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.


1 Answer 1


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 $f(t)=0$.

  • 1
    $\begingroup$ 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. $\endgroup$ Feb 11, 2017 at 18:32
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    $\begingroup$ $$|e^{iax}|= \left|\cos(ax) + i \sin(ax) \right| =\sqrt{\cos^{2}(ax) + \sin^{2}(ax)} = 1$$ 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 $R \to \infty$. $\endgroup$ Feb 11, 2017 at 21:30
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    $\begingroup$ @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. $\endgroup$ Feb 12, 2017 at 3:34
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    $\begingroup$ 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. $\endgroup$ Feb 12, 2017 at 3:34
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    $\begingroup$ @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 $$f(t) = \frac{1}{2 \pi} \left( \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega \color{red}- \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t- \pi)}}{w^{2}-1} \, d \omega \right) = \frac{1}{2\pi} \left(\pi \sin t - (- \pi \sin t) \right) = \sin t.$$ $\endgroup$ Feb 12, 2017 at 19:25

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