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

∫_0 ^∞ ((sinx)/x)dx

Commented byRohit@Thakur last updated on 29/Jun/20

Use Laplace Transform

Commented byRohit@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)

Commented byDwaipayan Shikari last updated on 29/Jun/20

Is there any other way?

Commented bysmridha 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).

Commented bysmridha 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).