Question-124141

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

Answered by Lordose last updated on 01/Dec/20

\int_{0}^{\infty} e^{- {ix}^{2}} dx

Set u = x \sqrt{i} \Rightarrow du = \sqrt{i} dx

\frac{1}{\sqrt{i}} \int_{0}^{\infty} e^{- u^{2}} du

\frac{1}{\sqrt{i}} \cdot \frac{\sqrt{π}}{2} = - \frac{i \sqrt{i π}}{2}

Answered by Bird last updated on 01/Dec/20

\int_{0}^{\infty} e^{- {i x}^{2}} d x = \int_{0}^{\infty} e^{- {(\sqrt{i} x)}^{2}} d x

=_{\sqrt{i} x = t} \int_{0}^{\infty} e^{- t^{2}} \frac{d t}{\sqrt{i}} = \frac{1}{e^{\frac{i π}{4}}} \times \frac{\sqrt{π}}{2}

= \frac{\sqrt{π}}{2} e^{- \frac{i π}{4}} = \frac{\sqrt{π}}{2} (c o s (\frac{π}{4}) - i s i n (\frac{π}{4}))

= \frac{\sqrt{π}}{2} (\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}) = \frac{\sqrt{2 π}}{4} - i \frac{\sqrt{2 π}}{4}

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