calculate ∫_0 ^∞ ((sin x)/x)dx


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Question Number 105545 by 4635 last updated on 29/Jul/20

$c a l c u l a t e \int_{0}^{\infty} \frac{\sin x}{x} d x$
	Answered by Ar Brandon last updated on 29/Jul/20
	$Let f(a)=∫_0 ^∞ ((sinx)/x)∙e^(−ax) dx⇒f ′(a)=−∫_0 ^∞ sinx∙e^(−ax) dx Let; p=∫_0 ^∞ cosx∙e^(−ax) dx , q=∫_0 ^∞ sinx∙e^(−ax) dx ⇒p−iq=∫_0 ^∞ (cosx−isinx)e^(−ax) dx=∫_0 ^∞ e^(−ix−ax) dx=∫_0 ^∞ e^(−(a+i)x) dx =−(1/(a+i))∙[e^(−(a+i)x) ]_0 ^∞ =(1/(a+i)) {lim_(x→∞) e^(−(a+i)x) =0} =((a−i)/(a^2 +1))=(a/(a^2 +1))−i((1/(a^2 +1))) ⇒ Im(p−iq)=Im{(a/(a^2 +1))−i((1/(a^2 +1)))}⇒q=(1/(a^2 +1)) ⇒f ′(a)=−q=−(1/(a^2 +1))⇒f(a)=−[Arctan(a)+C] But f(a)=∫_0 ^∞ ((sinx)/x)∙e^(−ax) dx , lim_(a→+∞) f(a)=0 ⇒lim_(a→+∞) [Arctan(a)+C]=0⇒C=−(π/2) f(0)=∫_0 ^∞ ((sinx)/x)dx=−[Arctan(0)+C]=(π/2) Hence ∫_0 ^∞ ((sinx)/x)dx=(π/2)$
	$Let f (a) = \int_{0}^{\infty} \frac{sinx}{x} \cdot e^{- ax} dx \Rightarrow f^{'} (a) = - \int_{0}^{\infty} sinx \cdot e^{- ax} dx$ $Let; p = \int_{0}^{\infty} cosx \cdot e^{- ax} dx, q = \int_{0}^{\infty} sinx \cdot e^{- ax} dx$ $\Rightarrow p - iq = \int_{0}^{\infty} (cosx - isinx) e^{- ax} dx = \int_{0}^{\infty} e^{- ix - ax} dx = \int_{0}^{\infty} e^{- (a + i) x} dx$ $= - \frac{1}{a + i} \cdot {[e^{- (a + i) x}]}_{0}^{\infty} = \frac{1}{a + i} {{\underset{x \to \infty}{\lim e}}^{- (a + i) x} = 0}$ $= \frac{a - i}{a^{2} + 1} = \frac{a}{a^{2} + 1} - i (\frac{1}{a^{2} + 1})$ $\Rightarrow I m (p - iq) = I m {\frac{a}{a^{2} + 1} - i (\frac{1}{a^{2} + 1})} \Rightarrow q = \frac{1}{a^{2} + 1}$ $\Rightarrow f^{'} (a) = - q = - \frac{1}{a^{2} + 1} \Rightarrow f (a) = - [Arctan (a) + C]$ $But f (a) = \int_{0}^{\infty} \frac{sinx}{x} \cdot e^{- ax} dx, \underset{a \to + \infty}{\lim f} (a) = 0$ $\Rightarrow \lim_{a \to + \infty} [Arctan (a) + C] = 0 \Rightarrow C = - \frac{π}{2}$ $f (0) = \int_{0}^{\infty} \frac{sinx}{x} dx = - [Arctan (0) + C] = \frac{π}{2}$ $Hence \int_{0}^{\infty} \frac{sinx}{x} dx = \frac{π}{2}$
	Answered by Aziztisffola last updated on 29/Jul/20
	$Using Laplace transforms. L{((sinx)/x)}=∫_s ^( ∞) L{sinx}dt=∫_s ^( ∞) (1/(1+t^2 ))dt =[arctan(t)]_s ^(t→∞) =(π/2)−arctan(s) then ∫_0 ^( ∞) e^(−sx) ((sin x)/x) dx=(π/2)−arctan(s) let s=0 ⇒∫_0 ^∞ ((sin x)/x)dx=(π/2)−arctan(0)=(π/2) Hence ∫_0 ^∞ ((sin x)/x)dx=(π/2)$
	$Using Laplace transforms .$ $L {\frac{\sin x}{x}} = \int_{s}^{\infty} L {\sin x} dt = \int_{s}^{\infty} \frac{1}{1 + t^{2}} dt$ $= {[\arctan (t)]}_{s}^{t \to \infty} = \frac{π}{2} - \arctan (s)$ $then \int_{0}^{\infty} e^{- s x} \frac{\sin x}{x} d x = \frac{π}{2} - \arctan (s)$ $let s = 0 \Rightarrow \int_{0}^{\infty} \frac{\sin x}{x} d x = \frac{π}{2} - \arctan (0) = \frac{π}{2}$ $Hence \int_{0}^{\infty} \frac{\sin x}{x} d x = \frac{π}{2}$
	Commented by 4635 last updated on 29/Jul/20

	$t h a n k y o u v e r y m u c h$
	Commented by Aziztisffola last updated on 29/Jul/20

	${you}^{'} re welcome$
	Answered by 1549442205PVT last updated on 30/Jul/20

	$Set I (k, λ) = \int_{0}^{\infty} e^{- kx} \frac{\sin λ x}{x} dx$ $Differentiate the integration I by λ we get$ $\frac{dI}{d λ} = \int_{0}^{\infty} e^{- kx} . \frac{1}{x} . {(\sin λ x)}^{'} dx = \int_{0}^{\infty} e^{- kx} \cos λ xdx$ $= \frac{1}{λ} \int_{0}^{\infty} e^{- kx} dsin λ x = \frac{e^{- kx} \sin λ x}{λ} ∣_{x = 0}^{\infty} + \frac{k}{λ} \int_{0}^{\infty} e^{- kx} \sin λ xdx$ $= \frac{e^{- kx} \sin λ x}{λ} ∣_{0}^{\infty} - \frac{k}{λ^{2}} \int_{0}^{\infty} e^{- kx} dcos λ x$ $= (\frac{e^{- kx} \sin λ x}{λ} - \frac{{ke}^{- kx} \cos λ x}{λ^{2}}) ∣_{0}^{\infty} - \frac{k^{2}}{λ^{2}} \int e^{- kx} \cos λ xdx$ $Thus, \frac{dI}{d λ} = (\frac{e^{- kx} \sin λ x}{λ} - \frac{{ke}^{- kx} \cos λ x}{λ^{2}}) ∣_{0}^{\infty} - \frac{k^{2}}{λ} . \frac{dI}{d λ}$ $\Rightarrow\Rightarrow (1 + \frac{k^{2}}{λ^{2}}) . \frac{dI}{d λ} = (\frac{e^{- kx} \sin λ x}{λ} - \frac{{ke}^{- kx} \cos λ x}{λ^{2}}) ∣_{0}^{\infty}$ $\frac{dI}{d λ} = \frac{e^{- kx} (λ \sin λ x - kcos λ x)}{k^{2} + λ^{2}} ∣_{0}^{\infty} = \frac{k}{λ^{2} + k^{2}}$ $\Rightarrow I = \tan^{- 1} (\frac{λ}{k}) (as \int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}))$ $Consequently, I (k, λ) = \int_{0}^{\infty} e^{- kx} \frac{\sin λ x}{x} dx = \tan^{- 1} (\frac{λ}{k})$ $It follows that$ $J = \int_{0}^{\infty} \frac{\sin λ x}{x} dx = \lim_{k \to 0} \int_{0}^{\infty} e^{- kx} \frac{\sin λ x}{x} dx$ $= {\underset{k \to 0}{\lim t a n}}^{- 1} (\frac{λ}{k}) = \frac{π}{2} (if λ > 0)$ $Therefore, K = \int_{0}^{\infty} \frac{sinx}{x} dx = \frac{π}{2}$
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