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According to Wikipedia, the Fourier transform $\hat f$ of an integrable function $f:\mathbb{R}\to\mathbb{C}$ is defined by

$$ \hat f\left(\xi\right)=\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi ix\xi}dx.\tag{1} $$

Consider $f\left(x\right)=e^{-x^2}$. Then, using $\left(1\right)$, I obtain

$$\hat f\left(\xi\right)=e^{-\left(\pi\xi\right)^2}\sqrt\pi.$$

However, WolframAlpha computes

$$ \mathcal F_x\left[e^{-x^2}\right]\left(\omega\right)=\frac{e^{-\frac{\omega^2}{4}}}{\sqrt 2}, $$

the back of my PDEs book yields

$$ \mathcal F\left(e^{-x^2}\right)\left(\xi\right)=\frac{e^{-\frac{\xi^2}{4}}}{2\sqrt\pi}, $$

and my professor asserts

$$ \mathcal F\left(e^{-x^2}\right)\left(\xi\right)=\left(2\pi\right)^\frac12e^{-\frac{\xi^2}2}. $$

What in the world is going on!?

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Indeed, there are many ways to define this transform, with different units for the frequency. It can be for example a frequency in Hz, or an angular speed in [rad/sec], or it can be unit-less for random signals processing to name one example. All these make the normalizing constant and the frequency variable itself to vary. See also the tables here: . Each has several columns. – yoki Nov 28 '12 at 21:59
Unfortunately there is no universally accepted convention for the Fourier transform. I prefer to use the one you have in equation $(1)$ since the inverse Fourier transform is easy to remember since it doesn't have factors in-front of the integral $$f(x) = \int_{\mathbb{R}} \hat{f}(\xi) \exp(2 \pi i \xi x) d \xi$$ – user17762 Nov 28 '12 at 22:01
up vote 2 down vote accepted

In different settings, the Fourier Transform uses different characters, $e^{-\lambda ix\xi}$, with different values of $\lambda$. Normalized so that $\mathcal{F}_\lambda(\mathcal{F}_\lambda(f))(x)=f(-x)$, the Fourier Transform is $$ \mathcal{F}_\lambda(f)(\xi)=\sqrt{\frac{\lambda}{2\pi}}\int_{-\infty}^\infty f(x)\,e^{-\lambda ix\xi}\,\mathrm{d}x\tag{1} $$ Applying $(1)$ to the function $e^{-\alpha x^2}$ yields $$ \begin{align} \mathcal{F}_\lambda\left(e^{-\alpha x^2}\right)(\xi) &=\sqrt{\frac{\lambda}{2\pi}}\int_{-\infty}^\infty e^{-\alpha x^2}\,e^{-\lambda ix\xi}\,\mathrm{d}x\\ &=\sqrt{\frac{\lambda}{2\pi}}\,e^{-\frac{\lambda^2}{4\alpha}\xi^2}\color{#C00000}{\int_{-\infty}^\infty e^{-\alpha(x+\frac{\lambda}{2\alpha} i\xi)^2}\,\mathrm{d}x}\\ &=\sqrt{\frac{\lambda}{2\pi}}\,e^{-\frac{\lambda^2}{4\alpha}\xi^2}\color{#C00000}{\int_{-\infty}^\infty e^{-\alpha x^2}\,\mathrm{d}x}\\ &=\sqrt{\frac\lambda{2\alpha}}\,e^{-\frac{\lambda^2}{4\alpha}\xi^2}\tag{2} \end{align} $$ Where the equality of the red integrals is assured by contour integration over a very wide rectangle.

Usually, either $\lambda=2\pi$, so that $\sqrt{\frac{\lambda}{2\pi}}=1$ in $(1)$, or $\lambda=1$, to simplify the character. Using $(2)$ with $\alpha=1$, we can determine which value of $\lambda$ is used in each case cited in the question.

In the first case, $\lambda=2\pi$ is used: $$ \mathcal{F}_{2\pi}\left(e^{-x^2}\right)(\xi)=\sqrt{\pi}\,e^{-\pi^2\xi^2}\tag{3} $$ Wolfram Alpha seems to use $\lambda=1$: $$ \mathcal{F}_1\left(e^{-x^2}\right)(\xi)=\sqrt{\tfrac12}\,e^{-\xi^2/4}\tag{4} $$ The PDE book uses $\lambda=1$, but seems to have chosen an asymmetric Fourier Transform, where the Fourier Transform takes the factor $\frac1{2\pi}$ instead of $\frac1{\sqrt{2\pi}}$ in $(1)$, and the Inverse Fourier Transform takes the factor $1$ instead of $\frac1{\sqrt{2\pi}}$. Thus, their answer is $(4)$ divided by $\sqrt{2\pi}$.

The form given by your professor would need to use $\lambda=\sqrt{2}$ and a factor of $\frac1{\sqrt{\pi}}$ instead of $\frac1{\sqrt[4]{2\pi^2}}$. My guess is that this is not what was intended. However, it looks somewhat like the pdf for Normal Distribution with variance $1$ and mean $0$.

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Shock horror: Math professors do get the details wrong occassionally. But they usually get the big picture right. – Harald Hanche-Olsen Nov 30 '12 at 23:19
@HaraldHanche-Olsen: true, and students miscopy as well; sometimes, both :-) – robjohn Nov 30 '12 at 23:29

What is going on is that there are different conventions for the placement of $2\pi$. You stick it as a factor in the forward Fourier transform, or the inverse transform, or you use a factor $\sqrt{2\pi}$ in both, or you put it in the exponent, or you heroically try to hide it from view altogether by changing the measure itself, as Rudin famously does in one of his books. The bottom line is, the theory of the Fourier transform needs $2\pi$ somewhere,and there is no universal agreement on where it should go.

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Oh, I see. Thank you. Also, do you happen to know which "version" of the Fourier transform my professor used to find $\left(2\pi\right)^\frac12e^{-\frac{\xi^2}{2}}$? – EsX_Raptor Nov 28 '12 at 22:15
@JosuéMolina: Not off the top of my head, but if you compute it as $a\int_{-\infty}^\infty e^{-x^2}e^{-bi\xi x}\,dx$, you can then easily check afterwards which combinations of $a$ and $b$ match his result. It's certainly a lot easier than repeating the calculation over and over with different choices of $a$ and $b$. – Harald Hanche-Olsen Nov 29 '12 at 9:27

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