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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.

Américo Tavares
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superAnnoyingUser
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2 Answers2

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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

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

Ron Gordon
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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?

     – superAnnoyingUser 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
  • $1-2ix-x^2+x^2 = (1-ix)^2$ and $-ix^2-iwx = -i(x+{w^2 \over 2}) + {iw^2 \over 4}$ – superAnnoyingUser Jan 25 '13 at 20:27
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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
  • So $\int_{-\infty}^{\infty} e^{-i x^2}e^{-i\omega x} dx = e^{iw^2/4}\int_{-\infty}^{\infty} e^{-i (x+w/2)^2}d(x+w/2)$ but the integral you sugested is still not in the picture. – superAnnoyingUser Jan 25 '13 at 20:45
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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
  • Thank you sir, I should have gotten it the first time. – superAnnoyingUser Jan 25 '13 at 22:10
  • One final question. We may conclude that

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

    but how does it follow that

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

     – superAnnoyingUser Jan 31 '13 at 20:33
  • $\sqrt{i} = e^{i \pi/4}$ (within a branch), so multiply both sides by $e^{-i \pi/4}$ (use same branch) which turns out to be $\sqrt{1/i}$. – Ron Gordon Jan 31 '13 at 20:51
  • I meant to ask how you change the lower bound in the definite integral from zero to negative infinity. How come $$\int_{-\infty}^0 dx : e^{-x^2} = \sqrt{i} \int_{-\infty}^0 dt e^{-i t^2}$$ – superAnnoyingUser Jan 31 '13 at 21:54
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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
  • This is about the integral over the arc in question. It is parametrized as $\gamma_R:z=Re^{i\phi}$ then $$\int_{\gamma_R} dze^{-z^2}= R \int_0^{\pi/4} d \phi e^{i \phi} e^{-R^2 \exp{(i 2 \phi)}} \not= i R \int_0^{\pi/4} d \phi e^{i \phi} e^{-R^2 \exp{(i 2 \phi)}}$$ Isn't that correct? – superAnnoyingUser Feb 01 '13 at 14:27
  • And one more thing. If you look at my first comment I say that $\int_{-\infty}^\infty dxe^-x^2 = \int_{-\infty}^\infty e^{-i x^2} (1-2 i x) dx = \int_{-\infty}^\infty e^-x^2 dx$ when actually $\int_{-\infty}^\infty dxe{^-x^2} = \int_{-\infty}^\infty e^{-i x^2} (1-2 i x) dx \not= \int_{-\infty}^\infty e^{-x^2} dx$ – superAnnoyingUser Feb 01 '13 at 14:45
  • @George: no; $dz = i R e^{i \phi}$. – Ron Gordon Feb 01 '13 at 14:46
  • But don't you have $\int_\gamma df(x) = \int_a^bdf(\gamma (t))$ since $$ \int_{\gamma} f^{'} (x)dx = \int_a^b f^{'} (\gamma (t))d\gamma (t)$$ – superAnnoyingUser Feb 01 '13 at 15:11
  • I think I'll have to consider the integral $\oint_{C_R} e^{-{1 \over 2 }z^2}dz$ insted, since the last question still stands. It may prove easyer to transform to the desired $$\int_{-\infty}^{\infty} e^{-ix^2} dx = \sqrt{\pi} e^{-i \frac{\pi}{4}} $$ Yet how come $\int_{-\infty}^{\infty} dx: e^{-i x^2} = \int_{-\infty}^{\infty} e^{-ix^2} dx$ – superAnnoyingUser Feb 02 '13 at 17:23
  • I'm sorry, man, but when you say $\int dx : f(x)$, do you in fact mean $\int f(x) dx$?

    Because $$df(x)=f'(x)dx$$

    hence $$dx f(x)=f(x)+xf'(x)dx$$

    and therefore $$\int dx : f(x) = \int f(x)+xf'(x))dx$$ which is where the misunderstanding might stem from.

     – superAnnoyingUser Feb 06 '13 at 17:36
  • @rlgordonma. I don't see why the 2nd integral approaches 0 as R goes to infinity (b/c when $\phi=\pi/4$ the magnitude of the integrand is 1). How do we fix that? – TCL Feb 17 '13 at 05:41
  • @TCL: even if we were to lose that exponential decay at just the endpoint of the integration region, I think the integral would still vanish by Riemann-Lebesgue. – Ron Gordon Feb 17 '13 at 08:20
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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.

mrf
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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
  • @mrf Note that there is a notion of improper Lebesgue integral of a function in $L^1_{\text{loc}}$ (in $\mathbf R^n$, not in an arbitrary measure space), and it’s required if you want to integrate things like $\sin x/x$. Amusingly, I first encountered it while reading about improper integrals of distributions (R. Wawak, Studia Math. 86 (1987), 205–220). – Alex Shpilkin Jan 28 '18 at 10:13