Q:

Inverse Fourier transform of a partial fraction decomposition?

For the function
$$\alpha(\omega)=\frac{R}{i\omega-\lambda}+\frac{R^*}{i\omega-\lambda^*},$$
where $R$ and $\lambda$ are both complex numbers, What is the simplest way to obtain the inverse Fourier transform (not going into Laplace transform):
$$h(t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} \alpha(\omega)\exp(i\omega t)\,\mathrm{d}\omega.$$
(The solution should be $h(t)=R\exp(\lambda t)+R^*\exp(\lambda^*t)$.)
and then go back to the frequency domain with
$$\alpha(\omega)=\int_{-\infty}^{+\infty} h(t)\exp(-i\omega t)\,\mathrm{d}t.$$

A:

$\def\i{\mathrm{i}}\def\e{\mathrm{e}}\def\d{\mathrm{d}}\def\Re{\mathop{\mathrm{Re}}}$First, for any $t > 0$, $r > |λ|$, by the residue theorem,$$
\int_{-r}^r \frac{\e^{\i zt}}{z + \i λ} \,\d z + \int\limits_{γ_r} \frac{\e^{\i zt}}{z + \i λ} \,\d z = \begin{cases}
2π\i · \e^{\i zt} \Biggr|_{z = -\i λ} = 2π\i \e^{λt}; & \Re(λ) < 0\\
0; & \Re(λ) > 0
\end{cases}
$$
where $γ_r = \{z = r\e^{\i θ} \mid 0 \leqslant θ \leqslant π\}$ with counterclockwise orientation. Note that$$
\lim_{|z| \to +\infty} \left| \frac{1}{z + \i λ} \right| = 0,
$$
making $r \to +\infty$ and by Jordan's lemma,$$
\int_{-\infty}^{+\infty} \frac{\e^{\i zt}}{z + \i λ} \,\d z = \begin{cases}
2π\i \e^{λt}; & \Re(λ) < 0\\
0; & \Re(λ) > 0
\end{cases}
$$
Analogously,$$
\int_{-\infty}^{+\infty} \frac{\e^{\i zt}}{z - \i λ} \,\d z = \begin{cases}
2π\i \e^{-λt}; & \Re(λ) > 0\\
0; & \Re(λ) < 0
\end{cases}
$$
Case 1: $\Re(λ) > 0$, then $\Re(\overline{λ}) = \Re(λ) > 0$. For $t > 0$,\begin{align*}
\int_{-\infty}^{+\infty} α(ω) \e^{\i ωt} \,\d ω &= \int_{-\infty}^{+\infty} \frac{R \e^{\i ωt}}{\i ω - λ} \,\d ω + \int_{-\infty}^{+\infty} \frac{\overline{R} \e^{\i ωt}}{\i ω - \overline{λ}} \,\d ω\\
&= -\i R \int_{-\infty}^{+\infty} \frac{\e^{\i zt}}{z + \i λ} \,\d z - \i \overline{R} \int_{-\infty}^{+\infty} \frac{\e^{\i zt}}{z + \i \overline{λ}} \,\d z\\
&= -\i R · 0 - \i \overline{R} · 0 = 0.
\end{align*}
For $t < 0$,\begin{align*}
\int_{-\infty}^{+\infty} α(ω) \e^{\i ωt} \,\d ω &= \int_{-\infty}^{+\infty} \frac{R \e^{\i ωt}}{\i ω - λ} \,\d ω + \int_{-\infty}^{+\infty} \frac{\overline{R} \e^{\i ωt}}{\i ω - \overline{λ}} \,\d ω\\
&= \int_{-\infty}^{+\infty} \frac{R \e^{\i ω'(-t)}}{-\i ω' - λ} \,\d ω' + \int_{-\infty}^{+\infty} \frac{\overline{R} \e^{\i ω'(-t)}}{-\i ω' - \overline{λ}} \,\d ω'\\
&= \i R \int_{-\infty}^{+\infty} \frac{\e^{\i z(-t)}}{z - \i λ} \,\d z + \i \overline{R} \int_{-\infty}^{+\infty} \frac{\e^{\i z(-t)}}{z - \i \overline{λ}} \,\d z\\
&= \i R · 2π\i \e^{-λ(-t)} + \i \overline{R} · 2π\i \e^{-\overline{λ}(-t)} = -2π (R \e^{λt} + \overline{R} \e^{\overline{λ}t}).
\end{align*}
Therfore,$$
h(t) = \frac{1}{2π} \int_{-\infty}^{+\infty} α(ω) \e^{\i ωt} \,\d ω = \begin{cases}
0; & t > 0\\
-(R \e^{λt} + \overline{R} \e^{\overline{λ}t}); & t < 0
\end{cases}
$$
Case 2: $\Re(λ) < 0$. Analogous calculation shows that$$
h(t) = \frac{1}{2π} \int_{-\infty}^{+\infty} α(ω) \e^{\i ωt} \,\d ω = \begin{cases}
R \e^{λt} + \overline{R} \e^{\overline{λ}t}; & t > 0\\
0; & t < 0
\end{cases}
$$

Now, if $\Re(λ) > 0$, then for any $ω \in \mathbb{R}$, $M > 0$,$$
\int_{-M}^0 \e^{λt} \e^{-\i ωt} \,\d t = \left. \frac{\e^{(λ - \i ω)t}}{λ - \i ω} \right|_{-M}^0 = -\frac{1}{\i ω - λ} (1 - \e^{-(λ - \i ω)M}).
$$
Note that $\Re(λ) > 0$ and $$
|\e^{-(λ - \i ω)M}| = \exp(\Re(-(λ - \i ω)M)) = \exp(-M \Re(λ)),
$$
making $M \to +\infty$,$$
\int_{-\infty}^0 \e^{λt} \e^{-\i ωt} \,\d t = -\frac{1}{\i ω - λ}.
$$
Analogously, if $\Re(λ) < 0$, then$$
\int_0^{+\infty} \e^{λt} \e^{-\i ωt} \,\d t = \frac{1}{\i ω - λ}.
$$
Case 1: $\Re(λ) > 0$, then $\Re(\overline{λ}) = \Re(λ) > 0$. For any $ω \in \mathbb{R}$,\begin{align*}
\int_{-\infty}^{+\infty} h(t) \e^{-\i ωt} \,\d t &= -\int_{-\infty}^0 (R \e^{λt} + \overline{R} \e^{\overline{λ}t}) \e^{-\i ωt} \,\d t\\
&= -R \int_0^{+\infty} \e^{λt} \e^{-\i ωt} \,\d t - \overline{R} \int_0^{+\infty} \e^{\overline{λ}t} \e^{-\i ωt} \,\d t\\
&= \frac{R}{\i ω - λ} + \frac{\overline{R}}{\i ω - \overline{λ}} = α(ω).
\end{align*}
Case 2: $\Re(λ) < 0$. Analogous calculation shows that$$
\int_{-\infty}^{+\infty} h(t) \e^{-\i ωt} \,\d t = α(ω).
$$

