1

I am trying to take the inverse Fourier transform of $$F(t) = \int_{-\infty}^\infty \frac{e^{i\omega t}}{\omega - i\epsilon} \; d\omega$$ where $\epsilon$ is a very small positive number.

I know the definition of a Fourier transform of a function $f$ on $\mathbb{R}$ is defined by $$\hat{f} (\xi) = \int_{-\infty}^\infty f(x) e^{-2\pi i x \xi} \; dx $$ for $\xi\in \mathbb{R},$ and its inverse Fourier transform is $$f(x) = \int_{-\infty}^\infty \hat{f}(\xi) e^{2\pi i x \xi} \; d\xi$$ for $x\in \mathbb{R}.$

However, I don't understand how to use this definition to actually compute the inverse Fourier transform. How should I proceed if I want to take the inverse Fourier transform of $F(t)$ defined above?

Mark Viola
  • 179,405

2 Answers2

2

Let $F(t)$ and $\hat F(\omega)$ be Fourier Transform pairs such that

$$\hat F(\omega)=\int_{-\infty}^\infty F(t)e^{-i\omega t}\,dt \tag 1$$

and

$$F(t)=\frac{1}{2\pi}\int_{-\infty}^\infty \hat F(\omega)e^{i\omega t}\,d\omega \tag 2$$

In the problem of interest, $F(t)=\int_{-\infty}^\infty\frac{e^{i\omega t}}{\omega -i\epsilon}\,d\omega$ and therefore $\hat F(\omega)=\frac{2\pi}{\omega -i \epsilon}$.

We can evaluate $F(t)$ using contour integration. We observe that the integrand has a simple pole at $\omega =i\epsilon$ in the upper-half plane.

We let $C_R$ be the contour comprised of the real-line segment $-R$ to $R$ and the semicircular arc centered at the origin with radius $R$ in the upper-half plane.

Then, for $t>0$

$$\oint_{C_R}\frac{e^{i\omega t}}{\omega -i\epsilon}\,d\omega=\int_{-R}^R\frac{e^{i\omega t}}{\omega -i\epsilon}\,d\omega+\int_0^{\pi}\frac{e^{iRe^{i\phi}t}}{Re^{i\phi}-i\epsilon}\,iRe^{i\phi}\,d\phi\tag3$$

The residue theorem guarantees that for $R>\epsilon$

$$\oint_{C_R}\frac{e^{i\omega t}}{\omega -i\epsilon}\,d\omega=2\pi i \text{Res}\left(\frac{e^{i\omega t}}{\omega -i\epsilon}, \omega =i\epsilon\right)=2\pi i e^{-\epsilon t}$$

Moreover, as $R\to \infty$, the integral on the right-hand side of $(3)$ approaches $0$.

Putting it all together yields

$$\int_{-\infty}^\infty\frac{e^{i\omega t}}{\omega -i\epsilon}\,d\omega=2\pi i e^{- \epsilon t}$$

for $t>0$.

For $t<0$, we change the contour so that the semi-circular arc is in the lower-half plane. In this case, there is no singularity and Cauchy's Integral Theorem guarantees that

$$\int_{-\infty}^\infty\frac{e^{i\omega t}}{\omega -i\epsilon}\,d\omega=0$$

for $t<0$.

Therefore,

$$\bbox[5px,border:2px solid #C0A000]{F(t)=\begin{cases}2\pi i e^{-\epsilon t}&,t>0\\\\0&,t<0\end{cases}}$$

Mark Viola
  • 179,405
  • Hi Mark - It's been a long time since I've done these contour integrals so I'm trying to make sense of the following: why can't I choose the return to be in the lower half plane, and then there is no pole so the contour integral is 0? Wouldn't I still get the integral over the return part to be 0? And then wouldn't that mean that I get 0 as the inverse transform? Something is wrong with this argument but I can't find it. – Joel May 22 '22 at 10:13
  • @Joel If one closes in the lower half plane, the exponential increases without bound as the radius of the semi-circle tends to infinity. – Mark Viola May 22 '22 at 13:16
  • thank you - of course this resolves it. – Joel May 23 '22 at 12:25
  • @Joel Interestingly for $t>0$, the integral over the semi-circle in the lower-half plane must converge to $F(t)$ as the radius of the semi-circle approaches infinity. – Mark Viola May 23 '22 at 13:01
1

I thought it might be instructive to present an approach that relies on real analysis only. To that end, we leverage the results in This answer and proceed

Note that we can write $F(t)$ as

$$\begin{align} F(t)&=\int_{-\infty}^\infty \frac{e^{i\omega t}}{\omega -i\epsilon}\,d\omega\\\\ &=\int_{-\infty}^\infty \frac{\omega +i\epsilon}{\omega^2+\epsilon^2}e^{i\omega t}\,d\omega\\\\ &=i\int_{-\infty}^\infty \frac{\omega \sin(\omega t)}{\omega^2+\epsilon^2}\,d\omega+i\epsilon \int_{-\infty}^\infty \frac{\cos(\omega t)}{\omega^2+\epsilon^2}\,d\omega\tag 1 \end{align}$$

In THIS ANSWER, I showed using real analysis only that

$$\int_{0}^\infty \frac{\cos(\omega x)}{x^2+1}\,dx=\frac{\pi}{2}e^{-|\omega|}\tag 2$$

Letting $\omega= \epsilon t$ and then enforcing the substitution $x= \omega/\epsilon$ in $(2)$, and exploiting even symmetry reveals

$$\int_{-\infty}^\infty \frac{\cos(\omega t)}{\omega^2+\epsilon^2}\,d\omega=\frac{\pi}{\epsilon}e^{-\epsilon|t|} \tag 3$$

Owing to the uniform convergence of $\int_{-\infty}^\infty \frac{\omega \sin(\omega t)}{\omega^2+\epsilon^2}\,d\omega$ for $t \ge t_0>0$, we see that

$$\frac{d}{d\omega}\int_{-\infty}^\infty \frac{\cos(\omega t)}{\omega^2+\epsilon^2}\,d\omega=-\int_{-\infty}^\infty \frac{\omega \sin(\omega t)}{\omega^2+\epsilon^2}\,d\omega=-\text{sgn}(t)\pi e^{-\epsilon |t|}\tag 4$$

Finally, using $(3)$ and $(4)$ in $(1)$, we find that

$$\bbox[5px,border:2px solid #C0A000]{F(t)=\begin{cases}2\pi i e^{-\epsilon t}&,t>0\\\\0&,t<0\end{cases}}$$

which agrees with the answer that relied on contour integration!

Mark Viola
  • 179,405