3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Is there a typo in the second exponential inside the integral? If $x \in \mathbb{R}^n$, then what does the argument mean? $\endgroup$ May 3 '13 at 22:29
  • 1
    $\begingroup$ Presumably it's a dot product. $\endgroup$ May 3 '13 at 22:32
  • $\begingroup$ This is Proposition 8.24 in Folland's Real Analysis $\endgroup$ May 3 '13 at 22:34
  • $\begingroup$ Do you know how to do this in $R^2$? $\endgroup$ May 3 '13 at 22:45
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.