Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 124141 by nico last updated on 01/Dec/20

Answered by Lordose last updated on 01/Dec/20

$$$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{ix}}^2}\mathrm{dx}$$$$\mathrm{Set}\mathrm{u}=\mathrm{x}\sqrt{\mathrm{i}}\Rightarrow \mathrm{du}=\sqrt{\mathrm{i}}\mathrm{dx}$$$$\frac{1}{\sqrt{\mathrm{i}}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}$$$$\frac{1}{\sqrt{\mathrm{i}}}\cdot \frac{\sqrt{\pi }}{2}=-\frac{\mathrm{i}\sqrt{\mathrm{i}\pi }}{2}$$

Answered by Bird last updated on 01/Dec/20

$${\int }_0^{\mathrm{\infty }}{e}^{-{ix}^2}dx={\int }_0^{\mathrm{\infty }}{e}^{-{\left(\sqrt{i}x\right)}^2}dx$$$${=}_{\sqrt{i}x=t}{\int }_0^{\mathrm{\infty }}{e}^{-t^2}\frac{dt}{\sqrt{i}}=\frac{1}{e^{\frac{i\pi }{4}}}\times \frac{\sqrt{\pi }}{2}$$$$=\frac{\sqrt{\pi }}{2}{e}^{-\frac{i\pi }{4}}=\frac{\sqrt{\pi }}{2}(cos\left(\frac{\pi }{4}\right)-isin\left(\frac{\pi }{4}\right))$$$$=\frac{\sqrt{\pi }}{2}(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})=\frac{\sqrt{2\pi }}{4}-i\frac{\sqrt{2\pi }}{4}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com