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Question Number 218662 by Nicholas666 last updated on 14/Apr/25
Prove: ∫^∞ _0 ((sin(x))/x) dx = (π/2)
$$\: \\ $$$$\:\:\:\:{Prove}:\:\:\:\:\underset{\mathrm{0}} {\int}^{\infty} \:\frac{{sin}\left({x}\right)}{{x}}\:{dx}\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$
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
$$\mathrm{LT}\left\{\mathrm{sin}\left({t}\right)\right\}=\frac{\mathrm{1}}{\mathrm{s}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{st}} \mathrm{sin}\left({t}\right)\mathrm{d}{t}=\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{sin}\left({t}\right)}{{t}}{e}^{−{st}} \mathrm{d}{t}=\int_{\:{s}} ^{\:\infty} \:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{t}=\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left({s}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{sin}\left({t}\right)}{{t}}\:\mathrm{d}{t}=\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{sin}\left({t}\right)}{{t}}{e}^{−{st}} \mathrm{d}{t}=\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\int_{{s}} ^{\infty} \:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{u} \\ $$$$=\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left({s}\right)\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{or}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left({t}\right)}{{t}}\:\mathrm{d}{t}=\mathrm{Si}\left({t}\right)+{C} \\ $$$$\therefore\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{Si}\left({t}\right)=\frac{\pi}{\mathrm{2}}\:\:\mathrm{Q}.\mathrm{E}.\mathrm{D} \\ $$
Commented by Nicholas666 last updated on 14/Apr/25
thank you sir
$${thank}\:{you}\:{sir} \\ $$
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) ✓
$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{dx}=\frac{\pi}{\mathrm{2}} \\ $$$$\:\Rightarrow{f}\left({t}\right)\left\{_{\mathrm{0},\mid{t}\mid>{a}\:\:} ^{\mathrm{1},\mid{t}\mid<{a}} \:,\:\:{a}>\mathrm{0}\right. \\ $$$$\Rightarrow\overset{\wedge} {{f}}\left(\omega\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\int_{\mathrm{0}} ^{\infty} {f}\left({t}\right){e}^{−{i}\omega{t}} {dt} \\ $$$$\:\:\:\overset{\wedge} {{f}}\left(\omega\right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\int_{−{a}\:\:} ^{{a}} \mathrm{1}{e}^{−{i}\omega{t}} {dt}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\left[\frac{{e}^{−{i}\omega{t}} }{−{i}\omega}\right]_{−{a}} ^{{a}} \\ $$$$\:\:\:\:\overset{\wedge} {{f}}\left(\omega\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\:\frac{{e}^{−\omega{a}} −{e}^{{i}\omega{a}} }{−{i}\omega}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\:\frac{−\mathrm{2}{isin}\left(\omega{a}\right)}{−{i}\omega}=\sqrt{\frac{\mathrm{2}}{\pi}}\:\frac{{sin}\left(\omega{a}\right)}{\omega} \\ $$$$\Rightarrow{f}\left({t}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\int_{−\infty} ^{\infty} \overset{\wedge} {{f}}\left(\omega\right){e}^{{i}\omega{t}} {d}\omega \\ $$$$\:\:\:\:\:\:{f}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{−\infty} ^{\infty} \sqrt{\frac{\mathrm{2}}{\pi}}\:\frac{{sin}\left(\omega{a}\right)}{\omega}\:{e}^{{i}\omega{t}} {d}\omega \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\pi}\int_{−\infty} ^{\infty} \frac{{sin}\left(\omega{a}\right)}{\omega}\left({cos}\left(\omega{t}\right)+{i}\:{sin}\left(\omega{t}\right){d}\omega\right. \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\pi}\int_{−\infty} ^{\infty} \frac{{sin}\left(\omega{a}\right)}{\omega}\left({cos}\left(\mathrm{0}\right)+{i}\:{sin}\left(\mathrm{0}\right)\right){d}\omega \\ $$$$\:\:\:\:\:{I}=\frac{\mathrm{1}}{\pi}\int_{−\infty} ^{\infty} \frac{{sin}\left(\omega{a}\right)}{\omega}\left(\mathrm{1}+\mathrm{0}\right){d}\omega=\frac{\mathrm{1}}{\pi}\int_{−\infty} ^{\infty} \frac{{sin}\left(\omega{a}\right)}{\omega}{d}\omega \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\pi}.\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\omega{a}\right)}{\omega}{d}\omega=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\omega{a}\right)}{\omega}{d}\omega \\ $$$$\Rightarrow{I}=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}/{a}}\:\frac{{dx}}{{a}}=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{dx}=\frac{\pi}{\mathrm{2}}\:\checkmark \\ $$$$ \\ $$
Answered by MrGaster last updated on 14/Apr/25
Commented by Nicholas666 last updated on 14/Apr/25
thank you sir...
$${thank}\:{you}\:{sir}… \\ $$
Answered by vnm last updated on 14/Apr/25
x>0, ∫_0 ^∞ e^(−tx) dt=−(1/x)e^(−tx) ∣_0 ^∞ =−(1/x)(0−1)=(1/x) ∫_0 ^∞ ((sin x)/x)dx=∫_0 ^∞ (∫_0 ^∞ e^(−tx) dt)_(1/x) sin x∙dx= ∫_0 ^∞ ∫_0 ^∞ e^(−tx) sin x∙dtdx= ∫_0 ^∞ (∫_0 ^∞ e^(−tx) sin x∙dx)dt= ∫_0 ^∞ (∫_0 ^∞ e^(−tx) ((e^(ix) −e^(−ix) )/(2i))dx)dt= (1/(2i))∫_0 ^∞ (∫_0 ^∞ e^((−t+i)x) dx−∫_0 ^∞ e^((−t−i)x) dx)dt= (1/(2i))∫_0 ^∞ ((1/(−t+i))e^((−t+i)x) ∣_0 ^∞ −(1/(−t−i))e^((−t−i)x) ∣_0 ^∞ )dt= (1/(2i))∫_0 ^∞ ((1/(−t+i))(0−1)+(1/(t+i))(0−1))dt= (1/(2i))∫_0 ^∞ ((1/(t−i))−(1/(t+i)))dt=(1/(2i))∫_0 ^∞ ((2i)/(t^2 +1))dt=∫_0 ^∞ (dt/(t^2 +1))= tan^(−1) t∣_0 ^∞ =(π/2)
$$ \\ $$$${x}>\mathrm{0},\:\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} {dt}=−\frac{\mathrm{1}}{{x}}{e}^{−{tx}} \mid_{\mathrm{0}} ^{\infty} =−\frac{\mathrm{1}}{{x}}\left(\mathrm{0}−\mathrm{1}\right)=\frac{\mathrm{1}}{{x}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:{x}}{{x}}{dx}=\int_{\mathrm{0}} ^{\infty} \underset{\mathrm{1}/{x}} {\underbrace{\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} {dt}\right)}}\mathrm{sin}\:{x}\centerdot{dx}= \\ $$$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} \mathrm{sin}\:{x}\centerdot{dtdx}= \\ $$$$\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} \mathrm{sin}\:{x}\centerdot{dx}\right){dt}= \\ $$$$\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} \frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}{dx}\right){dt}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} {e}^{\left(−{t}+{i}\right){x}} {dx}−\int_{\mathrm{0}} ^{\infty} {e}^{\left(−{t}−{i}\right){x}} {dx}\right){dt}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{−{t}+{i}}{e}^{\left(−{t}+{i}\right){x}} \mid_{\mathrm{0}} ^{\infty} −\frac{\mathrm{1}}{−{t}−{i}}{e}^{\left(−{t}−{i}\right){x}} \mid_{\mathrm{0}} ^{\infty} \right){dt}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{−{t}+{i}}\left(\mathrm{0}−\mathrm{1}\right)+\frac{\mathrm{1}}{{t}+{i}}\left(\mathrm{0}−\mathrm{1}\right)\right){dt}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{t}−{i}}−\frac{\mathrm{1}}{{t}+{i}}\right){dt}=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{i}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$\mathrm{tan}^{−\mathrm{1}} {t}\mid_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{2}} \\ $$
Commented by Nicholas666 last updated on 14/Apr/25
thank you Sir, great solution
$${thank}\:{you}\:{Sir},\:{great}\:{solution} \\ $$

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