How to evaluate real and imaginary part of $\Gamma\left(\frac{2}{3},-\frac{1}{3}\right)$

Question

I calculated the principal value of the following integral: $$PV\int_{0}^{\infty}\frac{t^{\frac{1}{3}}e^{-t}}{1-3t}dt=\left(\frac{\Im\left[\left(-1\right)^{\frac{5}{6}}\Gamma\left(\frac{2}{3},-\frac{1}{3}\right)\right]}{\sqrt[3]{3e}}-1\right)\Gamma\left(\frac{4}{3}\right)\approx -0.225197$$ I know that $$(-1)^{5/6}=-\frac{\sqrt{3}}{2}+\frac{i}{2}$$ But I don't know how to express the real and the imaginary part of $\Gamma\left(\frac{2}{3},-\frac{1}{3}\right)$ $$\Gamma\left(\frac{2}{3},-\frac{1}{3}\right)=\int_{-\frac{1}{3}}^{\infty}e^{-t}t^{2/3-1}\mathrm{d}t\approx 1.7682-0.717204 i$$

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Edit
A resolution like this would be fine too $$\Gamma\left(\frac{2}{3},-\frac{1}{3}\right)=\underbrace{\int_{a_1}^{b_1}f_1(t)\mathrm{d}t}_{\in\mathbb{R}}+i \underbrace{\int_{a_2}^{b_2}f_2(t)\mathrm{d}t}_{\in\mathbb{R}}$$

Answer 1

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}

Answer 2

Since

$$\Gamma\left(\frac23,-\frac13\right)=\Gamma\left(\frac23\right)-\left(-\frac13\right)^\frac23S,S=\sum_{n=0}^\infty\frac1{3^n\left(n+\frac23\right)n!}=\sqrt[3]{-1}3^\frac23\left(\Gamma\left(\frac13\right)-\Gamma\left(\frac23,-\frac13\right)\right)$$

we get

$$\operatorname{Re}\Gamma\left(\frac23,-\frac13\right)= \Gamma\left(\frac23\right)+\frac S{2\cdot3^\frac23}=\left(1-\frac{\sqrt[3]{-1}}2\right)\Gamma\left(\frac23\right)+\frac{\sqrt[3]{-1}}2\Gamma\left(\frac23,-\frac13\right)$$

and

$$\operatorname{Im} \Gamma\left(\frac23,-\frac13\right)=-\frac S{2\sqrt[6]3}=\frac{\sqrt[3]{-1}\sqrt3}2\gamma\left(\frac23,-\frac13\right)$$

with the lower incomplete gamma function. Both are shown here

Answer 3

I tried to avoid the hypergeometric function and arrived at the following result.

\begin{align} \Gamma\left(\frac{2}{3},-\frac{1}{3}\right) =& \frac{1}{\sqrt{3}} \left( \frac{2 \pi }{\Gamma\left(\frac{1}{3}\right)}-a\right)+ a \; I \\ a = & \frac{1}{4} \left(-3\; b-3^{5/6}\; e^{1/3}+\sqrt{3} \; \Gamma \left(\frac{2}{3}\right)\right) \\ b= & (-1)^{1/3} \sqrt{3}\; \Gamma \left(\frac{5}{3},-\frac{1}{3}\right)-\Gamma \left(\frac{2}{3}\right) I \\ \end{align}

Where $a$ is a real $(-0.7172039326..)$ equal to the imaginary part of $\Gamma(2/3,-1/3)$

And $b$ is a real $(0.5759692781..)$ defined by the incomplete gamma function cancelling its imaginary part.