Question Number 28694 by abdo imad last updated on 29/Jan/18
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}^{\mathrm{2}} } {cosx}\:{dx}\:\:{with}\:{t}>\mathrm{0}\:. \\ $$
Commented by abdo imad last updated on 29/Jan/18
$${let}\:{put}\:\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{tx}^{\mathrm{2}} } \:{cosx}\:{dx} \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{tx}^{\mathrm{2}} +{ix}} {dx}\:{because}\:\:\int_{−\infty} ^{+\infty} \:{e}^{−{tx}^{\mathrm{2}} } {sinxdx}=\mathrm{0}\:{but} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{e}^{−\left(\left(\sqrt{{t}}{x}\right)^{\mathrm{2}} \:−\mathrm{2}\sqrt{{t}}{x}\frac{{i}}{\mathrm{2}\sqrt{{t}}}\:\:+\left(\frac{{i}}{\mathrm{2}\sqrt{{t}}}\right)^{\mathrm{2}} \:−\left(\frac{{i}}{\mathrm{2}\sqrt{{t}}}\right)^{\mathrm{2}} \right)} {dx} \\ $$$$=\:\int_{−\:\infty} ^{+\infty} \:\:{e}^{−\left(\sqrt{{t}}{x}\:−\frac{{i}}{\mathrm{2}\sqrt{{t}}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}{t}}} {dx}\:\:\:\:\:\:\:\:\left({ch}.\sqrt{{t}}{x}\:−\frac{{i}}{\mathrm{2}\sqrt{{t}}}={u}\right) \\ $$$$=\:{e}^{−\frac{\mathrm{1}}{\mathrm{4}{t}}} \:\:\int_{−\infty} ^{+\infty} \:\:\:{e}^{−{u}^{\mathrm{2}} } \:\:\frac{{du}}{\:\sqrt{{t}}}=\frac{\sqrt{\pi}}{\:\sqrt{{t}}}\:{e}^{−\frac{\mathrm{1}}{\mathrm{4}{t}}} \:\:\:\:\left(\:\:\int_{−\infty} ^{+\infty} \:{e}^{−{u}^{\mathrm{2}} } {du}=\sqrt{\pi}\right){so} \\ $$$${I}=\:\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{{t}}}\:\:{e}^{−\frac{\mathrm{1}}{\mathrm{4}{t}}} \:. \\ $$