Finding the anti-Fourier’s transform (problem with integrals) I am having problems finding the integrals when calculating this anti-Fourier's transform.
$$ 
F(\omega) = 4(\omega ^2- 2\omega  +5)^{-1} 
$$
Then,
$$ f(t) = \frac{1}{j2 \pi} \left(\int_{-\infty }^{\infty } \dfrac{\mathrm{e}^{j \omega t}}{ \omega -2j-1} d\omega-\int_{-\infty }^{\infty } \dfrac{\mathrm{e}^{j \omega t}}{\omega +2j-1} d\omega\right)$$
How do we find this integral:
$$
\int_{-\infty }^{\infty } \dfrac{\mathrm{e}^{j \omega  t}}{\omega -2j-1} d\omega
$$
I tried to use some online calculator but I don't know why it uses some 'weird' u-substitution:
$u=jwt+j⋅(2j−1)w, dt=-j/w du$
Then,
$$e^{-j(2j-1)w}\int_{-\infty }^{\infty } \frac{e^{u}}{u} du$$
 A: Let's simplify things a bit.  Let $F(\omega)=\frac{4}{\omega^2-2\omega+5}=\frac{4}{(\omega-1)^2+4}$ and $f(t)$ be given by
$$\begin{align}
f(t)&=\mathscr{F}^{-1}\{F\}(t)\\\\
&=\frac2\pi \int_{-\infty}^\infty \frac1{(\omega-1)^2+4}e^{i\omega t}\,d\omega\\\\
&=\frac{2}{\pi}e^{it}\int_{-\infty}^\infty \frac{1}{\omega^2+4}e^{i\omega t}\,d\omega\tag1
\end{align}$$
Now the analysis boils down to evaluation of the integral on the right-hand side of $(1)$.  This can be accomplished in a number of ways.

METHODOLOGY $1$:  REAL ANALYSIS
Let $g(t)=\int_{-\infty}^\infty \frac{1}{\omega^2+4}e^{i\omega t}\,d\omega=2\int_0^\infty \frac{1}{\omega^2+4}\cos(\omega t)\,d\omega$.  One can justify differentiating under the integral to find that
$$\begin{align}
g'(t)&=2\int_0^\infty \frac{\omega}{\omega^2+4}\sin(\omega t)\,d\omega\\\\
&=2\int_0^\infty \frac{\omega^2+4-4}{\omega(\omega^2+4)}\sin(\omega t)\,d\omega\\\\
&=2\int_0^\infty \frac{\sin(\omega t)}{\omega}\,d\omega-8\int_0^\infty \frac{\sin(\omega t)}{\omega(\omega^2+4)}\,d\omega\\\\
&=\pi \text{sgn}(t)-8\int_0^\infty \int_0^t \frac{\cos(\omega t')}{\omega^2+4}\,dt'\,d\omega\tag2
\end{align}$$
For $t\ne 0$, we can differentiate $(2)$ to find that $g''(t)+4g(t)=0$, with $g(0)=\pi/2$ and $g'(0)=0$.  Therefore, we find that
$$g(t)=\frac\pi2e^{-2|t|}\tag3$$
Using $(3)$ in $(1)$ reveals
$$f(t)=e^{-2|t|+it}$$


METHODOLOGY $2$:  COMPLEX ANALYSIS
We can evaluate the integral on the right-hand side of $(1)$ using the residue theorem.  Proceeding, we have
$$\begin{align}
\int_{-\infty}^\infty \frac{e^{i\omega t}}{\omega^2 +4}\,d\omega&=2\pi i \text{Res}\left(\frac{e^{i\omega t}}{\omega^2 +4}, \omega =2i\text{sgn}(t)\right)\\\\
&=2\pi i \frac{e^{-2|t|}}{4i}\\\\
&=\frac\pi 2 e^{-2|t|}\tag4
\end{align}$$
Using $(4)$ in $(1)$ yields
$$f(t)=e^{-2|t|+it}$$
as expected!


METHODOLOGY $3$:  DISTRIBUTION THEORY
Let $h_-(t)=\int_{-\infty}^\infty \frac{e^{i\omega t}}{\omega-1- i2}\,d\omega$.  In distribution, we have
$$\begin{align}
h_-'(t)&=\int_{-\infty}^\infty \frac{i\omega e^{i\omega t}}{\omega-1- i2}\,d\omega\\\\
&=i\underbrace{\int_{-\infty}^\infty e^{i\omega t}\,d\omega}_{=2\pi \delta(t)} +i(1+ i2)h(t)
\end{align}$$
where $\delta$ is the Dirac Delta.  Therefore, in distribution we find that $h_-'(t) -(-2+i)h_-(t)=i2\pi \delta(t)$. Since $\lim_{t\to \pm \infty}h_-(t)=0$, we find that
$$h_-(t)=i2\pi  e^{- 2t+it}u(t)$$
Similarly, if $h_+(t)=\int_{-\infty}^\infty \frac{e^{i\omega t}}{\omega-1+ i2}\,d\omega$, we find that
$$h_+(t)=-i2\pi e^{2t+it}u(-t)$$
where $h_-(t)-h_+(t)=i2\pi e^{-2|t|+it}$.
Multiplying by $\frac{1}{i2\pi}$ recovers the expected result!
