Finding inverse Fourier transform of $\frac{1}{(1+iw)^2}$

Question

Find the inverse Fourier transform of the function $$ \frac{1}{(1+iw)^2} $$

So I know the inverse is given by the integral $$ \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \frac{e^{iwx}}{(1+iw)^2}dw $$ but I don't know how to go about solving this.

Answer 1

A good strategy for determining (Inverse) Fourier Transforms is to use Fourier Transform theorems and Fourier Transform table lookups to the fullest extent possible, before attempting integration.

Observe that

$$\dfrac{1}{(1+i\omega)^2} = i\dfrac{d}{d\omega} \dfrac{1}{1+i\omega}$$

and note that the Fourier Transform has a derivative theorem

$$\mathscr{F}\left\{x^n f(x)\right\}=\mathscr{F}\left\{x^n \mathscr{F}^{-1}\left\{ F(\omega)\right\}\right\}= i^n\dfrac{d^n}{d\omega^n} F(\omega)$$

A table lookup of Fourier Transforms yields

$$\mathscr{F}\left\{e^{-ax}H(x)\right\} = \dfrac{1}{\sqrt{2\pi}}\cdot\dfrac{1}{a+i\omega}$$

where $H(x)$ is the Heaviside unit step function. (That forward transform is easy to compute and verify for yourself.)

You can now say that

$$\begin{align*}\mathscr{F}^{-1}\left\{\dfrac{1}{(1+i\omega)^2}\right\} &= \mathscr{F}^{-1}\left\{ i\dfrac{d}{d\omega} \dfrac{1}{1+i\omega}\right\}\\ \\ &= x\mathscr{F}^{-1}\left\{ \dfrac{1}{1+i\omega}\right\}\\ \\ &= \sqrt{2\pi}x\mathscr{F}^{-1}\left\{\dfrac{1}{\sqrt{2\pi}}\cdot \dfrac{1}{1+i\omega}\right\}\\ \\ &= \sqrt{2\pi}xe^{-x}H(x)\\ \end{align*}$$