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Question Number 86830 by Ar Brandon last updated on 31/Mar/20
Prove that ∫_0 ^∞ ((sin x)/x)dx = (π/2)
$${Prove}\:\:{that}\:\int_{\mathrm{0}} ^{\infty} \frac{{sin}\:{x}}{{x}}{dx}\:=\:\frac{\pi}{\mathrm{2}} \\ $$
Answered by mind is power last updated on 31/Mar/20
let f(z)=(e^(iz) /z) ∀ε,R>0 ε<R we have ∫_C_ε f(z)dz=2iπRes(f,0) C_ε =]−R,−ε[∪[εe^(iθ) ,θ∈[−π,0]]]ε,+R[ ∪Re^(iθ) =2iπ ∫_ε ^R (−f(−z)+f(z))dz=2iπ−∫_(−π) ^0 f(εe^(iθ) ).iεe^(iθ) dθ(∫_0 ^π ie^(iRcos(θ)−Rsin(θ)) ∫_ε ^R 2i((sin(x))/x)dx=2iπ−∫_0 ^π i((e^(iε(cos(θ)+isin(θ))) /(εe^(iθ) ))).εe^(iθ) dθ−∫_0 ^π ie^(iRcos(θ)−Rsin(θ)) dθ 2i∫_0 ^R ((sin(x))/x)dx=2iπ−lim_(ε→0) ∫_0 ^π ie^(iεe^(iθ) ) dθ−i∫_0 ^π e^(iRcos(θ)−Rsin(θ)) dθ=iπ ⇒2i∫_0 ^∞ ((sin(z))/z)dz=iπ⇒∫_0 ^(+∞) ((sin(z))/z)=(π/2)
$${let}\:{f}\left({z}\right)=\frac{{e}^{{iz}} }{{z}} \\ $$$$\forall\epsilon,{R}>\mathrm{0}\:\:\:\epsilon<{R}\:{we}\:{have} \\ $$$$\int_{{C}_{\epsilon} } {f}\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left({f},\mathrm{0}\right) \\ $$$$\left.{C}_{\epsilon} =\right]−{R},−\epsilon\left[\cup\left[\epsilon{e}^{{i}\theta} ,\theta\in\left[−\pi,\mathrm{0}\right]\right]\right]\epsilon,+{R}\left[\:\cup{Re}^{{i}\theta} =\mathrm{2}{i}\pi\right. \\ $$$$\int_{\epsilon} ^{{R}} \left(−{f}\left(−{z}\right)+{f}\left({z}\right)\right){dz}=\mathrm{2}{i}\pi−\int_{−\pi} ^{\mathrm{0}} {f}\left(\epsilon{e}^{{i}\theta} \right).{i}\epsilon{e}^{{i}\theta} {d}\theta\left(\int_{\mathrm{0}} ^{\pi} {ie}^{{iRcos}\left(\theta\right)−{Rsin}\left(\theta\right)} \right. \\ $$$$\int_{\epsilon} ^{{R}} \mathrm{2}{i}\frac{{sin}\left({x}\right)}{{x}}{dx}=\mathrm{2}{i}\pi−\int_{\mathrm{0}} ^{\pi} {i}\left(\frac{{e}^{{i}\epsilon\left({cos}\left(\theta\right)+{isin}\left(\theta\right)\right)} }{\epsilon{e}^{{i}\theta} }\right).\epsilon{e}^{{i}\theta} {d}\theta−\int_{\mathrm{0}} ^{\pi} {ie}^{{iRcos}\left(\theta\right)−{Rsin}\left(\theta\right)} {d}\theta \\ $$$$\mathrm{2}{i}\int_{\mathrm{0}} ^{{R}} \frac{{sin}\left({x}\right)}{{x}}{dx}=\mathrm{2}{i}\pi−\underset{\epsilon\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{\mathrm{0}} ^{\pi} {ie}^{{i}\epsilon{e}^{{i}\theta} } {d}\theta−{i}\int_{\mathrm{0}} ^{\pi} {e}^{{iRcos}\left(\theta\right)−{Rsin}\left(\theta\right)} {d}\theta={i}\pi \\ $$$$\Rightarrow\mathrm{2}{i}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({z}\right)}{{z}}{dz}={i}\pi\Rightarrow\int_{\mathrm{0}} ^{+\infty} \frac{{sin}\left({z}\right)}{{z}}=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 01/Apr/20
Thanks Sir, but i think you applied a theory which i don′t yet know. Can you please emphasize more on this?
$${Thanks}\:\:{Sir},\:\:{but}\:\:{i}\:\:{think}\:\:{you}\:\:{applied}\:\:{a}\:\:{theory}\:\:{which}\:\:{i}\:\:{don}'{t}\:\:{yet}\:\:{know}. \\ $$$${Can}\:\:{you}\:\:{please}\:\:{emphasize}\:\:{more}\:\:{on}\:\:{this}? \\ $$
Commented by mathmax by abdo last updated on 01/Apr/20
sir ar brondon try to study complex anslysis step by step and that the cocept bevome eazy for you...
$${sir}\:{ar}\:{brondon}\:{try}\:{to}\:{study}\:{complex}\:{anslysis}\:{step}\:{by}\:{step}\:{and} \\ $$$${that}\:{the}\:{cocept}\:{bevome}\:{eazy}\:{for}\:{you}… \\ $$
Commented by Ar Brandon last updated on 02/Apr/20
OK thanks so much.
$${OK}\:\:{thanks}\:\:{so}\:\:{much}. \\ $$
Answered by TANMAY PANACEA. last updated on 31/Mar/20
f(a)=∫_0 ^∞ ((sinx)/x)×e^(−ax) dx ((df(a))/da)=∫_0 ^∞ ((sinx)/x)×e^(−ax) ×−x dx (−1)(df/da)=∫_0 ^∞ e^(−ax) ×sinx dx q=∫_0 ^∞ e^(−ax) ×sinx dx p=∫_0 ^∞ e^(−ax) ×cosx dx p+iq=∫_0 ^∞ e^(−ax) ×e^(ix) dx =∫_0 ^∞ e^(−x(a−i)) dx =∣(e^(−x(a−i)) /(a−i))∣_0 ^∞ =(1/(a−i))=((a+i)/(a^2 +1)) so q=∫_0 ^∞ e^(−ax) sinx dx=(1/(a^2 +1)) now ((−df(a))/da)=(1/(a^2 +1)) ∫df=∫((−da)/(a^2 +1)) f(a)=−tan^(−1) (a)+c now look f(a)=∫_0 ^∞ ((sinx)/x)×e^(−ax) dx when a→∞ then e^(−ax) →0 so f(a)=0 hence.. 0=−tan^(−1) (∞)+c c=(π/2) f(a)=−tan^(−1) (a)+(π/2) f(a)=∫_0 ^∞ ((sinx)/x)×e^(−ax) dx=−tan^(−1) (a)+(π/2) now look f(0)=∫_0 ^∞ ((sinx)/x)×e^(−0) dx=(π/2) ∫_0 ^∞ ((sinx)/x)=(π/2)
$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}×{e}^{−{ax}} {dx} \\ $$$$\frac{{df}\left({a}\right)}{{da}}=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}×{e}^{−{ax}} ×−{x}\:{dx} \\ $$$$\left(−\mathrm{1}\right)\frac{{df}}{{da}}=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} ×{sinx}\:{dx} \\ $$$$\:\:{q}=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} ×{sinx}\:{dx} \\ $$$${p}=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} ×{cosx}\:{dx} \\ $$$${p}+{iq}=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} ×{e}^{{ix}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}\left({a}−{i}\right)} {dx} \\ $$$$=\mid\frac{{e}^{−{x}\left({a}−{i}\right)} }{{a}−{i}}\mid_{\mathrm{0}} ^{\infty} =\frac{\mathrm{1}}{{a}−{i}}=\frac{{a}+{i}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$${so}\:{q}=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} {sinx}\:{dx}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$${now}\:\frac{−{df}\left({a}\right)}{{da}}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int{df}=\int\frac{−{da}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$${f}\left({a}\right)=−{tan}^{−\mathrm{1}} \left({a}\right)+{c} \\ $$$${now}\:{look}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}×{e}^{−{ax}} \:{dx} \\ $$$${when}\:{a}\rightarrow\infty\:\:{then}\:{e}^{−{ax}} \rightarrow\mathrm{0}\:\:\:{so}\:{f}\left({a}\right)=\mathrm{0} \\ $$$${hence}.. \\ $$$$\mathrm{0}=−{tan}^{−\mathrm{1}} \left(\infty\right)+{c} \\ $$$${c}=\frac{\pi}{\mathrm{2}} \\ $$$${f}\left({a}\right)=−{tan}^{−\mathrm{1}} \left({a}\right)+\frac{\pi}{\mathrm{2}} \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}×{e}^{−{ax}} {dx}=−{tan}^{−\mathrm{1}} \left({a}\right)+\frac{\pi}{\mathrm{2}} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{look}} \\ $$$$\boldsymbol{{f}}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}×{e}^{−\mathrm{0}} \:{dx}=\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}}=\frac{\pi}{\mathrm{2}} \\ $$
Commented by Ar Brandon last updated on 01/Apr/20
got it, thanks
$${got}\:\:{it},\:\:{thanks} \\ $$$$ \\ $$

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