Question Number 28999 by abdo imad last updated on 03/Feb/18
$${prove}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} }{\:\sqrt{{t}}}{dt}=\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{ix}} }{\:\sqrt{{x}}}{dx}. \\ $$
Commented by abdo imad last updated on 04/Feb/18
$$\:{the}\:{ch}.\:\sqrt{{x}}={t}\:{give}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{ix}} }{\:\sqrt{{x}}}{dx}=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{it}^{\mathrm{2}} } }{{t}}\:\mathrm{2}{tdt} \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\sqrt{{i}}{t}\right)^{\mathrm{2}} } {dt}\:\:\:{the}\:{ch}.\sqrt{{i}}{t}={u} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{u}^{\mathrm{2}} } \:\frac{{du}}{\:\sqrt{{i}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\left({ch}.\sqrt{{i}}{t}={u}\right) \\ $$$$\:\frac{\mathrm{2}}{\:\sqrt{{i}}}\:\frac{\sqrt{\pi}}{\mathrm{2}}=\:\frac{\sqrt{\pi}}{\:\sqrt{{i}}}\:\:\:\:\:{but}\:\:\:\:\:\:\:{i}={e}^{{i}\frac{\pi}{\mathrm{2}}} \Rightarrow\sqrt{{i}}=\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{ix}} }{\:\sqrt{{x}}}{dx}=\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:{e}^{−\frac{\pi}{\mathrm{4}}} \:\:\:\:\sqrt{\pi}=\sqrt{\pi}\:\:\:\:{from}\:{another}\:{side} \\ $$$${the}\:{ch}.\:\sqrt{{t}}={u}\:{give}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} }{\:\sqrt{{t}}}{dt}=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{u}^{\mathrm{2}} } }{{u}}\:\left(\mathrm{2}{u}\right){du} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}^{\mathrm{2}} } {du}=\mathrm{2}\:\frac{\sqrt{\pi}}{\mathrm{2}}=\sqrt{\pi}\:\:\:\:\:\:\:{so} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} }{\:\sqrt{{t}}}{dt}=\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{ix}} }{\:\sqrt{{x}}}{dx}. \\ $$$$ \\ $$