# Fourier Transform of $\dfrac{1}{x^2+2x+2}$

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.