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I am reading a script and I found the following statement:

$$ \int_{-\infty}^\infty \frac{1}{\sigma \sqrt{2\pi}} \exp \left(\frac{-x^2}{2 \sigma^2 }\right) \exp(i \, x\, \xi) \,dx = \exp\left(-\frac{1}{2} \xi^2 \sigma^2\right) $$

I tried to check it with wolframalpha for $\sigma = \xi = 1$ but it can't solve the left-hand integral. Is this statement even true? I have the feeling that something important is missing.

Michael Hardy
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Adam
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1 Answers1

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$$\int_{-\infty}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{{\frac{-x^{2}}{2\sigma^{2}}}}e^{ix\xi}dx=\int_{-\infty}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{\frac{-1}{2\sigma^{2}}(x^{2}-2ix\xi\sigma^{2})}dx=\int_{-\infty}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{\frac{-1}{2\sigma^{2}}((x-i\xi\sigma^{2})^{2})+\xi^{2}\sigma^{4}}dx$$

$$=e^{\frac{-1}{2}\xi^{2}\sigma^{2}}\int_{-\infty}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{\frac{-1}{2\sigma^{2}}(x-i\xi\sigma^{2})^{2}}dx$$

You can verify through a contour integral estimate that:

$$\int_{-\infty}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{\frac{-1}{2\sigma^{2}}(x-i\xi\sigma^{2})^{2}}dx=\int_{-\infty}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{\frac{-1}{2\sigma^{2}}x^{2}}dx$$

$$=\int_{-\infty}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{-(\frac{x}{\sigma\sqrt{2}})^{2}}dx=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty} e^{-x^{2}}dx=1.$$

user71352
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