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Question Number 101011 by Dwaipayan Shikari last updated on 29/Jun/20

∫_0 ^∞ ((sinx)/x)dx

∫0∞sinxxdx

Commented by Rohit@Thakur last updated on 29/Jun/20

Use Laplace Transform

UseLaplaceTransform

Commented by Rohit@Thakur last updated on 29/Jun/20

We can evaluate it using Laplace transform L{f(t)}= F(s) = ∫_0 ^∞ e^(−st) f(t) dt ∴L{((Sint)/t)} = ∫_0 ^∞ e^(−st ) ((Sint)/t)dt = tan^(−1) ((1/s)) F(0) = ∫_0 ^∞ ((Sintdt)/t) = lim tan^(−1) ((1/s)) s→0 ∴∫_0 ^∞ ((Sint dt)/t) = (π/2)

WecanevaluateitusingLaplacetransformL{f(t)}=F(s)=∫0∞e−stf(t)dt∴L{Sintt}=∫0∞e−stSinttdt=tan−1(1s)F(0)=∫0∞Sintdtt=limtan−1(1s)s→0∴∫0∞Sintdtt=π2

Commented by Dwaipayan Shikari last updated on 29/Jun/20

Is there any other way?

Isthereanyotherway?

Commented by smridha last updated on 29/Jun/20

let′s do it by Feynman Technique let I(a)=∫_0 ^∞ ((sin(x))/x).e^(−ax) dx so I(0)=∫_0 ^∞ ((sin(x))/x) dx and I(∞)=0[aslim_(a→∞) e^(−ax) =0] now diff:both sides wrt a we get (d/da)[I(a)]=∫_0 ^∞ ((sin(x))/x).(∂/∂a)(e^(−ax) )dx I^′ (a)=−∫_0 ^∞ sin(x).e^(−ax) dx=−(1/(a^2 +1)) now integrating both sides wrt a ∫_0 ^∞ I^′ (a)da=−∫_0 ^∞ (da/(a^2 +1))=−[tan^(−1) (a)]_0 ^∞ =−(𝛑/2) I(∞)−I(0)=−(𝛑/2) 0−∫_0 ^∞ ((sin(x))/x)dx=−(𝛑/2) so ∫_0 ^∞ ((sin(x))/x)dx=(𝛑/2).

let′sdoitbyFeynmanTechniqueletI(a)=∫0∞sin(x)x.e−axdxsoI(0)=∫0∞sin(x)xdxandI(∞)=0[aslima→∞e−ax=0]nowdiff:bothsideswrtawegetdda[I(a)]=∫0∞sin(x)x.∂∂a(e−ax)dxI′(a)=−∫0∞sin(x).e−axdx=−1a2+1nowintegratingbothsideswrta∫0∞I′(a)da=−∫0∞daa2+1=−[tan−1(a)]0∞=−π2I(∞)−I(0)=−π20−∫0∞sin(x)xdx=−π2so∫0∞sin(x)xdx=π2.

Commented by smridha last updated on 29/Jun/20

another way I=∫_0 ^∞ ((sinx)/x)dx by Cauchy principal value theorem we get I=p∫_(−∞) ^(+∞) (e^(ix) /(2ix))dx now two semi−infinite segments costituting the pricipal value integral (i)semicircle C_R of redius(R⇒∞) (ii)semicircle C_r of redius (r⇒0) is taken clockwise now ∮(e^(iz) /(2iz))dz=I+∫_c_r (e^(iz) /(2iz))dz+∫_c_R (e^(iz) /(2iz))dz=0 by Jordan′s lemma the integral over C_R vanishes. the clockwise path C_r half way around the pole z=0 contributes half the value of full circuit so the residue around the pole z=0 is (1/(2i)) so ∫_c_r =−𝛑i.(1/(2i))=−(𝛑/2) so I=(𝛑/2).

anotherwayI=∫0∞sinxxdxbyCauchyprincipalvaluetheoremwegetI=p∫−∞+∞eix2ixdxnowtwosemi−infinitesegmentscostitutingthepricipalvalueintegral(i)semicircleCRofredius(R⇒∞)(ii)semicircleCrofredius(r⇒0)istakenclockwisenow∮eiz2izdz=I+∫creiz2izdz+∫cReiz2izdz=0byJordan′slemmatheintegraloverCRvanishes.theclockwisepathCrhalfwayaroundthepolez=0contributeshalfthevalueoffullcircuitsotheresiduearoundthepolez=0is12iso∫cr=−πi.12i=−π2soI=π2.

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