# Fourier Transform calculation

I am trying to calculate the Fourier Transform of $$f(x)=\exp(-\frac{|x|^2}{2}).$$ Thus, I am looking at the integral $$\hat{f}(u)=\int_{\mathbb{R}^n} \exp(-\frac{|x|^2}{2}) \cdot \exp(ix\cdot u) dx.$$ I can't figure out how to evaluate this integral. Am I trying the wrong approach to calculate the transform or should I be able the integral. Note the integral is a Lebesgue integral.

Thanks.

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Is there a typo in the second exponential inside the integral? If $x \in \mathbb{R}^n$, then what does the argument mean? – Sammy Black May 3 '13 at 22:29
Presumably it's a dot product. – Sharkos May 3 '13 at 22:32
This is Proposition 8.24 in Folland's Real Analysis – Corey Harris May 3 '13 at 22:34
Do you know how to do this in $R^2$? – Mhenni Benghorbal May 3 '13 at 22:45

Just see the special case first in $R^3$, that's $x=(x_1,x_2,x_3)$ and $u=(u_1,u_2,u_3)$, then you can handle the general case. So, we have

$$f(u_1,u_2,u_3)= \int_{{R}^3} e^{-|x|^2}e^{-ix.u}dx$$

$$=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x_1^2+x_2^2+x_3^2)}e^{-i(x_1u_1+x_2u_2+x_3 u_3)}dx_1dx_2dx_3$$

$$= \int_{-\infty}^{\infty}e^{-(x_1^2+ix_1u_1)}dx_1\int_{-\infty}^{\infty}e^{-(x_2^2+ix_2u_2)}dx_2 \int_{-\infty}^{\infty}e^{-(x_3^2+ix_3u_3)}dx_3$$

$$= \prod_{k=1}^{3}\int_{-\infty}^{\infty}e^{-(x_k^2+ix_ku_k)}dx_k = \prod_{k=1}^{3}\sqrt{\pi}e^{-\frac{1}{4}u_k^2 } = {\pi}^{\frac{3}{2}}e^{-\frac{1}{4}(u_1^2+u_2^2+u_3^2)}= {\pi}^{\frac{3}{2}}e^{-\frac{1}{4}|u|^2}.$$

Now, you can figure out the general case easily. Note that, for evaluating the above integrals, we first competed the square then use the Gaussian integral.

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Hint: $\exp(x)\exp(y)=\exp(x+y)$, and complete the square.

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