Numerical evidence strongly suggests that:
$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, \sigma > 0$$
and
$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot (1-\frac11\mathrm{e})\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, -1 < \sigma < 0$$
and
$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot \left(1-0\,\mathrm{e}\right)\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, -2 < \sigma < -1$$
and
$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot \left(1-\frac12\,\mathrm{e}\right)\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, -3 < \sigma < -2$$
and
$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot \left(1-\frac13\,\mathrm{e}\right)\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, -4 < \sigma < -3$$
and
$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot \left(1-\frac38\,\mathrm{e}\right)\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, -5 < \sigma < -4$$
where the rational factors before $\mathrm{e}$ seem to be OEIS: A053557, i.e.:
$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot \left(1-\mathrm{e}\sum_{k=0}^n \frac{(-1)^k}{k!}\,\right)\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, -(n+1) < \sigma < -n, n \in \mathbb{N}$$
I expect this is already known in the math literature, however would like to understand how this could be derived. Tried many sources on the web e.g. the Wiki and Wolfram pages and also OEIS: A061382 however without success. Does anyone know how this is done or would be able to provide a link to a proof?
P.S.: A well-known and maybe related result is:
$$\int_{-\infty}^{\infty}\frac{\cos(x)}{(x^2+1)} \,\mathrm{d}x = 2\cdot\frac{\pi}{\mathrm{e}}$$