Inverse fourier transform of $1/(1+s^2)$

Hoi, I want to have the inverse fourier transform $\mathcal{F}^{-1}(\frac{1}{1+s^2})$.

So I thought about using some properties of fourier-transform. But knowing the answer I must make some sort of mistakes in my reasoning, but i dont understand what im doing wrong:

I know the answer is : $$ce^{-|x|}$$ and according to wolframalpha $c= \sqrt{\pi/2}$.

But i got this: Some calculations give:

$$\frac{1}{1+s^2} = \frac{1}{1-is}\cdot \frac{1}{1 +is} = \mathcal{F}(H(t)e^{-t})\cdot \mathcal{F}(H(-t)e^{t}) = \mathcal{F}[H(t)e^{-t}\ast H(-t)e^{t}]$$

That is according some properties of Fourer transform: $F(g \ast f) = F(g)F(f)$

So that would then imply the answer is $$H(t)e^{-t}\ast H(-t)e^{t}$$

But that doesnt give me the right answer...what is my big error. I get calculating this convolution: $\frac{1}{2}e^{-x}$

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Is it $\frac{1}{1+s^2}$ or $\frac{1}{(1+s)^2}$? –  Mhenni Benghorbal Apr 1 '13 at 13:34
$1/(1+s^2)$. Sorry. Otherwise none of what I said wouldve made sense :P –  DinkyDoe Apr 1 '13 at 13:47
You can use residue theorem. –  Mhenni Benghorbal Apr 1 '13 at 18:56
for what exactly do i use it then. In line of the reasoning above I wouldnt need it, so what is my error? –  DinkyDoe Apr 1 '13 at 19:30
The answer should be $\frac{1}{2}e^{-t}H(t) + \frac{1}{2}e^{t}H(-t).$ –  Mhenni Benghorbal Apr 2 '13 at 14:15

In your calculation of the convolution I'm sure that at some point you get a $\sqrt{x}^2$, and you're saying that's equal to $x$, when it's actually $|x|$, that is what that answer is telling you. It's a very common mistake, check you're calculations agains to see if that is the problem.

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ok, do u see a problem with my reasoning here then? Cause im almost sure the convolution itself is the right answer. But see under here, what mistake do I make then... –  DinkyDoe Apr 1 '13 at 12:38

There should be a mistake in your computation of convolution, because the map $t\mapsto H(t)e^{-t}\star H(-t)e^{t}$ is even (convolution is commutative).

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Im sorry: meant to say $H(t)e^{-t}\star H(-t)e^{t}$ –  DinkyDoe Apr 1 '13 at 12:28
So, there should be a mistake. I got that. But is my reasoning correct to conclude that $H(t)e^{-t}\star H(-t)e^{t}$ is the right answer? And if so...see under here. What am I misunderstanding here? –  DinkyDoe Apr 1 '13 at 12:57

Ok, thanks for your feed-back but my computation goes as follows...

and i dont see the mistake:

\begin{align*}\int_{\mathbb{R}}H(\tau)e^{-\tau}H(\tau-x)e^{(x-\tau)}d\tau &= \int_{\mathbb{R}}H(\tau)H(\tau-x)e^{(x-2\tau)}d\tau \\ & = \int_{x}^{\infty}e^{(x-2\tau)}d\tau \\ &=\frac{1}{2}e^{-x} \end{align*}

So where am i going wrong? (Also that constant $c$ doesnt even look like the $\sqrt{\pi/2}$)

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