Fourier inverse transform of a given integral

Question

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?

Answer 1

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 $\hat F(\omega)=\frac{2\pi}{\omega -i \epsilon}$.

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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,

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

Answer 2

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

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

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

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In THIS ANSWER, I showed using real analysis only that

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

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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

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

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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!