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: $\frac{1}{2}e^{-x}$
Best Answer
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 $$ F^{-1}\left(\frac{1}{1+s^2}\right) = \sqrt{\frac{\pi}{2}}e^{-\left | x \right |}\, . $$