For contour integrals with integration limits $0$ and $\infty$, it is an elementary to trick to plug in a $\ln z$; i.e. we instead consider the contour integral:
$$\oint_C \frac{z\ln z}{(z-\gamma)(z^2+\Omega^2)(e^z-1)}dz$$
where $C$ is the contour as shown in the image:
I will elaborate.
Let $$f(z)= \frac{z}{(z-\gamma)(z^2+\Omega^2)(e^z-1)}dz$$ and let $J$ be your integral.
$$I_2+I_4=\int^\infty_0 f(t)\ln t dt$$
$$I_8+I_6=\int^0_\infty f(t)\ln(te^{2\pi i})dt=\int^0_\infty f(t)\ln t+f(t)\cdot 2\pi i dt$$
Then, $$I_2+I_4+I_6+I_8=-2\pi i\int^\infty_0 f(t)dt =-2\pi i J$$
Moreover, $I_1$ and $I_5$ vanish in the limit.
Also,
$$I_3=-\frac12\cdot 2\pi i\cdot\ln\gamma \lim_{z\to \gamma}f(z)(z-\gamma)$$ The negative sign is due to the clockwise direction of $I_3$ and because $I_3$ is a semicircle so there is a $\frac12$ factor.
Similarly,
$$I_7=-\frac12\cdot2\pi i\cdot(\ln\gamma+2\pi i)\lim_{z\to\gamma}f(z)(z-\gamma)$$
Just to sum things up a little bit, we have
$$-2\pi iJ=2\pi i\sum\text{Res} -I_3-I_7$$
The calculation of the residue is quite straightforward.
At $+ i\Omega$, the residue is
$$(\ln\Omega+\frac{i\pi}2)\lim_{z\to i\Omega}f(z)(z-i\Omega)$$
At $-i\Omega$, the residue is
$$(\ln\Omega+\frac{3\pi i}2)\lim_{z\to -i\Omega}f(z)(z+i\Omega)$$
At $2\pi in$ where $n>0$, the residue is
$$(\ln 2\pi n+\frac{i\pi}2)\lim_{z\to2\pi in}f(z)(z-2\pi i n)$$
At $2\pi in$ where $n<0$, the residue is
$$(\ln 2\pi n+\frac{3i\pi}2)\lim_{z\to2\pi in}f(z)(z-2\pi i n)$$
Note that $J$ is real, so considering only the real part of the residue is sufficient.
Hint.
As $\hat f(s) = \frac{1}{1+e^s}$ the residues for $\hat f(s)$ are located at $1+e^s = 0$ or over the imaginary axis at $(2k+1)\pi i,\ \ k\in \mathbb{Z}$ and $Res[\hat f((2k+1)\pi i)] = -1$ then
$$
\hat f(s) =-\lim_{n\to \infty}\sum_{k=0}^{k=n}\frac{2s}{s^2+(2k+1)^2\pi^2}
$$
also
$$
f(t) = -2\lim_{n\to\infty}\sum_{k=0}^{k=n}\cos\left((2k+1)\pi t\right)
$$
Follows the graphics for $\int_0^T f(t)dt$ with $n = 20$
Best Answer
I found my own method that doesn't involve complex analysis or Fourier transforms. Just a bit of trickery with commuting derivatives and limits with integrals, which could actually be justified with convergence theorems from measure theory that I will neglect:
First, convert the integral to a different form:
$\begin{align} I(x)&=\operatorname{PV}\int_{-\infty}^{+\infty}\frac{e^{-t^2}}{t-x}dt\\\\ &=\lim_{\epsilon\to 0}\left(\int_{-\infty}^{x-\epsilon}\frac{e^{-t^2}}{t-x}dt +\int_{x+\epsilon}^{\infty}\frac{e^{-t^2}}{t-x}dt\right)\\\\ &=\lim_{\epsilon\to 0}\left(-\int_{\epsilon}^{\infty}\frac{e^{-(t-x)^2}}{t}dt +\int_{\epsilon}^{\infty}\frac{e^{-(t+x)^2}}{t}dt\right)\\\\ &=\int_0^\infty\frac{-e^{-(t-x)^2}+e^{-(t+x)^2}}{t}dt\\\\ &=-2e^{-x^2}\int_0^\infty\frac{\sinh(2xt)}{t}e^{-t^2}dt\\\\ \end{align}$
Now the remaining integral can be solved by differentiating under the integral sign:
$\begin{align} J(x)&=\int_0^\infty\frac{\sinh(2xt)}{t}e^{-t^2}dt\\\\ J'(x)&=2\int_0^\infty\cosh(2xt)e^{-t^2}dt\\\\ &=\sqrt\pi\,e^{x^2} \end{align}$
where the previous step comes from expanding the $\cosh$ as exponentials and completing the squares to obtain simple Gaussians. Then we have:
$\begin{align} J(x)&=J(0)+\int_0^x J'(t)dt\\\\ &=0+\sqrt\pi\int_0^xe^{t^2}dt\\\\ &=\frac{\pi}{2}\operatorname{erfi}(x) \end{align}$
Finally we have
$\begin{align} I(x)&=-\pi e^{-x^2}\operatorname{erfi}(x)\\\\ &=-2\sqrt\pi F(x) \end{align}$