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$\mathrm{Re}(i) = 0$, but the fourier transform of $f(x) = e^{-ix^2}$ is $g(\alpha) = \sqrt{\pi\over i}\times e^{i\alpha^2 \over 4}$, is it not?

Is there an easy to show that it is so, knowing the fourier transform of $h(x) = e^{-x^2}$? (It is $k(\alpha) = \sqrt\pi\times e^{-\alpha^2 \over 4}$.)

The non-unitary, angular frequency Fourier transform is taken into consideration.

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Is there any textbook available online, for free, that would include this problem? –  Student Jan 23 '13 at 21:16
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2 Answers

up vote 7 down vote accepted

Not sure of a textbook, but I can tell you that a simple way to show that

$$\int_{-\infty}^{\infty} dx \: e^{-i x^2} = \sqrt{\frac{\pi}{i}} = \sqrt{\pi} e^{-i \frac{\pi}{4}} $$

is to consider the integral

$$\oint_{C_R} dz \: e^{-z^2} $$

where $C_R$ consists of the interval $[0,R]$ along the $\Re{z}$ axis, a circular arc of radius $R$ centered at the origin, with endpoints at $(R,0)$ and $(R,R)/\sqrt{2}$, and the line segment from $(R,R)/\sqrt{2}$ to the origin. Note that there are no poles inside of $C_R$, for any value of $R$. Then take the limit as $R \rightarrow \infty$ and note that the integral along the circular arc vanishes. Apply Cauchy's Integral Theorem, and the desired result is shown.

Note that this analysis applies to the Fourier transform of such a function as well, as all the transform piece does is shift the center of the quadratic in the exponential. The piece that is the transform is factored outside of the integral and doesn't change this analysis.

EDIT

To be explicit, we can write

$$\begin{align} \oint_{C_R} dz \: e^{-z^2} &= 0 \\ &= \int_0^R dx \: e^{-x^2} + i R \int_0^{\pi/4} d \phi e^{i \phi} e^{-R^2 \exp{(i 2 \phi)}} + e^{i \pi/4} \int_R^0 dt \: e^{-i t^2} \end{align} $$

The 2nd integral vanishes as $R \rightarrow \infty$ because the exponential term in the exponent does not change sign within the integration region. We may then conclude that

$$\int_0^{\infty} dx \: e^{-x^2} = \sqrt{i} \int_0^{\infty} dt e^{-i t^2}$$

or

$$\int_{-\infty}^{\infty} dt \: e^{-i t^2} = \sqrt{\frac{1}{i}} \int_{-\infty}^{\infty} dx \: e^{-x^2} = \sqrt{\frac{\pi}{i}} $$

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I'm not sure that I follow. I need to calculate $\int_{-\infty}^{\infty} e^{-i x^2}e^{-i\omega x} dx = \int_{-\infty}^{\infty} e^{-i x^2}\cos(\omega x)dx$ since $\sin(x)$ is uneven. Now $\int_{-\infty}^{\infty} dx \: e^{-i x^2} = \int_{-\infty}^{\infty} e^{-i x^2}(1-2ix) dx = \sqrt{\pi\over i} $ may be true, but so what? –  Student Jan 24 '13 at 11:22
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Complete the square in the first integral. –  Ron Gordon Jan 24 '13 at 11:44
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I meant $-i (x^2 + \omega x ) = -i (x + \omega/2)^2 + i \omega^2/4$ –  Ron Gordon Jan 25 '13 at 20:29
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Now let $x \leftarrow x + \omega/2$. This integral turns out to be the integral over the line segment from $(R,R)/\sqrt{2}$ to the origin, and when added to the integral of $e^{-x^2}$ over the real axis, equals zero by Cauchy's integral theorem, as I stated in the answer. I'll clarify. –  Ron Gordon Jan 25 '13 at 20:47
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The integrands in both integrals are even, so I essentially multiplied both sides of the equation by 2. –  Ron Gordon Jan 31 '13 at 21:55
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The Fourier transform of $f(x) = \exp(-ix^2)$ doesn't exist in the usual sense. Note that $f \notin L^1$.

However, the function $f$ is bounded and continuous and, as such, can be identified with a tempered distribution, so $\hat f$ exists, also as a tempered distribution. Working out the details, you do in fact end up with

$$\hat f(\omega) = \sqrt{\frac{\pi}{2}} (1-i) e^{i\omega^2/4}$$

which is the same as what you have written, with a suitable interpretation of $\sqrt{i}$. The approach outlined by rlgordonma is probably the easiest way to get to the result.

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You are right that $f\neq L^1$, but still the Fourier transform exists as an improper integral (no distribution theory needed). –  Fabian Jan 23 '13 at 22:29
    
@Fabian Yes, but it feels wrong to talk about Fourier transforms in terms of Riemann integrals. The required integrals don't exist as Lebesgue integrals. (The approach suggested by rlgordonma that I endorsed is distribution theory free.) –  mrf Jan 23 '13 at 22:36
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