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I have no understanding of getting an Inverse Fourier Transform. I also if somebody has material of solving Inverse Fourier Transform would be great.

The problem is: $displaystyle F = frac1{w^2 + 9}$

One way to do this is to set up the inverse Fourier transform formula:

$$f(t)=frac{1}{2pi} int_{-infty}^infty F(omega)e^{iomega t}domega=frac{1}{2pi} int_{-infty}^infty frac{e^{iomega t}}{omega^2+9}domega$$

However, this integration turns out to be rather difficult. Therefore, we instead use a table of Fourier transforms, which tells us:

$$f(t)=e^{-a|t|} iff F(omega)=frac{2a}{omega^2+a^2}$$

Now, for $a=3$, this table matches the form of our $F(omega)$ very closely:

$$f(t)=e^{-3|t|} iff F(omega)=frac{6}{omega^2+9}$$

At this point, we just need to get rid of the $6$ somehow. Since the Fourier transform is linear, this means if we divide $F(omega)$ by $6$, then $f(t)$ also gets divided by $6$. Thus, we have:

$$f(t)=frac 1 6e^{-3|t|} iff F(omega)=frac{1}{omega^2+9}$$

Therefore, the inverse Fourier transform of $frac{1}{omega^2+9}$ is $frac 1 6 e^{-3|t|}$.