The Question:

(i) Determine the Fourier Transform of

$$f(x) = \frac{1}{a^2+x^2} \qquad a>0$$

(ii) Hence determine the Fourier Transform of

$$g(x) = \frac{1}{x^2+2x+2}$$

My Attempt:

(i) I got $\hat f(s) = \frac \pi a e^{-a|s|}$ which should be correct

(ii) Observe that

$$g(x) = \frac{1}{x^2+2x+2} = \frac{1}{(x+1)^2+1} \implies g(x-1)=\frac{1}{x^2+1^2}$$

So, using the previous part of the question with $a=1$, we get

$$\hat g(s-1) = \pi e^{-|s|} \implies \hat g(s) = \pi e^{-|s+1|}$$

But, according to Wolfram Alpha Fourier Transform Calculator, the correct answer should be

$$\hat g(s) = \pi e^{-s(1+i)}\big(e^{2s}H(-s)+H(s)\big)$$

where $H$ is the Heaviside Step Function.

Can someone explain to me what I have done wrong and/or the correct way to tackle this question? Thanks.


Let $f$ be a nice function on $\Bbb R$, and let $g(x)=f(x+1)$. Then $$\sqrt{2\pi}\,\hat g(s)=\int_{-\infty}^\infty g(x)e^{isx}\,dx =\int_{-\infty}^\infty f(x+1)e^{isx}\,dx =\int_{-\infty}^\infty f(x)e^{is(x-1)}\,dx=e^{-isx}\sqrt{2\pi}\,\hat f(s).$$ In your example, $\hat f(s)=\pi e^{-|s|}$, so that $\hat g(s)=\pi e^{-is}e^{-s}$ for $s>0$ and $\hat g(s)=\pi e^{-is}e^s$ for $s<0$. I think this is what Wolfie is telling you.


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.