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Question Number 141859 by iloveisrael last updated on 24/May/21

 ∫_0 ^∞  ((4e^(−x^2 ) )/((2x^2 +1)^2 )) dx

04ex2(2x2+1)2dx

Commented by Dwaipayan Shikari last updated on 24/May/21

(√π)

π

Answered by qaz last updated on 24/May/21

∫_0 ^∞ ((4e^(−x^2 ) )/((2x^2 +1)^2 ))dx  =4∫_0 ^∞ e^(−x^2 ) (1/(Γ(2)))∫_0 ^∞ ue^(−u(2x^2 +1)) dudx  =4∫_0 ^∞ ue^(−u) du∫_0 ^∞ e^(−x^2 ) e^(−2ux^2 ) dx  =4∫_0 ^∞ ue^(−u) du∫_0 ^∞ e^(−(2u+1)x^2 ) dx  =4∫_0 ^∞ ue^(−u) ((Γ((1/2)))/(2(2u+1)^(1/2) ))du  =(√π)∫_0 ^∞ ((2ue^(−u) )/( (√(2u+1))))du  =−(√π)((√(2u+1))/e^u )∣_0 ^∞   =(√π)

04ex2(2x2+1)2dx=40ex21Γ(2)0ueu(2x2+1)dudx=40ueudu0ex2e2ux2dx=40ueudu0e(2u+1)x2dx=40ueuΓ(12)2(2u+1)12du=π02ueu2u+1du=π2u+1eu0=π

Commented by qaz last updated on 24/May/21

use ((Γ(((a+1)/c)))/(c∙b^((a+1)/c) ))=∫_0 ^∞ x^a e^(−bx^c ) dx

useΓ(a+1c)cba+1c=0xaebxcdx

Commented by Dwaipayan Shikari last updated on 24/May/21

∫_0 ^∞ x^a e^(−bx^c ) dx   bx^c =u  =(1/(b^((a+1)/c) c))∫_0 ^∞ u^(((a+1)/c)−1) e^(−u) du=((Γ(((a+1)/c)))/(b^((a+1)/c) c))

0xaebxcdxbxc=u=1ba+1cc0ua+1c1eudu=Γ(a+1c)ba+1cc

Answered by mindispower last updated on 24/May/21

=∫_0 ^∞ (e^(−x^2 ) /((x^2 +(1/2))))dx  let∫_0 ^∞ (e^(−tx^2 ) /((x^2 +a^2 )))dx=h(t)  h′(t)−a^2 h(t)=−∫_0 ^∞ e^(−tx^2 ) dx,x=(y/( (√t)))⇒dx=(dy/( (√t)))  =−∫_0 ^∞ e^(−y^2 ) .(dy/( (√t)))=−((√π)/(2(√t)))  h(t)=ke^(a^2 t)   ⇒k=−(√π)(1/a).∫e^(−a^2 t) .d(√(a^2 t))=−(√π)∫e^(−u^2 ) .(du/a)  erf(x)=(2/( (√π)))∫_0 ^x e^(−t^2 ) dt  =−(π/(2a))erf(a(√t))+g(a)=k  =(−(π/(2a))erf(a(√t))+g(a))e^(a^2 t) =h(t)  erf(x)=(2/( (√π)))∫_0 ^x e^(−t^2 ) dt  h(0)=∫_0 ^∞ (dx/(x^2 +a^2 ))=(π/(2a))=g(a)  h(t)=(π/(2a))(1−erf(a(√t)))e^(a^2 t) =g(a,t)∣_(t=1)   (−(π/(2a^2 ))(1−erf(a))−(2/( (√π)))e^(−a^2 ) .(π/(2a)))e^a^2  +π(1−erf(a))e^a^2    erf(a)((π/(2a^2 ))−π)+π−(π/(2a^2 ))−((√π)/a)=∫_0 ^∞ (e^(−x^2 ) /(x^2 +a^2 ))dx=Ψ(a)  Ψ′(a)=∫_0 ((−2ae^(−x^2 ) )/((x^2 +a^2 )^2 ))dx  ⇒−((Ψ′(a))/(2a))=∫_0 ^∞ (e^(−x^2 ) /((x^2 +a^2 )^2 ))dx  =((erf(a))/(−2a))((π/(2a^2 ))−π)−(1/(2a))(π−(π/(2a^2 )))+((√π)/(2a^2 ))∣a^2 =(1/2)  =(√π)=∫_0 ^∞ (e^(−x^2 ) /((x^2 +(1/2))^2 ))dx  withe this we can evaluat ∫_0 ^∞ (e^(−x^2 ) /((x^2 +s^2 )^n )),s∈R,n∈N

=0ex2(x2+12)dxlet0etx2(x2+a2)dx=h(t)h(t)a2h(t)=0etx2dx,x=ytdx=dyt=0ey2.dyt=π2th(t)=kea2tk=π1a.ea2t.da2t=πeu2.duaerf(x)=2π0xet2dt=π2aerf(at)+g(a)=k=(π2aerf(at)+g(a))ea2t=h(t)erf(x)=2π0xet2dth(0)=0dxx2+a2=π2a=g(a)h(t)=π2a(1erf(at))ea2t=g(a,t)t=1(π2a2(1erf(a))2πea2.π2a)ea2+π(1erf(a))ea2erf(a)(π2a2π)+ππ2a2πa=0ex2x2+a2dx=Ψ(a)Ψ(a)=02aex2(x2+a2)2dxΨ(a)2a=0ex2(x2+a2)2dx=erf(a)2a(π2a2π)12a(ππ2a2)+π2a2a2=12=π=0ex2(x2+12)2dxwithethiswecanevaluat0ex2(x2+s2)n,sR,nN

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