Question Number 4758 by madscientist last updated on 05/Mar/16
$$\int_{−\infty} ^{\infty} {e}^{−{x}^{\mathrm{2}\:} } \:{dx}\:=\:\sqrt{\pi\:} \\ $$$${is}\:{this}\:{true},\:{if}\:{so}\:{how}? \\ $$
Answered by Yozzii last updated on 05/Mar/16
$${Let}\:{f}\left({x}\right)={e}^{−{x}^{\mathrm{2}} } \:\left({x}\in\mathbb{R}\right).\:{Since}\:{f}\left(−{x}\right)={e}^{−\left(−{x}\right)^{\mathrm{2}} } ={e}^{−{x}^{\mathrm{2}} } ={f}\left({x}\right) \\ $$$${then}\:{f}\left({x}\right)\:{is}\:{an}\:{even}\:{function}.\:{In}\:{general}, \\ $$$${for}\:{g}\left({u}\right)\:{being}\:{an}\:{even}\:{function},\:\int_{−{a}} ^{{a}} {g}\left({u}\right){du}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {g}\left({u}\right){du} \\ $$$${if}\:{the}\:{integral}\:{exists}\:{wholly}\:{on}\:{u}\in\left[−{a},{a}\right]. \\ $$$${Now}\:{let}\:{I}\left({a}\right)=\int_{−{a}} ^{{a}} {f}\left({x}\right){dx}.\:{Then},\:{I}\left({a}\right)=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx} \\ $$$${since}\:{f}\left({x}\right)\:{is}\:{even}.\:{Let}\:{us}\:{now}\:{determine} \\ $$$${l}=\underset{{a}\rightarrow\infty} {\mathrm{lim}}{I}\left({a}\right)=\underset{{a}\rightarrow\infty} {\mathrm{lim}2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}.\Rightarrow{l}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}. \\ $$$${Substitute}\:{u}={x}^{\mathrm{2}} \:{where}\:{x}\geqslant\mathrm{0}.\Rightarrow{x}=\sqrt{{u}}. \\ $$$${Also},\:{du}=\mathrm{2}{xdx}=\mathrm{2}\sqrt{{u}}{dx}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}{u}^{−\mathrm{1}/\mathrm{2}} {du}={dx}. \\ $$$${At}\:{x}=\mathrm{0},{u}=\mathrm{0}^{\mathrm{2}} =\mathrm{0}\:{and}\:{as}\:{x}\rightarrow\infty,{x}^{\mathrm{2}} \rightarrow\infty \\ $$$$\Rightarrow{u}\rightarrow\infty.\:{l}\:{hence}\:{becomes}\: \\ $$$${l}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \mathrm{0}.\mathrm{5}{u}^{−\mathrm{1}/\mathrm{2}} {e}^{−{u}} {du}=\int_{\mathrm{0}} ^{\infty} {u}^{−\mathrm{1}/\mathrm{2}} {e}^{−{u}} {du}. \\ $$$${The}\:{gamma}\:{function}\:\Gamma\:{is}\:{given}\:{by}\:{the} \\ $$$${form}\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt}\:\left({x}>\mathrm{0}\right).\:{So}\:{truly} \\ $$$${l}=\Gamma\left(\mathrm{1}/\mathrm{2}\right).\:{From}\:{the}\:{equation}\:{for}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$$\Gamma\left({x}\right)\Gamma\left(\mathrm{1}−{x}\right)=\frac{\pi}{{sin}\pi{x}},\:{let}\:{x}=\mathrm{1}/\mathrm{2}. \\ $$$$\Rightarrow\Gamma^{\mathrm{2}} \left(\mathrm{0}.\mathrm{5}\right)=\frac{\pi}{{sin}\mathrm{0}.\mathrm{5}\pi}=\pi\Rightarrow\Gamma\left(\mathrm{0}.\mathrm{5}\right)=\pm\sqrt{\pi}. \\ $$$${Since}\:{f}\left({x}\right)>\mathrm{0}\:{for}\:{all}\:{x}\geqslant\mathrm{0},\:{the}\:{integral} \\ $$$${ought}\:{to}\:{be}\:{non}−{negative}.\:{Hence},\:\Gamma\left(\mathrm{0}.\mathrm{5}\right)=\sqrt{\pi} \\ $$$${only}.\Rightarrow\:{l}=\sqrt{\pi}.\:{But}\:{l}\:{is}\:{equivalent}\:{to}\:{the} \\ $$$${form}\:\int_{−\infty} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}.\:{So},\:\int_{−\infty} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\sqrt{\pi}.\: \\ $$$${The}\:{mathematical}\:{statement}\:{is}\:{true}. \\ $$$$ \\ $$$$ \\ $$