Question Number 33705 by math khazana by abdo last updated on 22/Apr/18
$${let}\:\:\alpha>\mathrm{0}\:\:{find}\:{the}\:{fourier}\:{transform}\:{of} \\ $$$${f}\left({t}\right)\:=\:{e}^{−{a}^{\mathrm{2}} {t}^{\mathrm{2}} } \\ $$
Commented by math khazana by abdo last updated on 22/Apr/18
$${F}\left({f}\left({x}\right)\right)=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\:\int_{−\infty} ^{+\infty} \:\:{f}\left({t}\right)\:{e}^{−{ixt}} {dt}\:\:. \\ $$
Answered by sma3l2996 last updated on 22/Apr/18
$${F}\left({f}\left({t}\right)\right)\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\int_{−\infty} ^{\infty} {e}^{−{a}^{\mathrm{2}} {t}^{\mathrm{2}} −{ixt}} {dt}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\int_{−\infty} ^{+\infty} {e}^{−\left({a}^{\mathrm{2}} {t}^{\mathrm{2}} +{ixt}\right)} {dx} \\ $$$${a}^{\mathrm{2}} {t}^{\mathrm{2}} +{ixt}={a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{2}×{at}×\frac{{ix}}{\mathrm{2}{a}}+\left(\frac{{ix}}{\mathrm{2}{a}}\right)^{\mathrm{2}} −\left(\frac{{ix}}{\mathrm{2}{a}}\right)^{\mathrm{2}} =\left({at}+\frac{{ix}}{\mathrm{2}{a}}\right)^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$${let}\:\:{u}={at}+\frac{{ix}}{\mathrm{2}{a}}\Rightarrow{dt}=\frac{{du}}{{a}} \\ $$$${F}\left({f}\left({t}\right)\right)\left({x}\right)=\frac{\mathrm{1}}{{a}\sqrt{\mathrm{2}\pi}}\int_{−\infty} ^{\infty} {e}^{−{u}^{\mathrm{2}} −\frac{{x}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }} {du}=\frac{{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }} }{{a}\sqrt{\mathrm{2}\pi}}\int_{−\infty} ^{+\infty} {e}^{−{u}^{\mathrm{2}} } {du} \\ $$$${we}\:{know}\:{that}\:\:\int_{−\infty} ^{+\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\sqrt{\pi} \\ $$$${so} \\ $$$${F}\left({f}\left({t}\right)\right)\left({x}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}{a}}{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }} \\ $$