Prove: ∫^∞ _0 ((sin(x))/x) dx = (π/2)


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Question Number 218662 by Nicholas666 last updated on 14/Apr/25

$P r o v e : {\int_{0}}^{\infty} \frac{s i n (x)}{x} d x = \frac{π}{2}$
	Answered by SdC355 last updated on 14/Apr/25
	$LT{sin(t)}=(1/(s^2 +1)) ∫_0 ^( ∞) e^(−st) sin(t)dt=(1/(s^2 +1)) ∫_0 ^( ∞) ((sin(t))/t)e^(−st) dt=∫_( s) ^( ∞) (1/(u^2 +1)) dt=(π/2)−tan^(−1) (s) ∫_0 ^( ∞) ((sin(t))/t) dt=lim_(s→0) ∫_0 ^( ∞) ((sin(t))/t)e^(−st) dt=lim_(s→0) ∫_s ^∞ (1/(u^2 +1)) du =lim_(s→0) ((π/2)−tan^(−1) (s))=(π/2) or ∫_0 ^∞ ((sin(t))/t) dt=Si(t)+C ∴lim_(t→∞) Si(t)=(π/2) Q.E.D$
	$LT {\sin (t)} = \frac{1}{s^{2} + 1}$ $\int_{0}^{\infty} e^{- s t} \sin (t) d t = \frac{1}{s^{2} + 1}$ $\int_{0}^{\infty} \frac{\sin (t)}{t} e^{- s t} d t = \int_{s}^{\infty} \frac{1}{u^{2} + 1} d t = \frac{π}{2} - \tan^{- 1} (s)$ $\int_{0}^{\infty} \frac{\sin (t)}{t} d t = \lim_{s \to 0} \int_{0}^{\infty} \frac{\sin (t)}{t} e^{- s t} d t = \lim_{s \to 0} \int_{s}^{\infty} \frac{1}{u^{2} + 1} d u$ $= \lim_{s \to 0} (\frac{π}{2} - \tan^{- 1} (s)) = \frac{π}{2}$ $or \int_{0}^{\infty} \frac{\sin (t)}{t} d t = Si (t) + C$ $∴ \lim_{t \to \infty} Si (t) = \frac{π}{2} Q . E . D$
	Commented by Nicholas666 last updated on 14/Apr/25

	$t h a n k y o u s i r$
	Answered by Nicholas666 last updated on 14/Apr/25
	$∫_0 ^∞ ((sin(x))/x)dx=(π/2) ⇒f(t){_(0,∣t∣>a ) ^(1,∣t∣<a) , a>0 ⇒f^∧ (ω)=(1/( (√(2π))))∫_0 ^∞ f(t)e^(−iωt) dt f^∧ (ω)(1/( (√(2π))))∫_(−a ) ^a 1e^(−iωt) dt=(1/( (√(2π))))[(e^(−iωt) /(−iω))]_(−a) ^a f^∧ (ω)=(1/( (√(2π)))) ((e^(−ωa) −e^(iωa) )/(−iω))=(1/( (√(2π)))) ((−2isin(ωa))/(−iω))=(√(2/π)) ((sin(ωa))/ω) ⇒f(t)=(1/( (√(2π))))∫_(−∞) ^∞ f^∧ (ω)e^(iωt) dω f(t)=(1/(2π))∫_(−∞) ^∞ (√(2/π)) ((sin(ωa))/ω) e^(iωt) dω =(1/π)∫_(−∞) ^∞ ((sin(ωa))/ω)(cos(ωt)+i sin(ωt)dω ⇒I=(1/π)∫_(−∞) ^∞ ((sin(ωa))/ω)(cos(0)+i sin(0))dω I=(1/π)∫_(−∞) ^∞ ((sin(ωa))/ω)(1+0)dω=(1/π)∫_(−∞) ^∞ ((sin(ωa))/ω)dω ⇒I=(1/π).2∫_0 ^∞ ((sin(ωa))/ω)dω=(2/π)∫_0 ^∞ ((sin(ωa))/ω)dω ⇒I=(2/π)∫_0 ^∞ ((sin(x))/(x/a)) (dx/a)=(2/π)∫_0 ^∞ ((sin(x))/x)dx ⇒∫_0 ^∞ ((sin(x))/x)dx=(π/2) ✓$
	$\int_{0}^{\infty} \frac{s i n (x)}{x} d x = \frac{π}{2}$ $\Rightarrow f (t) {_{0, ∣ t ∣> a}^{1, ∣ t ∣< a}, a > 0$ $\Rightarrow \overset{\land}{f} (ω) = \frac{1}{\sqrt{2 π}} \int_{0}^{\infty} f (t) e^{- i ω t} d t$ $\overset{\land}{f} (ω) \frac{1}{\sqrt{2 π}} \int_{- a}^{a} 1 e^{- i ω t} d t = \frac{1}{\sqrt{2 π}} {[\frac{e^{- i ω t}}{- i ω}]}_{- a}^{a}$ $\overset{\land}{f} (ω) = \frac{1}{\sqrt{2 π}} \frac{e^{- ω a} - e^{i ω a}}{- i ω} = \frac{1}{\sqrt{2 π}} \frac{- 2 i s i n (ω a)}{- i ω} = \sqrt{\frac{2}{π}} \frac{s i n (ω a)}{ω}$ $\Rightarrow f (t) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} \overset{\land}{f} (ω) e^{i ω t} d ω$ $f (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} \sqrt{\frac{2}{π}} \frac{s i n (ω a)}{ω} e^{i ω t} d ω$ $= \frac{1}{π} \int_{- \infty}^{\infty} \frac{s i n (ω a)}{ω} (c o s (ω t) + i s i n (ω t) d ω$ $\Rightarrow I = \frac{1}{π} \int_{- \infty}^{\infty} \frac{s i n (ω a)}{ω} (c o s (0) + i s i n (0)) d ω$ $I = \frac{1}{π} \int_{- \infty}^{\infty} \frac{s i n (ω a)}{ω} (1 + 0) d ω = \frac{1}{π} \int_{- \infty}^{\infty} \frac{s i n (ω a)}{ω} d ω$ $\Rightarrow I = \frac{1}{π} . 2 \int_{0}^{\infty} \frac{s i n (ω a)}{ω} d ω = \frac{2}{π} \int_{0}^{\infty} \frac{s i n (ω a)}{ω} d ω$ $\Rightarrow I = \frac{2}{π} \int_{0}^{\infty} \frac{s i n (x)}{x / a} \frac{d x}{a} = \frac{2}{π} \int_{0}^{\infty} \frac{s i n (x)}{x} d x$ $\Rightarrow \int_{0}^{\infty} \frac{s i n (x)}{x} d x = \frac{π}{2} ✓$
	Answered by MrGaster last updated on 14/Apr/25

	Commented by Nicholas666 last updated on 14/Apr/25

	$t h a n k y o u s i r . . .$
	Answered by vnm last updated on 14/Apr/25

	$x > 0, \int_{0}^{\infty} e^{- t x} d t = - \frac{1}{x} e^{- t x} ∣_{0}^{\infty} = - \frac{1}{x} (0 - 1) = \frac{1}{x}$ $\int_{0}^{\infty} \frac{\sin x}{x} d x = \int_{0}^{\infty} \underset{1 / x}{\underset{⏟}{(\int_{0}^{\infty} e^{- t x} d t)}} \sin x \cdot d x =$ $\int_{0}^{\infty} \int_{0}^{\infty} e^{- t x} \sin x \cdot d t d x =$ $\int_{0}^{\infty} (\int_{0}^{\infty} e^{- t x} \sin x \cdot d x) d t =$ $\int_{0}^{\infty} (\int_{0}^{\infty} e^{- t x} \frac{e^{i x} - e^{- i x}}{2 i} d x) d t =$ $\frac{1}{2 i} \int_{0}^{\infty} (\int_{0}^{\infty} e^{(- t + i) x} d x - \int_{0}^{\infty} e^{(- t - i) x} d x) d t =$ $\frac{1}{2 i} \int_{0}^{\infty} (\frac{1}{- t + i} e^{(- t + i) x} ∣_{0}^{\infty} - \frac{1}{- t - i} e^{(- t - i) x} ∣_{0}^{\infty}) d t =$ $\frac{1}{2 i} \int_{0}^{\infty} (\frac{1}{- t + i} (0 - 1) + \frac{1}{t + i} (0 - 1)) d t =$ $\frac{1}{2 i} \int_{0}^{\infty} (\frac{1}{t - i} - \frac{1}{t + i}) d t = \frac{1}{2 i} \int_{0}^{\infty} \frac{2 i}{t^{2} + 1} d t = \int_{0}^{\infty} \frac{d t}{t^{2} + 1} =$ $\tan^{- 1} t ∣_{0}^{\infty} = \frac{π}{2}$
	Commented by Nicholas666 last updated on 14/Apr/25

	$t h a n k y o u S i r, g r e a t s o l u t i o n$
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