# Calculating the inverse Fourier transform of $\frac{2i \sin(\pi \omega)}{\omega^{2}-1}$ using the definition

I applied the FT to a piecewise function defined as:

$$f(t) = \begin{cases} \sin(t), & \text{-π≤t≤π} \\ 0, & \text{otherwise} \end{cases}$$

and got $F[f(t)]= \frac{2i\sin(\pi\omega)}{\omega^2-1}$

I thought it'd be a interesting to attempt to applying the inverse FT to try to get back to the piecewise function. However I am an engineer and have limited knowledge of complex variables.

When applying the definition of the inverse Fourier transform I get:

$$f(t)= \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{2i\sin(\pi\omega)}{\omega^2-1} e^{i\omega t} dt.$$

Is anyone able to help me do this? or give me a hint of where to start? I have no idea how to choose the contour and which direction I should be closing it.

I assume there is a typo and you meant $$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty}\frac{2i \sin(\pi \omega)}{\omega^{2}-1} \, e^{i t \omega} \, d {\color{red}{\omega}}. \tag{1}$$

And I assume the definition of the Fourier transform you're using is $$\mathcal{F}[f(t)](\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} \, dt.$$

We can express $(1)$ as the difference of two Cauchy principal value integrals.

Specifically, \begin{align} f(t) &= \frac{1}{2\pi} \int_{-\infty}^{\infty}\frac{2i \sin(\pi \omega)}{\omega^{2}-1} \, e^{i t \omega} \, d {\color{red}{\omega}} \\ &= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{e^{i \pi \omega} - e^{- i \pi \omega}}{w^{2}-1} \, e^{i t \omega } \, d \omega \\ &= \frac{1}{2 \pi} \left( \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega - \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t- \pi)}}{w^{2}-1} \, d \omega \right). \end{align}

If $a \ge 0$, the magnitude of the function $e^{iaz}$ is bounded in the upper half-plane.

And if $a \le 0$, the magnitude of $e^{iaz}$ is bounded in the lower half-plane.

So if ${\color{red}{t \ge - \pi}}$, we can evaluate the first integral by integrating the function $$h(z) = \frac{e^{i z(t+ \pi)}}{z^{2}-1}$$ around a closed semicircular contour in the upper half-plane that has small half-circle indentations around the simple poles at $z=-1$ and $z=1$.

Letting the radii of the indentations go to zero and the radius of the big arc go to infinity, we get $$\operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega - i \pi \, \text{Res} [h(z), -1] - \, i \pi \, \text{Res} [h(z), 1] =0,$$

which implies that \begin{align} \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega &= i \pi \, \text{Res} [h(z), 1] + \, i \pi \, \text{Res} [h(z), -1] \\ &= i \pi \left( \frac{e^{i (t+ \pi)}}{2} - \frac{e^{-i (t+ \pi)}}{2} \right) \\ &= - \pi \sin (t+ \pi) \\ &=\pi \sin t. \end{align}

I used fact that if a function $f(z)$ has a simple pole at $z_{0}$, then $$\lim_{r \to 0} \int_{C_{r}} f(z) \, dz = i \alpha \, \text{Res}[f(z), z_{0}],$$ where $C_{r}$ is an arc of the circle $|z- z_{0}|=r$ of angle $\alpha$. (This is sometimes referred to as the fractional residue theorem.)

And if ${\color{red}{ t \le \pi}}$, we can evaluate the second integral by integrating the function $$g(z) = \frac{e^{i z(t- \pi)}}{z^{2}-1}$$ around a similar contour in the lower half-plane.

This causes us to detour around the poles in the opposite direction (i.e., counterclockwise.)

We end up with \begin{align} \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t- \pi)}}{w^{2}-1} \, d \omega &= {\color{red}{-}}i \pi \, \text{Res} [g(z), 1] {\color{red}{-}}i \pi \, \text{Res} [g(z), -1] \\ &= -i \pi \left( \frac{e^{i (t- \pi)}}{2} - \frac{e^{-i (t- \pi)}}{2} \right) \\ &= \pi \sin (t- \pi) \\ &=-\pi \sin t. \end{align}

So if ${\color{red}{|t| \le \pi}}$, $$\frac{1}{2\pi} \int_{-\infty}^{\infty}\frac{2i \sin(\pi \omega)}{\omega^{2}-1} \, e^{i t \omega} \, d \omega= \frac{1}{2 \pi} \left(\pi \sin t + \pi \sin t \right) = \sin t.$$

If $t< - \pi$, both $h(z)$ and $g(z)$ have to be integrated in the lower-half plane.

And if $t > \pi$, both $h(z)$ and $g(z)$ have to be integrated in the upper-half plane.

In both cases, this results in $f(t)=0$.

• Alternatively, you could just use Jordan's lemma. And it doesn't really matter in which direction you go around the contour. If we went in the opposite direction, we would get $$\operatorname{PV} \int_{\infty}^{-\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega + i \pi \, \text{Res} [f(z), -1] + \, i \pi \, \text{Res} [f(z), 1] =0,$$ which leads to the same result. Feb 11, 2017 at 18:32
• $$|e^{iax}|= \left|\cos(ax) + i \sin(ax) \right| =\sqrt{\cos^{2}(ax) + \sin^{2}(ax)} = 1$$ And if $a \ge 0$, then along a big arc of radius $R$ in the upper half-plane , we have $$\left|\frac{e^{iaz}}{z^{2}-1}\right| \le \frac{1}{\left|{(Re^{it})^{2}-1}\right|} \le \frac{1}{\left||R^{2}e^{2it}|-|-1| \right|} = \frac{1}{R^{2}-1},$$ where $0 \le t \le \pi$. (The second inequality is just the reverse triangle inequality.) Combine this with the fact that the length of the big arc is $\pi R$, and then apply the estimation lemma to show that the integral along the big arc vanishes as $R \to \infty$. Feb 11, 2017 at 21:30
• @JoeyWheeler The simple poles, which are on the real axis, are not inside the contour. We're avoiding them with small half-circle indentations, and then letting the radii of these half-circles go to zero. The general picture is something like THIS, except, in our case, there are only two poles on the real axis and no poles inside the contour. Feb 12, 2017 at 3:34
• As I mentioned in my answer, there is a theorem about this that is sometimes called the fractional residue theorem. The reason it's $i \pi$ and not $2 \pi i$ is because they're half-circles and not entire circles. Feb 12, 2017 at 3:34
• @JoeyWheeler If $|t| \le \pi$, the first PV integral evaluates to $\pi \sin t$ and the second PV integral evaluates to $- \pi \sin t$. But $$f(t) = \frac{1}{2 \pi} \left( \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega \color{red}- \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t- \pi)}}{w^{2}-1} \, d \omega \right) = \frac{1}{2\pi} \left(\pi \sin t - (- \pi \sin t) \right) = \sin t.$$ Feb 12, 2017 at 19:25