Inverse Fourier transform of $ \frac1{a+\mathrm{j}w} $

Question

The Fourier transform of $f(t)$ is defined as $$F(\mathrm{j}w)=\int_{-\infty}^{\infty}f(t)\,\mathrm{e}^{-\mathrm{j}wt}\,\mathrm{d}t,$$ while the corresponding inverse Fourier transform is defined as $$f(t)=\frac1{2\pi}\int_{-\infty}^{\infty}F(\mathrm{j}w)\,\mathrm{e}^{\mathrm{j}wt}\,\mathrm{d}w.$$

Let $f(t)=\mathrm{e}^{-at}h(t)$, $a>0$, where $h(t)$ is the Heaviside function and $a$ is a real constant.

Fourier transform of this function is $$F(\mathrm{j}w)=\int_0^\infty f(t)\,\mathrm{e}^{-\mathrm{j}wt}\,\mathrm{d}t=\int_0^\infty\mathrm{e}^{-at}\mathrm{e}^{-\mathrm{j}wt}\,\mathrm{d}t=\frac1{a+\mathrm{j}w}.$$ How can I calculate the inverse Fourier transform of $\dfrac1{a+\mathrm{j}w}$, $$f(t)=\frac1{2\pi}\int_{-\infty}^\infty\frac1{a+\mathrm{j}w}\mathrm{e}^{\mathrm{j}wt}\,\mathrm{d}w\,?$$ Although $\dfrac1{a+\mathrm{j}w}$ doesn't look complicated, there is no way I can calculate this integral. Generally, I have problems calculating inverse FTs. Any suggestions? Thanks in advance.

Answer 1

METHOD $1$:

We can proceed to evaluate the integral if we invoke Generalized Functions. To that end, we write the inverse Fourier Transform representation for $f$ as

$$f(t)=\frac{1}{2\pi}\int_{- \infty}^{ \infty}\frac{e^{j\omega t}}{a+j\omega}\,d\omega \tag 1$$

Then, note that the derivative $f'$ is given by

$$f'(t)=\frac{1}{2\pi}\int_{- \infty}^{ \infty}\frac{j\omega\,e^{j\omega t}}{a+j\omega}\,d\omega \tag 2$$

Adding $(2)$ and $a$ times $(1)$ reveals that

$$\begin{align} f'(t) +af(t)&=\frac{1}{2\pi}\int_{- \infty}^{ \infty}e^{j\omega t}\,d\omega \\\\ &=\delta(t) \tag 3 \end{align}$$

where $\delta(t)$ is the Dirac Delta Distribution. Solution to the ODE expressed in $(3)$ is

$$f(t)=e^{-at}h(t)$$

as was to be shown!

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METHOD $2$:

We begin by writing the integral representation for $f(t)$ as the Cauchy Principal Value

$$f(t)=\frac{1}{2\pi}\lim_{R\to \infty}\int_{-R}^{ R}\frac{e^{j\omega t}}{a+j\omega}\,d\omega \tag 4$$

Note that we need not evaluate the integral in terms of its Cauchy Principal Value. Inasmuch as the improper integral converges, its Cauchy Principal Value converges to the same value.

Now, we define a new function $f_R(t)$ as

$$f_R(t)=\oint_C \frac{e^{jzt}}{a+jz}\,dz$$

where the closed contour $C$ is comprised of the real line segment from $z=-R$ to $z=+R$ and for $t>0$ ($t<0$) the semicircle $C_R$ of radius $R$ in the upper (lower) half plane.

Jordan's Lemma guarantees that as $R\to \infty$, the contribution of the integral from the integration over $C_R$ goes to zero. Therefore, we have $f(t)=\lim_{R\to \infty}f_R(t)$.

Now, we note that since $a>0$ that the only singularity of $\frac{e^{jzt}}{a+jz}$ is a $z=ja$. Thus, from the Residue Theorem we have

$$f(t)= \begin{cases} 2\pi i\text{Res}\left(\frac{e^{jzt}}{a+jz},z=ja\right)=e^{-at}&, t>0\\\\ 0&,t<0 \end{cases}$$

as expected!