$1.$ Note that $\operatorname{Im}(\log \Gamma(z)) =\arg \Gamma(z)$.
$2.$ The general asymptotic expansion of the logarithm of the gamma function is
$$\tag{1}
\log \Gamma (z+h) \sim \left( {z+h - \frac{1}{2}} \right)\log z - z + \frac{1}{2}\log (2\pi ) + \sum\limits_{k = 2}^\infty {\frac{{B_{k}(h) }}{{k(k - 1)z^{k - 1} }}} ,
$$
as $z\to \infty$ in $|\arg z|\leq \pi -\delta$ ($<\pi$) with any fixed complex $h$. Here $B_k(h)$ denotes the Bernoulli polynomials. Taking $h=x$ and $z=\text{i}y$ we find
$$\tag{2}
\arg \Gamma (x+\text{i}y) \sim y\log y-y +\left(x-\frac{1}{2}\right)\!\frac{\pi }{2} + \sum\limits_{k = 1}^\infty {\frac{{( - 1)^k B_{2k}(x) }}{{2k(2k - 1)y^{2k - 1} }}} ,
$$
as $y\to +\infty$ with fixed real $x$. Note that by the Schwarz reflection principle $\arg \Gamma (x-\text{i}y)=-\arg \Gamma (x+\text{i}y)$.
I shall add that there is a Stokes phenomenon happening on the imaginary axis for the logarithm of the Gamma function. This brings in exponentially small corrections to $(1)$, namely
\begin{align*}
\log \Gamma (z+h) \sim \left( {z+h - \frac{1}{2}} \right)\log z - z + \frac{1}{2}\log (2\pi ) & - \log (1 - {\rm e}^{\pm 2\pi {\rm i}(z + h)} )\\ & + \sum\limits_{k = 2}^\infty {\frac{{B_{k}(h) }}{{k(k - 1)z^{k - 1} }}} ,
\end{align*}
as $z\to \infty$ in $\frac{\pi}{2}<\pm \arg z \leq \pi -\delta$ ($<\pi$) with any fixed complex $h$. Thus, $(2)$ is most useful when $x\geq 0$, whereas if $x<0$, one may use
$$
\arg \Gamma (x+\text{i}y) \sim y\log y-y +\left(x-\frac{1}{2}\right)\!\frac{\pi }{2} +\! \sum\limits_{n = 1}^\infty {\frac{{\mathrm{e}^{ - 2\pi ny} }}{n}\sin (2\pi nx)} +\! \sum\limits_{k = 1}^\infty {\frac{{( - 1)^k B_{2k}(x) }}{{2k(2k - 1)y^{2k - 1} }}} ,
$$
as $y\to +\infty$.
For $\Re(s) >0$, the Mellin transform of $e^{-x}$ is $\Gamma(s)$.
Therefore, for any value of $\sigma >0$, the Mellin inversion theorem states $$e^{-x} = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma+ i \infty} x^{-s} \, \Gamma(s) \, \mathrm ds = \frac{1}{2 \pi} \int_{-\infty}^{\infty} x^{-(\sigma +it)} \, \Gamma(\sigma +it) \, \mathrm dt.$$
Letting $x=1$, we get $$ e^{-1} = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \Gamma(\sigma +it) \, \mathrm dt , \quad \sigma >0. $$
Now suppose that $f(x) = \sum_{k=0}^{\infty} \phi(k) (-x)^{k} $ is a function such that $\phi$ satisfies the conditions of Ramanujan's master theorem and
$$\int_{0}^{\infty} x^{s-1} \sum_{k=0}^{\infty} \phi(k) (-x)^{k} \mathrm dx = \frac{\pi}{\sin (\pi s)} \, \phi(-s)$$ for $0 < \Re(s) < \delta$.
Then for a positive integer $N$, $$\int_{0}^{\infty} x^{s-1} \sum_{k=N}^{\infty} \phi(k) (-x)^{k} \, \mathrm dx= \frac{\pi}{\sin (\pi s)} \, \phi(-s) $$ for $-N < \Re(s) < -N+1$.
This is Theorem 8.1 in the paper Ramanujan's Master Theorem by Amdeberhan, Espinosa, et al.
The proof given in the paper is to apply Ramanujan's master theorem to the function $\sum_{k=0}^{\infty} \phi(k+N) (-x)^{k}$ and then shift the parameter $s$.
So for $-1 < \Re(s) <0$, the Mellin transform of $$e^{-x}-1= \sum_{n=1}^{\infty} \frac{1}{\Gamma(n+1)} (-x)^{n}$$ is also $\Gamma(s)$.
Therefore, for any value of $\sigma$ between $-1$ and $0$, the Mellin inversion theorem states $$ e^{-x}-1= \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma+ i \infty} x^{-s} \, \Gamma(s) \, \mathrm ds = \frac{1}{2 \pi} \int_{-\infty}^{\infty} x^{-(\sigma +it)} \, \Gamma(\sigma +it) \, \mathrm dt. $$
Letting $x=1$, we have $$e^{-1}-1 = \frac{1}{2\pi} \int_{-\infty}^{\infty} \Gamma(\sigma +it) \, \mathrm dt, \quad -1 < \sigma <0. $$
Similarly, $$e^{-x}-1 + x \bigg|_{x=1} =e^{-1}= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \Gamma(\sigma +it) \, \mathrm dt, \quad -2 < \sigma <-1, $$
$$e^{-x}-1 + x - \frac{x^{2}}{2} \bigg|_{x=1} =e^{-1} - \frac{1}{2}= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \Gamma(\sigma +it) \, \mathrm dt, \quad -3 < \sigma <-2, $$
and so on.
Best Answer
Can be converted to hypegeometric ... \begin{align} \int_0^\infty \!{{{\rm e}^{-t}}{\frac {1}{\sqrt [3]{t}}}}\,{\rm d}t &=\Gamma\left(\frac23\right) \qquad\text{[real]} \\ \int_{-1/3}^0 \!{{{\rm e}^{-t}}{\frac {1}{\sqrt [3]{t}}}}\,{\rm d}t &= -{\frac { \left( i{3}^{{5/6}}-\sqrt [3]{3} \right) {{\rm e}^{{ 1/3}}}}{20\, }\left( {\mbox{$_1$F$_1$}(1;\,{\frac{8}{3}};\,-{\frac{1}{3}})}-5 \right) } \\&= {\frac { \left(\sqrt [3]{3} \right) {{\rm e}^{{ 1/3}}}}{20\, }\left( {\mbox{$_1$F$_1$}(1;\,{\frac{8}{3}};\,-{\frac{1}{3}})}-5 \right) } \qquad\text{[real]} \\&\qquad - i{\frac { \left( {3}^{{5/6}} \right) {{\rm e}^{{ 1/3}}}}{20\, }\left( {\mbox{$_1$F$_1$}(1;\,{\frac{8}{3}};\,-{\frac{1}{3}})}-5 \right) } \qquad\text{[purely imaginary]} \end{align}