 ) dx =_((√i)x−(1/(2(√i)))=t) e^(i/4) ∫_(−(1/(2(√i)))) ^(+∞) e^(−t^2 ) (dt/(√i)) =e^(−((iπ)/4)) e^(i/4) ∫_(−(1/2)e^(−((iπ)/4)) ) ^(+∞) e^(−t^2 ) dt =e^((−(π/4)+(1/4))i) {((√π)/2)−∫_0 ^(−(1/2)e^(−t^2 ) ) dt}...be continued...$
	$I = \int_{0}^{\infty} e^{- x} \cos (x^{2}) dx for all a > 0 x \to e^{- x} \cos (x^{2}) is continue on [0, a] so$ $integrable on [0, a] at [a, + \infty [\lim_{x \to + \infty} e^{- x} \cos (x^{2}) dx = 0 \Rightarrow$ $\int_{a}^{+ \infty} e^{- x} \cos (x^{2}) dx cv \Rightarrow \int_{0}^{\infty} e^{- x} \cos (x^{2}) dx converge .$ $the vslue I = Re (\int_{0}^{\infty} e^{- x + {ix}^{2}} dx) and$ $\int_{0}^{\infty} e^{- x + {ix}^{2}} dx = \int_{0}^{\infty} e^{{(\sqrt{i} x)}^{2} - 2 \sqrt{i} x \times \frac{1}{2 \sqrt{i}} + {(\frac{1}{2 \sqrt{i}})}^{2} - {(\frac{1}{2 \sqrt{i}})}^{2}} dx$ $= \int_{0}^{\infty} e^{{(\sqrt{i} x - \frac{1}{2 \sqrt{i}})}^{2} - \frac{1}{4 i}} dx = e^{\frac{i}{4}} \int_{0}^{\infty} e^{{(\sqrt{i} x - \frac{1}{2 \sqrt{i}})}^{2}} dx =_{\sqrt{i} x - \frac{1}{2 \sqrt{i}} = t} e^{\frac{i}{4}} \int_{- \frac{1}{2 \sqrt{i}}}^{+ \infty} e^{- t^{2}} \frac{dt}{\sqrt{i}}$ $= e^{- \frac{i π}{4}} e^{\frac{i}{4}} \int_{- \frac{1}{2} e^{- \frac{i π}{4}}}^{+ \infty} e^{- t^{2}} dt = e^{(- \frac{π}{4} + \frac{1}{4}) i} {\frac{\sqrt{π}}{2} - \int_{0}^{- \frac{1}{2} e^{- t^{2}}} dt} . . . be continued . . .$
	Commented by 675480065 last updated on 08/Jun/20

	$t h a n k s t e a c h e r$
	Commented by abdomathmax last updated on 08/Jun/20

	$you are welcome .$
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