∫_0 ^∞ ((cosx)/(x^2 +1))dx


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Question Number 103721 by Dwaipayan Shikari last updated on 16/Jul/20

$\int_{0}^{\infty} \frac{c o s x}{x^{2} + 1} d x$
	Answered by Ar Brandon last updated on 16/Jul/20

	$I = \int_{0}^{\infty} \frac{cosx}{x^{2} + 1} dx$ $Let f (a) = \int_{0}^{\infty} \frac{cosx}{x^{2} + 1} \cdot e^{- a (x^{2} + 1)} dx \Rightarrow f^{'} (a) = - \int_{0}^{\infty} cosx \cdot e^{- a (x^{2} + 1)} dx$ $Let p = \int_{0}^{\infty} cosx \cdot e^{- a (x^{2} + 1)} dx, q = \int_{0}^{\infty} sinx \cdot e^{- a (x^{2} + 1)} dx$ $\Rightarrow p + qi = \int_{0}^{\infty} e^{xi - a (x^{2} + 1)} dx = \int_{0}^{\infty} e^{- ({ax}^{2} - xi + a)} dx$ $W h a t c a n w e d o n e x t ?$
	Commented by Aziztisffola last updated on 16/Jul/20

	$linearite of integral and use \int_{- \infty}^{+ \infty} e^{- x^{2}} dx = \sqrt{π}$
	Commented by abdomathmax last updated on 17/Jul/20
	$let complete the work of sir brandon we have f^′ (a) =−∫_0 ^∞ cosx e^(−a(x^2 +1)) dx ⇒−f^′ (a) =Re(∫_0 ^∞ e^(ix−ax^2 −a) dx) and ∫_0 ^∞ e^(ix−ax^2 −a) dx =(1/2)e^(−a) ∫_(−∞) ^∞ e^(−ax^2 +ix) dx (a>0) =(1/2)e^(−a) ∫_(−∞) ^∞ e^(−a(x^2 −((2ix)/(2a)) +((i/(2a)))^2 −((i/(2a)))^2 )) dx =(1/2)e^(−a) ∫_0 ^∞ e^(−a({x−(i/(2a))}^2 +(1/(4a^2 )))) dx =(1/2)e^(−a) .e^(−(1/(4a ))) ∫_(−∞) ^∞ e^(−a(x−(i/(2a)))^2 ) dx (√a)(x−(1/(2a)))=u =(1/2)e^(−(a+(1/(4a)))) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/(√a)) =(1/(2(√a))) e^(−(a+(1/(4a)))) (√π) ⇒f^′ (a) =−(1/2)(√(π/a))e^(−(a+(1/(4a)))) ⇒ f(a) =−(1/2)∫_0 ^a ((√π)/(√t)) e^(−(t+(1/(4t)))) dt +c =_((√t)=u) −((√π)/2) ∫_0 ^(√a) (1/u) e^(−(u^2 +(1/(4u^2 )))) (2u)du +c f(a)=c−(√π)∫_0 ^(√a) e^(−(u^2 +(1/(4u^2 )))) du lim_(a→+∞) f(a) =0 ⇒c =(√π)∫_0 ^∞ e^(−(u^2 +(1/(4u^2 )))) du ⇒ f(a) =(√(π )) ( ∫_0 ^∞ e^(−(u^2 +(1/(4u^2 )))) du−∫_0 ^(√a) e^(−(u^2 +(1/(4u^2 )))) du =(√π) ∫_(√a) ^(+∞) e^(−(u^2 +(1/(4u^2 ))) du) I =lim_(a→0) f(a) =(√π)∫_0 ^∞ e^(−(u^2 +(1/(4u^2 )))) du ...be continued...$
	$let complete the work of sir brandon$ $we have f^{^{'}} (a) = - \int_{0}^{\infty} cosx e^{- a (x^{2} + 1)} dx$ $\Rightarrow - f^{^{'}} (a) = Re (\int_{0}^{\infty} e^{ix - {ax}^{2} - a} dx) and$ $\int_{0}^{\infty} e^{ix - {ax}^{2} - a} dx = \frac{1}{2} e^{- a} \int_{- \infty}^{\infty} e^{- {ax}^{2} + ix} dx (a > 0)$ $= \frac{1}{2} e^{- a} \int_{- \infty}^{\infty} e^{- a (x^{2} - \frac{2 ix}{2 a} + {(\frac{i}{2 a})}^{2} - {(\frac{i}{2 a})}^{2})} dx$ $= \frac{1}{2} e^{- a} \int_{0}^{\infty} e^{- a ({x - \frac{i}{2 a}}^{2} + \frac{1}{{4 a}^{2}})} dx$ $= \frac{1}{2} e^{- a} . e^{- \frac{1}{4 a}} \int_{- \infty}^{\infty} e^{- a {(x - \frac{i}{2 a})}^{2}} dx \sqrt{a} (x - \frac{1}{2 a}) = u$ $= \frac{1}{2} e^{- (a + \frac{1}{4 a})} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{du}{\sqrt{a}}$ $= \frac{1}{2 \sqrt{a}} e^{- (a + \frac{1}{4 a})} \sqrt{π} \Rightarrow f^{^{'}} (a) = - \frac{1}{2} \sqrt{\frac{π}{a}} e^{- (a + \frac{1}{4 a})} \Rightarrow$ $f (a) = - \frac{1}{2} \int_{0}^{a} \frac{\sqrt{π}}{\sqrt{t}} e^{- (t + \frac{1}{4 t})} dt + c$ $=_{\sqrt{t} = u} - \frac{\sqrt{π}}{2} \int_{0}^{\sqrt{a}} \frac{1}{u} e^{- (u^{2} + \frac{1}{{4 u}^{2}})} (2 u) du + c$ $f (a) = c - \sqrt{π} \int_{0}^{\sqrt{a}} e^{- (u^{2} + \frac{1}{{4 u}^{2}})} du$ $\lim_{a \to + \infty} f (a) = 0 \Rightarrow c = \sqrt{π} \int_{0}^{\infty} e^{- (u^{2} + \frac{1}{{4 u}^{2}})} du \Rightarrow$ $f (a) = \sqrt{π} (\int_{0}^{\infty} e^{- (u^{2} + \frac{1}{{4 u}^{2}})} du - \int_{0}^{\sqrt{a}} e^{- (u^{2} + \frac{1}{{4 u}^{2}})} du$ $= \sqrt{π} \int_{\sqrt{a}}^{+ \infty} e^{- (u^{2} + \frac{1}{{4 u}^{2}}) du}$ $I = \lim_{a \to 0} f (a) = \sqrt{π} \int_{0}^{\infty} e^{- (u^{2} + \frac{1}{{4 u}^{2}})} du$ $. . . be continued . . .$
	Answered by abdomathmax last updated on 16/Jul/20

	$this integral is solved at the platform ?$ $let A = \int_{0}^{\infty} \frac{cosx}{x^{2} + 1} dx \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{cosx}{x^{2} + 1} dx$ $= Re (\int_{- \infty}^{+ \infty} \frac{e^{ix}}{x^{2} + 1} dx) let φ (z) = \frac{e^{iz}}{z^{2} + 1}$ $we have \lim_{z \to + \infty} ∣ z φ (z) ∣ = 0 and$ $φ (z) = \frac{e^{iz}}{(z - i) (z + i)} residus tbeorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, i) = 2 i π \times \frac{e^{- 1}}{2 i}$ $= \frac{π}{e} \Rightarrow 2 A = \frac{π}{e} \Rightarrow A = \frac{π}{2 e}$
	Commented by Ar Brandon last updated on 17/Jul/20
	Well-done Sir ��
	Commented by Dwaipayan Shikari last updated on 17/Jul/20

	$Thanking all of you$
	Commented by mathmax by abdo last updated on 17/Jul/20

	$you are welcome sir .$
	Commented by Dwaipayan Shikari last updated on 01/Aug/20
	https://t.me/c/1457744274/126
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