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: $\frac{1}{2}e^{-x}$
 A: The Solution can be easily obtained by using Leibniz differentiation under integral sign, and obtaining a differential equation. Solving the D.E gives an equation with constants $C_1$ and $C_2$, or here they are equal hence $C$. This $C$ is same as the "c" obtained from wolframalpha ,
$Ce^{-\left | x \right |}$    ,
giving $C= \sqrt{\frac{\pi}{2}}$   ,
hence
$$
F^{-1}\left(\frac{1}{1+s^2}\right) =
\sqrt{\frac{\pi}{2}}e^{-\left | x \right |}\, .
$$
A: 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). 
A: 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.
A: In your Calculation, x shouldn't be the lower border in the integral. It must be H(x)*x. Because, if x < 0, the integral is cut, then the lower border is 0. And in the end it is (x-2H(x)*x) = -|x| ... :)
A: It has been a while, but since no answer was given, I decided to explain the mistake. The problem is indeed with the convolution.
You must take into consideration the cases $x>0$ and $x<0$. For the first case, the convolution has $$H(x)e^{-x}\ast H(-x)e^x =\frac{e^{-x}}{2} = \frac{e^{-|x|}}{2},$$ as an answer, while for the case $x<0$ you have $$H(x)e^{-x}\ast H(-x)e^x =\frac{e^x}{2} = \frac{e^{-(-x)}}{2} = \frac{e^{-|x|}}{2}.$$
