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Statement

\begin{equation} \int_{\mathbb R} \exp \left( -2\pi (\frac{x}{\sqrt{2}})^2 \right) \exp\left( -i2 \pi \frac {x}{\sqrt{2}} \cdot f \right) dx = \exp \left( -\pi f^{2} / 2 \right) \end{equation}

My attempt based on J.G.'s tips

Call the integral I(f).

\begin{equation} I(f) = \int_{\mathbb R} \left( -\pi x^{2} - i \sqrt{2} \pi x f \right) dx \end{equation}

Integration by parts: $dv = dx$, $u = I(f)$, $v = x$, and $du = \delta I / \delta f$ so

\begin{equation} \int_{\mathbb R} u dv = I(f) \cdot x - \int_{\mathbb R} x \frac{\delta I}{\delta f} \end{equation}

How can you get the second power in the final result in the integration?

Léo Léopold Hertz 준영
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2 Answers2

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We are computing here the Fourier transform of a Gaussian. It is a basic fact of Fourier theory that the result is again a Gaussian, appearing as a function of the "frequency variable". This can be proven using residue calculus, or applying Green's theorem to a suitable setup in ${\mathbb R}^2$. In this way one obtains, e.g., $$\int_{-\infty}^\infty e^{-x^2/2}\>\cos(\lambda x)\>dx=\sqrt{2\pi}\>e^{-\lambda^2/2}\ .$$ The rest is symmetry and scaling variables.

Christian Blatter
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Call this integral $I\left( f\right)$ so $I\left( 0\right)=\int_{\mathbb{R}}\exp\left(-\pi x^2\right)=1$ and $$\frac{\partial I}{\partial f}=-i\pi\sqrt{2}\int_\mathbb{R}x \exp\left(-\pi x^2-i\pi\sqrt{2}fx\right)dx.$$Rewrite $I,\,\frac{\partial I}{\partial f}$ using $x = y - \frac{if}{\sqrt{2}}$ to prove that $\frac{\partial I}{\partial f} = -\pi fI$. (You'll need to use integration by parts. If you get stuck, try emulating the method of the top answer here.) Thus $I = \exp\left(-\tfrac{\pi f^2}{2}\right)$.

Community
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J.G.
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  • I am thinking if $I'(f) = f I(f)$ here. How are you getting $I'(f) = f \int x (orig) dx$, particularly the new $x$ inside the integral on the right-hand-side? – Léo Léopold Hertz 준영 Feb 05 '16 at 11:41
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    @Masi You must have a sign error somewhere. I recommend you write $I,,\frac{\partial I}{\partial f}$ as real numbers, as my link does. (In both cases the imaginary part is an odd function that integrates to 0.) Then $\frac{\partial I}{\partial f}$ is the integral of $uv'$ where, up to multiplicative constants, $u = \sin \pi\sqrt{2}fx,,v = \exp\left(-\pi x^2\right)$. – J.G. Feb 05 '16 at 11:47