I would like to calculate the Fourier's inverse transform of the following function
$$F(\omega)=\frac{\omega-ia}{\omega-ib}, \ \ \ \ \ \ (1)$$
where a and b are real and positive.
Hence, I should evaluate the following integral:
$$f(t)=\int_{-\infty}^{+\infty}\frac{\omega-ia}{\omega-ib}e^{i\omega t} d\omega$$.
If a=b, then $f(t)=\delta(t)$.
If $a\neq b$, I can use residue theorem. The function has only one imaginary pole, at $\omega=ib$.
Hence, for $t<0$ I get $f(t)=0$. On the other hand, if $t>0$ I get:
$$f(t)=2\pi \left(a-b\right)e^{-bt}$$
Putting everything together, I have that
$$f(t)=2\pi \left(a-b\right)e^{-bt}\theta(t),$$
where $\theta(t)$ is the Heaviside Function.
However, I read on the books that the Fourier transform of $e^{-\alpha t}\theta(t)$ (with $\alpha$ real and positive) is:
$$\frac{1}{\alpha+i\omega}.$$
But then, according to the latter relation, the Fourier transform of $f(t)=2\pi \left(a-b\right)e^{-bt}$ would read:
$$F(\omega)=\frac{a-b}{b-i\omega}=\frac{ib-ia}{\omega-ib}. \ \ \ \ \ \ (2)$$
The latter expression is different with respect to the relation (1).
Why do I get this difference?
Can you help me?
Thank you very much for your help.
Best Regards.
Note that $F(\omega)=\frac{\omega-ia}{\omega-ib}=1-\frac{i(a-b)}{\omega-ib}$. Then,
$$\mathscr{F^{-1}}\{F\}(t)=\delta(t)-\frac{i(a-b)}{2\pi}\int_{-\infty}^\infty \frac{e^{i\omega t}}{\omega-ib}\,d\omega$$
From the residue theorem
$$\begin{align} \int_{-\infty}^\infty \frac{e^{i\omega t}}{\omega-ib}\,d\omega=\begin{cases}2\pi i e^{-bt}&,t>0\\\\0&,t<0\end{cases} \end{align}$$
Hence, we find that
$$\mathscr{F^{-1}}\{F\}(t)=\delta(t)+(a-b)e^{-bt}H(t)$$
Alternatively, we know that if $G(\omega)=\omega F(\omega)$, then
$$\mathscr{F^{-1}}\{G\}(t)=-i\frac{dg(t)}{dt}$$
where the derivative is in the sense of distributions. Hence, since the derivative of the Heaviside function is the Dirac Delta distribution, we find that
$$\begin{align} \mathscr{F^{-1}}\{F\}(t)&=\left(-i\frac{d}{dt}-ia\right)\left( ie^{-bt}H(t)\right)\\\\ &=\delta(t)+(a-b)e^{-bt}H(t) \end{align}$$
as expected!