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Question Number 31105 by abdo imad last updated on 02/Mar/18
prove that ∫_0 ^x e^(−t^2 ) dt =((√π)/2) −(e^(−x^2 ) /( (√π))) ∫_0 ^∞ (e^(−x^2 t^2 ) /(1+t^2 )) dt with x>0
provethat∫0xe−t2dt=π2−e−x2π∫0∞e−x2t21+t2dtwithx>0
Commented by abdo imad last updated on 05/Mar/18
let put f(x)= ∫_0 ^∞ (e^(−x^2 t^2 ) /(1+t^2 ))dt with x>0 after verifying the conditions of derivality of f on ]0,+∞[ we get f^′ (x)= −∫_0 ^∞ ((2t^2 x e^(−x^2 t^2 ) )/(1+t^2 ))dt= −2x ∫_0 ^∞ ((t^2 e^(−x^2 t^2 ) )/(1+t^2 ))dt =−2x∫_0 ^∞ (((1+t^2 −1)e^(−x^2 t^2 ) )/(1+t^2 ))dt=−2x ∫_0 ^∞ e^(−x^2 t^2 ) dt +2x∫_0 ^∞ (e^(−x^2 t^2 ) /(1+t^2 ))dt =2x f(x)−2x ∫_0 ^∞ e^(−(xt)^2 ) dt=2xf(x)−2x ∫_0 ^∞ e^(−u^2 ) (du/x) =2xf(x)−(√π) ⇒ f^′ (x) −2xf(x)=−(√π) e.h⇒f^′ −2xf=0⇒(f^′ /f)=2x⇒ln∣f∣= x^2 ⇒f(x)=k e^x^2 mvc method⇒k^′ e^x^2 + 2kx e^x^2 −2xk e^x^2 =−(√π) ⇒ k^′ e^x^2 =−(√π) ⇒k^′ =−(√π) e^(−x^2 ) ⇒k(x)= ∫_0 ^x −(√π) e^(−t^2 ) dt +λ λ=k(0)=f(0)=(π/2) ⇒k(x)=(π/2) −(√π) ∫_0 ^x e^(−t^2 ) dt ⇒ ∫_0 ^∞ (e^(−x^2 t^2 ) /(1+t^2 ))dt=((π/2) −(√π) ∫_0 ^x e^(−t^2 ) dt)e^x^2 ⇒ e^(−x^2 ) ∫_0 ^∞ (e^(−x^2 t^2 ) /(1+t^2 ))dt= (π/2) −(√π) ∫_0 ^x e^(−t^2 ) dt ⇒ (√π) ∫_0 ^x e^(−t^2 ) dt= (π/2) − e^(−x^2 ) ∫_0 ^∞ (e^(−x^2 t^2 ) /(1+t^2 ))dt ⇒ ∫_0 ^x e^(−t^2 ) dt= ((√π)/2) −(e^(−x^2 ) /( (√π))) ∫_0 ^∞ (e^(−x^2 t^2 ) /(1+t^2 ))dt .
letputf(x)=∫0∞e−x2t21+t2dtwithx>0afterverifyingtheconditionsofderivalityoffon]0,+∞[wegetf′(x)=−∫0∞2t2xe−x2t21+t2dt=−2x∫0∞t2e−x2t21+t2dt=−2x∫0∞(1+t2−1)e−x2t21+t2dt=−2x∫0∞e−x2t2dt+2x∫0∞e−x2t21+t2dt=2xf(x)−2x∫0∞e−(xt)2dt=2xf(x)−2x∫0∞e−u2dux=2xf(x)−π⇒f′(x)−2xf(x)=−πe.h⇒f′−2xf=0⇒f′f=2x⇒ln∣f∣=x2⇒f(x)=kex2mvcmethod⇒k′ex2+2kxex2−2xkex2=−π⇒k′ex2=−π⇒k′=−πe−x2⇒k(x)=∫0x−πe−t2dt+λλ=k(0)=f(0)=π2⇒k(x)=π2−π∫0xe−t2dt⇒∫0∞e−x2t21+t2dt=(π2−π∫0xe−t2dt)ex2⇒e−x2∫0∞e−x2t21+t2dt=π2−π∫0xe−t2dt⇒π∫0xe−t2dt=π2−e−x2∫0∞e−x2t21+t2dt⇒∫0xe−t2dt=π2−e−x2π∫0∞e−x2t21+t2dt.

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