Computing an Inverse Fourier Transform

Question

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: $\displaystyle F = \frac1{w^2 + 9}$

Answer 1

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

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

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 $\frac 1 6 e^{-3|t|}$.