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Question Number 28999 by abdo imad last updated on 03/Feb/18

$$provethat{\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{\sqrt{t}}dt=e^{i\frac{\pi }{4}}{\int }_0^{\mathrm{\infty }}\frac{e^{-ix}}{\sqrt{x}}dx.$$

Commented by abdo imad last updated on 04/Feb/18

$$thech.\sqrt{x}=tgive{\int }_0^{\mathrm{\infty }}\frac{e^{-ix}}{\sqrt{x}}dx={\int }_0^{\mathrm{\infty }}\frac{e^{-{it}^2}}{t}2tdt$$$$=2{\int }_0^{\mathrm{\infty }}{e}^{-{\left(\sqrt{i}t\right)}^2}dtthech.\sqrt{i}t=u$$$$=2{\int }_0^{\mathrm{\infty }}{e}^{-u^2}\frac{du}{\sqrt{i}}(ch.\sqrt{i}t=u)$$$$\frac{2}{\sqrt{i}}\frac{\sqrt{\pi }}{2}=\frac{\sqrt{\pi }}{\sqrt{i}}buti=e^{i\frac{\pi }{2}}\Rightarrow \sqrt{i}=e^{i\frac{\pi }{4}}\Rightarrow$$$$e^{i\frac{\pi }{4}}{\int }_0^{\mathrm{\infty }}\frac{e^{-ix}}{\sqrt{x}}dx=e^{i\frac{\pi }{4}}{e}^{-\frac{\pi }{4}}\sqrt{\pi }=\sqrt{\pi }fromanotherside$$$$thech.\sqrt{t}=ugive{\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{\sqrt{t}}dt={\int }_0^{\mathrm{\infty }}\frac{e^{-u^2}}{u}\left(2u\right)du$$$$=2{\int }_0^{\mathrm{\infty }}{e}^{-u^2}du=2\frac{\sqrt{\pi }}{2}=\sqrt{\pi }so$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-t}}{\sqrt{t}}dt=e^{i\frac{\pi }{4}}{\int }_0^{\mathrm{\infty }}\frac{e^{-ix}}{\sqrt{x}}dx.$$$$$$

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