Question Number 103721 by Dwaipayan Shikari last updated on 16/Jul/20

∫_0 ^∞ ((cosx)/(x^2 +1))dx

Answered by Ar Brandon last updated on 16/Jul/20

I=∫_0 ^∞ ((cosx)/(x^2 +1))dx  Let f(a)=∫_0 ^∞ ((cosx)/(x^2 +1))∙e^(−a(x^2 +1)) dx⇒f′(a)=−∫_0 ^∞ cosx∙e^(−a(x^2 +1)) dx  Let p=∫_0 ^∞ cosx∙e^(−a(x^2 +1)) dx , q=∫_0 ^∞ sinx∙e^(−a(x^2 +1)) dx  ⇒p+qi=∫_0 ^∞ e^(xi−a(x^2 +1)) dx=∫_0 ^∞ e^(−(ax^2 −xi+a)) dx  What can we do next ?

Commented byAziztisffola last updated on 16/Jul/20

linearite of integral and use ∫_(−∞) ^(+∞) e^(−x^2 ) dx=(√π)

Commented byabdomathmax last updated on 17/Jul/20

let complete the work of sir brandon  we have f^′ (a) =−∫_0 ^∞  cosx e^(−a(x^2  +1)) dx  ⇒−f^′ (a) =Re(∫_0 ^∞  e^(ix−ax^2 −a) dx) and  ∫_0 ^∞  e^(ix−ax^2 −a) dx =(1/2)e^(−a) ∫_(−∞) ^∞  e^(−ax^2 +ix)  dx    (a>0)  =(1/2)e^(−a)  ∫_(−∞) ^∞  e^(−a(x^2  −((2ix)/(2a))   +((i/(2a)))^2 −((i/(2a)))^2 )) dx  =(1/2)e^(−a)  ∫_0 ^∞   e^(−a({x−(i/(2a))}^2 +(1/(4a^2 ))))  dx  =(1/2)e^(−a)  .e^(−(1/(4a )))  ∫_(−∞) ^∞   e^(−a(x−(i/(2a)))^2 ) dx        (√a)(x−(1/(2a)))=u  =(1/2)e^(−(a+(1/(4a))))   ∫_(−∞) ^(+∞)  e^(−u^2 )  (du/(√a))  =(1/(2(√a))) e^(−(a+(1/(4a))))  (√π) ⇒f^′ (a) =−(1/2)(√(π/a))e^(−(a+(1/(4a))))  ⇒  f(a) =−(1/2)∫_0 ^a  ((√π)/(√t)) e^(−(t+(1/(4t))))  dt +c  =_((√t)=u)    −((√π)/2) ∫_0 ^(√a)   (1/u) e^(−(u^2  +(1/(4u^2 ))))  (2u)du +c  f(a)=c−(√π)∫_0 ^(√a)  e^(−(u^2  +(1/(4u^2 ))))  du  lim_(a→+∞) f(a) =0 ⇒c =(√π)∫_0 ^∞  e^(−(u^2 +(1/(4u^2 )))) du ⇒  f(a) =(√(π )) ( ∫_0 ^∞  e^(−(u^2  +(1/(4u^2 )))) du−∫_0 ^(√a)  e^(−(u^2  +(1/(4u^2 )))) du  =(√π) ∫_(√a) ^(+∞)   e^(−(u^2 +(1/(4u^2 ))) du)   I =lim_(a→0) f(a) =(√π)∫_0 ^∞  e^(−(u^2  +(1/(4u^2 )))) du  ...be continued...

Answered by abdomathmax last updated on 16/Jul/20

this integral is solved at the platform?  let A =∫_0 ^∞  ((cosx)/(x^2  +1))dx ⇒2A =∫_(−∞) ^(+∞)  ((cosx)/(x^2  +1))dx  =Re(∫_(−∞) ^(+∞)  (e^(ix) /(x^2  +1))dx) let ϕ(z) =(e^(iz) /(z^2 +1))  we have lim_(z→+∞) ∣zϕ(z)∣ =0 and   ϕ(z) =(e^(iz) /((z−i)(z+i))) residus tbeorem give  ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ Res(ϕ,i) =2iπ×(e^(−1) /(2i))  =(π/e) ⇒ 2A =(π/e) ⇒ A =(π/(2e))

Commented byAr Brandon last updated on 17/Jul/20

Well-done Sir 👏👏

Commented byDwaipayan Shikari last updated on 17/Jul/20

Thanking all of you

Commented bymathmax by abdo last updated on 17/Jul/20

you are welcome sir.

Commented byDwaipayan Shikari last updated on 01/Aug/20

https://t.me/c/1457744274/126