Question Number 31105 by abdo imad last updated on 02/Mar/18
$${prove}\:{that}\:\int_{\mathrm{0}} ^{{x}} \:\:\:{e}^{−{t}^{\mathrm{2}} } {dt}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:−\frac{{e}^{−{x}^{\mathrm{2}} } }{\:\sqrt{\pi}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}^{\mathrm{2}} {t}^{\mathrm{2}} } }{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:{with}\:{x}>\mathrm{0} \\ $$
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 .](https://www.tinkutara.com/question/Q31297.png)
$${let}\:{put}\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}^{\mathrm{2}} {t}^{\mathrm{2}} } }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{with}\:{x}>\mathrm{0}\:{after}\:{verifying}\:{the} \\ $$$$\left.{conditions}\:{of}\:{derivality}\:{of}\:{f}\:{on}\:\right]\mathrm{0},+\infty\left[\:{we}\:{get}\right. \\ $$$${f}^{'} \left({x}\right)=\:−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{t}^{\mathrm{2}} {x}\:{e}^{−{x}^{\mathrm{2}} {t}^{\mathrm{2}} } }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\:−\mathrm{2}{x}\:\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} \:{e}^{−{x}^{\mathrm{2}} {t}^{\mathrm{2}} } }{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=−\mathrm{2}{x}\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \:−\mathrm{1}\right){e}^{−{x}^{\mathrm{2}} {t}^{\mathrm{2}} } }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=−\mathrm{2}{x}\:\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} {t}^{\mathrm{2}} } {dt}\:+\mathrm{2}{x}\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{x}^{\mathrm{2}} {t}^{\mathrm{2}} } }{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{2}{x}\:{f}\left({x}\right)−\mathrm{2}{x}\:\int_{\mathrm{0}} ^{\infty} {e}^{−\left({xt}\right)^{\mathrm{2}} } {dt}=\mathrm{2}{xf}\left({x}\right)−\mathrm{2}{x}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}^{\mathrm{2}} } \frac{{du}}{{x}} \\ $$$$=\mathrm{2}{xf}\left({x}\right)−\sqrt{\pi}\:\Rightarrow\:{f}^{'} \left({x}\right)\:−\mathrm{2}{xf}\left({x}\right)=−\sqrt{\pi} \\ $$$${e}.{h}\Rightarrow{f}^{'} \:−\mathrm{2}{xf}=\mathrm{0}\Rightarrow\frac{{f}^{'} }{{f}}=\mathrm{2}{x}\Rightarrow{ln}\mid{f}\mid=\:{x}^{\mathrm{2}} \:\Rightarrow{f}\left({x}\right)={k}\:{e}^{{x}^{\mathrm{2}} } \\ $$$${mvc}\:{method}\Rightarrow{k}^{'} \:{e}^{{x}^{\mathrm{2}} } \:+\:\mathrm{2}{kx}\:{e}^{{x}^{\mathrm{2}} } \:−\mathrm{2}{xk}\:{e}^{{x}^{\mathrm{2}} } =−\sqrt{\pi}\:\Rightarrow \\ $$$${k}^{'} {e}^{{x}^{\mathrm{2}} } \:=−\sqrt{\pi}\:\Rightarrow{k}^{'} =−\sqrt{\pi}\:{e}^{−{x}^{\mathrm{2}} } \:\Rightarrow{k}\left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} −\sqrt{\pi}\:{e}^{−{t}^{\mathrm{2}} } {dt}\:+\lambda \\ $$$$\lambda={k}\left(\mathrm{0}\right)={f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}\:\Rightarrow{k}\left({x}\right)=\frac{\pi}{\mathrm{2}}\:−\sqrt{\pi}\:\int_{\mathrm{0}} ^{{x}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}^{\mathrm{2}} {t}^{\mathrm{2}} } }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\left(\frac{\pi}{\mathrm{2}}\:−\sqrt{\pi}\:\int_{\mathrm{0}} ^{{x}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\right){e}^{{x}^{\mathrm{2}} } \:\Rightarrow \\ $$$${e}^{−{x}^{\mathrm{2}} \:} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}^{\mathrm{2}} {t}^{\mathrm{2}} } }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\:\frac{\pi}{\mathrm{2}}\:−\sqrt{\pi}\:\int_{\mathrm{0}} ^{{x}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:\Rightarrow \\ $$$$\sqrt{\pi}\:\:\int_{\mathrm{0}} ^{{x}} \:\:{e}^{−{t}^{\mathrm{2}} } {dt}=\:\frac{\pi}{\mathrm{2}}\:−\:{e}^{−{x}^{\mathrm{2}} } \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}^{\mathrm{2}} {t}^{\mathrm{2}} } }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\:{e}^{−{t}^{\mathrm{2}} } {dt}=\:\frac{\sqrt{\pi}}{\mathrm{2}}\:−\frac{{e}^{−{x}^{\mathrm{2}} } }{\:\sqrt{\pi}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{x}^{\mathrm{2}} {t}^{\mathrm{2}} } }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:. \\ $$